Related
Is it possible to forward-declare a function in Python? I want to sort a list using my own cmp function before it is declared.
print "\n".join([str(bla) for bla in sorted(mylist, cmp = cmp_configs)])
I've put the definition of cmp_configs method after the invocation. It fails with this error:
NameError: name 'cmp_configs' is not defined
Is there any way to "declare" cmp_configs method before it's used?
Sometimes, it is difficult to reorganize code to avoid this problem. For instance, when implementing some forms of recursion:
def spam():
if end_condition():
return end_result()
else:
return eggs()
def eggs():
if end_condition():
return end_result()
else:
return spam()
Where end_condition and end_result have been previously defined.
Is the only solution to reorganize the code and always put definitions before invocations?
Wrap the invocation into a function of its own so that
foo()
def foo():
print "Hi!"
will break, but
def bar():
foo()
def foo():
print "Hi!"
bar()
will work properly.
The general rule in Python is that a function should be defined before its usage, which does not necessarily mean it needs to be higher in the code.
If you kick-start your script through the following:
if __name__=="__main__":
main()
then you probably do not have to worry about things like "forward declaration". You see, the interpreter would go loading up all your functions and then start your main() function. Of course, make sure you have all the imports correct too ;-)
Come to think of it, I've never heard such a thing as "forward declaration" in python... but then again, I might be wrong ;-)
If you don't want to define a function before it's used, and defining it afterwards is impossible, what about defining it in some other module?
Technically you still define it first, but it's clean.
You could create a recursion like the following:
def foo():
bar()
def bar():
foo()
Python's functions are anonymous just like values are anonymous, yet they can be bound to a name.
In the above code, foo() does not call a function with the name foo, it calls a function that happens to be bound to the name foo at the point the call is made. It is possible to redefine foo somewhere else, and bar would then call the new function.
Your problem cannot be solved because it's like asking to get a variable which has not been declared.
I apologize for reviving this thread, but there was a strategy not discussed here which may be applicable.
Using reflection it is possible to do something akin to forward declaration. For instance lets say you have a section of code that looks like this:
# We want to call a function called 'foo', but it hasn't been defined yet.
function_name = 'foo'
# Calling at this point would produce an error
# Here is the definition
def foo():
bar()
# Note that at this point the function is defined
# Time for some reflection...
globals()[function_name]()
So in this way we have determined what function we want to call before it is actually defined, effectively a forward declaration. In python the statement globals()[function_name]() is the same as foo() if function_name = 'foo' for the reasons discussed above, since python must lookup each function before calling it. If one were to use the timeit module to see how these two statements compare, they have the exact same computational cost.
Of course the example here is very useless, but if one were to have a complex structure which needed to execute a function, but must be declared before (or structurally it makes little sense to have it afterwards), one can just store a string and try to call the function later.
If the call to cmp_configs is inside its own function definition, you should be fine. I'll give an example.
def a():
b() # b() hasn't been defined yet, but that's fine because at this point, we're not
# actually calling it. We're just defining what should happen when a() is called.
a() # This call fails, because b() hasn't been defined yet,
# and thus trying to run a() fails.
def b():
print "hi"
a() # This call succeeds because everything has been defined.
In general, putting your code inside functions (such as main()) will resolve your problem; just call main() at the end of the file.
There is no such thing in python like forward declaration. You just have to make sure that your function is declared before it is needed.
Note that the body of a function isn't interpreted until the function is executed.
Consider the following example:
def a():
b() # won't be resolved until a is invoked.
def b():
print "hello"
a() # here b is already defined so this line won't fail.
You can think that a body of a function is just another script that will be interpreted once you call the function.
Sometimes an algorithm is easiest to understand top-down, starting with the overall structure and drilling down into the details.
You can do so without forward declarations:
def main():
make_omelet()
eat()
def make_omelet():
break_eggs()
whisk()
fry()
def break_eggs():
for egg in carton:
break(egg)
# ...
main()
# declare a fake function (prototype) with no body
def foo(): pass
def bar():
# use the prototype however you see fit
print(foo(), "world!")
# define the actual function (overwriting the prototype)
def foo():
return "Hello,"
bar()
Output:
Hello, world!
No, I don't believe there is any way to forward-declare a function in Python.
Imagine you are the Python interpreter. When you get to the line
print "\n".join([str(bla) for bla in sorted(mylist, cmp = cmp_configs)])
either you know what cmp_configs is or you don't. In order to proceed, you have to
know cmp_configs. It doesn't matter if there is recursion.
You can't forward-declare a function in Python. If you have logic executing before you've defined functions, you've probably got a problem anyways. Put your action in an if __name__ == '__main__' at the end of your script (by executing a function you name "main" if it's non-trivial) and your code will be more modular and you'll be able to use it as a module if you ever need to.
Also, replace that list comprehension with a generator express (i.e., print "\n".join(str(bla) for bla in sorted(mylist, cmp=cmp_configs)))
Also, don't use cmp, which is deprecated. Use key and provide a less-than function.
Import the file itself. Assuming the file is called test.py:
import test
if __name__=='__main__':
test.func()
else:
def func():
print('Func worked')
TL;DR: Python does not need forward declarations. Simply put your function calls inside function def definitions, and you'll be fine.
def foo(count):
print("foo "+str(count))
if(count>0):
bar(count-1)
def bar(count):
print("bar "+str(count))
if(count>0):
foo(count-1)
foo(3)
print("Finished.")
recursive function definitions, perfectly successfully gives:
foo 3
bar 2
foo 1
bar 0
Finished.
However,
bug(13)
def bug(count):
print("bug never runs "+str(count))
print("Does not print this.")
breaks at the top-level invocation of a function that hasn't been defined yet, and gives:
Traceback (most recent call last):
File "./test1.py", line 1, in <module>
bug(13)
NameError: name 'bug' is not defined
Python is an interpreted language, like Lisp. It has no type checking, only run-time function invocations, which succeed if the function name has been bound and fail if it's unbound.
Critically, a function def definition does not execute any of the funcalls inside its lines, it simply declares what the function body is going to consist of. Again, it doesn't even do type checking. So we can do this:
def uncalled():
wild_eyed_undefined_function()
print("I'm not invoked!")
print("Only run this one line.")
and it runs perfectly fine (!), with output
Only run this one line.
The key is the difference between definitions and invocations.
The interpreter executes everything that comes in at the top level, which means it tries to invoke it. If it's not inside a definition.
Your code is running into trouble because you attempted to invoke a function, at the top level in this case, before it was bound.
The solution is to put your non-top-level function invocations inside a function definition, then call that function sometime much later.
The business about "if __ main __" is an idiom based on this principle, but you have to understand why, instead of simply blindly following it.
There are certainly much more advanced topics concerning lambda functions and rebinding function names dynamically, but these are not what the OP was asking for. In addition, they can be solved using these same principles: (1) defs define a function, they do not invoke their lines; (2) you get in trouble when you invoke a function symbol that's unbound.
Python does not support forward declarations, but common workaround for this is use of the the following condition at the end of your script/code:
if __name__ == '__main__': main()
With this it will read entire file first and then evaluate condition and call main() function which will be able to call any forward declared function as it already read the entire file first. This condition leverages special variable __name__ which returns __main__ value whenever we run Python code from current file (when code was imported as a module, then __name__ returns module name).
"just reorganize my code so that I don't have this problem." Correct. Easy to do. Always works.
You can always provide the function prior to it's reference.
"However, there are cases when this is probably unavoidable, for instance when implementing some forms of recursion"
Can't see how that's even remotely possible. Please provide an example of a place where you cannot define the function prior to it's use.
Now wait a minute. When your module reaches the print statement in your example, before cmp_configs has been defined, what exactly is it that you expect it to do?
If your posting of a question using print is really trying to represent something like this:
fn = lambda mylist:"\n".join([str(bla)
for bla in sorted(mylist, cmp = cmp_configs)])
then there is no requirement to define cmp_configs before executing this statement, just define it later in the code and all will be well.
Now if you are trying to reference cmp_configs as a default value of an argument to the lambda, then this is a different story:
fn = lambda mylist,cmp_configs=cmp_configs : \
"\n".join([str(bla) for bla in sorted(mylist, cmp = cmp_configs)])
Now you need a cmp_configs variable defined before you reach this line.
[EDIT - this next part turns out not to be correct, since the default argument value will get assigned when the function is compiled, and that value will be used even if you change the value of cmp_configs later.]
Fortunately, Python being so type-accommodating as it is, does not care what you define as cmp_configs, so you could just preface with this statement:
cmp_configs = None
And the compiler will be happy. Just be sure to declare the real cmp_configs before you ever invoke fn.
Python technically has support for forward declaration.
if you define a function/class then set the body to pass, it will have an empty entry in the global table.
you can then "redefine" the function/class later on to implement the function/class.
unlike c/c++ forward declaration though, this does not work from outside the scope (i.e. another file) as they have their own "global" namespace
example:
def foo(): pass
foo()
def foo(): print("FOOOOO")
foo()
foo is declared both times
however the first time foo is called it does not do anything as the body is just pass
but the second time foo is called. it executes the new body of print("FOOOOO")
but again. note that this does not fix circular dependancies. this is because files have their own name and have their own definitions of functions
example 2:
class bar: pass
print(bar)
this prints <class '__main__.bar'> but if it was declared in another file it would be <class 'otherfile.foo'>
i know this post is old, but i though that this answer would be useful to anyone who keeps finding this post after the many years it has been posted for
One way is to create a handler function. Define the handler early on, and put the handler below all the methods you need to call.
Then when you invoke the handler method to call your functions, they will always be available.
The handler could take an argument nameOfMethodToCall. Then uses a bunch of if statements to call the right method.
This would solve your issue.
def foo():
print("foo")
#take input
nextAction=input('What would you like to do next?:')
return nextAction
def bar():
print("bar")
nextAction=input('What would you like to do next?:')
return nextAction
def handler(action):
if(action=="foo"):
nextAction = foo()
elif(action=="bar"):
nextAction = bar()
else:
print("You entered invalid input, defaulting to bar")
nextAction = "bar"
return nextAction
nextAction=input('What would you like to do next?:')
while 1:
nextAction = handler(nextAction)
I am trying to create a dynamic method executor, where I have a list that will always contain two elements. The first element is the name of the file, the second element is the name of the method to execute.
How can I achieve this?
My below code unfortunately doesn't work, but it will give you an good indication of what I am trying to achieve.
from logic.intents import CenterCapacity
def method_executor(event):
call_reference = ['CenterCapacity', 'get_capacity']
# process method call
return call_reference[0].call_reference[1]
Thanks!
You can use __import__ to look up the module by name and then then use getattr to find the method. For example if the following code is in a file called exec.py then
def dummy(): print("dummy")
def lookup(mod, func):
module = __import__(mod)
return getattr(module, func)
if __name__ == "__main__":
lookup("exec","dummy")()
will output
dummy
Addendum
Alternatively importlib.import_module can be used, which although a bit more verbose, may be easier to use.
The most important difference between these two functions is that import_module() returns the specified package or module (e.g. pkg.mod), while __import__() returns the top-level package or module (e.g. pkg).
def lookup(mod, func):
import importlib
module = importlib.import_module(mod)
return getattr(module, func)
starting from:
from logic.intents import CenterCapacity
def method_executor(event):
call_reference = ['CenterCapacity', 'get_capacity']
# process method call
return call_reference[0].call_reference[1]
Option 1
We have several options, the first one is using a class reference and the getattr. For this we have to remove the ' around the class and instantiate the class before calling a reference (you do not have to instantiate the class when the method is a staticmethod.)
def method_executor(event):
call_reference = [CenterCapacity, 'get_capacity'] # We now store a class reference
# process method call
return getattr(call_reference[0](), call_reference[1])
option 2
A second option is based on this answer. It revolves around using the getattr method twice. We firstly get module using sys.modules[__name__] and then get the class from there using getattr.
import sys
def method_executor(event):
call_reference = ['CenterCapacity', 'get_capacity']
class_ref = getattr(sys.modules[__name__], call_reference[0])
return getattr(class_ref, call_reference[1])
Option 3
A third option could be based on a full import path and use __import__('module.class'), take a look at this SO post.
(Note: This answer assumes that the necessary imports have already happened, and you just need a mechanism to invoke the functions of the imported modules. If you also want the import do be done by some program code, I will have to add that part, using importlib library)
You can do this:
globals()[call_reference[0]].__dict__[call_reference[1]]()
Explanation:
globals() returns a mapping between global variable names and their referenced objects. The imported module's name counts as one of these global variables of the current module.
Indexing this mapping object with call_reference[0] returns the module object containing the function to be called.
The module object's __dict__ maps each attribute-name of the module to the object referenced by that attribute. Functions defined in the module also count as attributes of the module.
Thus, indexing __dict__ with the function name call_reference[1] returns the function object.
Is it possible to forward-declare a function in Python? I want to sort a list using my own cmp function before it is declared.
print "\n".join([str(bla) for bla in sorted(mylist, cmp = cmp_configs)])
I've put the definition of cmp_configs method after the invocation. It fails with this error:
NameError: name 'cmp_configs' is not defined
Is there any way to "declare" cmp_configs method before it's used?
Sometimes, it is difficult to reorganize code to avoid this problem. For instance, when implementing some forms of recursion:
def spam():
if end_condition():
return end_result()
else:
return eggs()
def eggs():
if end_condition():
return end_result()
else:
return spam()
Where end_condition and end_result have been previously defined.
Is the only solution to reorganize the code and always put definitions before invocations?
Wrap the invocation into a function of its own so that
foo()
def foo():
print "Hi!"
will break, but
def bar():
foo()
def foo():
print "Hi!"
bar()
will work properly.
The general rule in Python is that a function should be defined before its usage, which does not necessarily mean it needs to be higher in the code.
If you kick-start your script through the following:
if __name__=="__main__":
main()
then you probably do not have to worry about things like "forward declaration". You see, the interpreter would go loading up all your functions and then start your main() function. Of course, make sure you have all the imports correct too ;-)
Come to think of it, I've never heard such a thing as "forward declaration" in python... but then again, I might be wrong ;-)
If you don't want to define a function before it's used, and defining it afterwards is impossible, what about defining it in some other module?
Technically you still define it first, but it's clean.
You could create a recursion like the following:
def foo():
bar()
def bar():
foo()
Python's functions are anonymous just like values are anonymous, yet they can be bound to a name.
In the above code, foo() does not call a function with the name foo, it calls a function that happens to be bound to the name foo at the point the call is made. It is possible to redefine foo somewhere else, and bar would then call the new function.
Your problem cannot be solved because it's like asking to get a variable which has not been declared.
I apologize for reviving this thread, but there was a strategy not discussed here which may be applicable.
Using reflection it is possible to do something akin to forward declaration. For instance lets say you have a section of code that looks like this:
# We want to call a function called 'foo', but it hasn't been defined yet.
function_name = 'foo'
# Calling at this point would produce an error
# Here is the definition
def foo():
bar()
# Note that at this point the function is defined
# Time for some reflection...
globals()[function_name]()
So in this way we have determined what function we want to call before it is actually defined, effectively a forward declaration. In python the statement globals()[function_name]() is the same as foo() if function_name = 'foo' for the reasons discussed above, since python must lookup each function before calling it. If one were to use the timeit module to see how these two statements compare, they have the exact same computational cost.
Of course the example here is very useless, but if one were to have a complex structure which needed to execute a function, but must be declared before (or structurally it makes little sense to have it afterwards), one can just store a string and try to call the function later.
If the call to cmp_configs is inside its own function definition, you should be fine. I'll give an example.
def a():
b() # b() hasn't been defined yet, but that's fine because at this point, we're not
# actually calling it. We're just defining what should happen when a() is called.
a() # This call fails, because b() hasn't been defined yet,
# and thus trying to run a() fails.
def b():
print "hi"
a() # This call succeeds because everything has been defined.
In general, putting your code inside functions (such as main()) will resolve your problem; just call main() at the end of the file.
There is no such thing in python like forward declaration. You just have to make sure that your function is declared before it is needed.
Note that the body of a function isn't interpreted until the function is executed.
Consider the following example:
def a():
b() # won't be resolved until a is invoked.
def b():
print "hello"
a() # here b is already defined so this line won't fail.
You can think that a body of a function is just another script that will be interpreted once you call the function.
Sometimes an algorithm is easiest to understand top-down, starting with the overall structure and drilling down into the details.
You can do so without forward declarations:
def main():
make_omelet()
eat()
def make_omelet():
break_eggs()
whisk()
fry()
def break_eggs():
for egg in carton:
break(egg)
# ...
main()
# declare a fake function (prototype) with no body
def foo(): pass
def bar():
# use the prototype however you see fit
print(foo(), "world!")
# define the actual function (overwriting the prototype)
def foo():
return "Hello,"
bar()
Output:
Hello, world!
No, I don't believe there is any way to forward-declare a function in Python.
Imagine you are the Python interpreter. When you get to the line
print "\n".join([str(bla) for bla in sorted(mylist, cmp = cmp_configs)])
either you know what cmp_configs is or you don't. In order to proceed, you have to
know cmp_configs. It doesn't matter if there is recursion.
You can't forward-declare a function in Python. If you have logic executing before you've defined functions, you've probably got a problem anyways. Put your action in an if __name__ == '__main__' at the end of your script (by executing a function you name "main" if it's non-trivial) and your code will be more modular and you'll be able to use it as a module if you ever need to.
Also, replace that list comprehension with a generator express (i.e., print "\n".join(str(bla) for bla in sorted(mylist, cmp=cmp_configs)))
Also, don't use cmp, which is deprecated. Use key and provide a less-than function.
Import the file itself. Assuming the file is called test.py:
import test
if __name__=='__main__':
test.func()
else:
def func():
print('Func worked')
TL;DR: Python does not need forward declarations. Simply put your function calls inside function def definitions, and you'll be fine.
def foo(count):
print("foo "+str(count))
if(count>0):
bar(count-1)
def bar(count):
print("bar "+str(count))
if(count>0):
foo(count-1)
foo(3)
print("Finished.")
recursive function definitions, perfectly successfully gives:
foo 3
bar 2
foo 1
bar 0
Finished.
However,
bug(13)
def bug(count):
print("bug never runs "+str(count))
print("Does not print this.")
breaks at the top-level invocation of a function that hasn't been defined yet, and gives:
Traceback (most recent call last):
File "./test1.py", line 1, in <module>
bug(13)
NameError: name 'bug' is not defined
Python is an interpreted language, like Lisp. It has no type checking, only run-time function invocations, which succeed if the function name has been bound and fail if it's unbound.
Critically, a function def definition does not execute any of the funcalls inside its lines, it simply declares what the function body is going to consist of. Again, it doesn't even do type checking. So we can do this:
def uncalled():
wild_eyed_undefined_function()
print("I'm not invoked!")
print("Only run this one line.")
and it runs perfectly fine (!), with output
Only run this one line.
The key is the difference between definitions and invocations.
The interpreter executes everything that comes in at the top level, which means it tries to invoke it. If it's not inside a definition.
Your code is running into trouble because you attempted to invoke a function, at the top level in this case, before it was bound.
The solution is to put your non-top-level function invocations inside a function definition, then call that function sometime much later.
The business about "if __ main __" is an idiom based on this principle, but you have to understand why, instead of simply blindly following it.
There are certainly much more advanced topics concerning lambda functions and rebinding function names dynamically, but these are not what the OP was asking for. In addition, they can be solved using these same principles: (1) defs define a function, they do not invoke their lines; (2) you get in trouble when you invoke a function symbol that's unbound.
Python does not support forward declarations, but common workaround for this is use of the the following condition at the end of your script/code:
if __name__ == '__main__': main()
With this it will read entire file first and then evaluate condition and call main() function which will be able to call any forward declared function as it already read the entire file first. This condition leverages special variable __name__ which returns __main__ value whenever we run Python code from current file (when code was imported as a module, then __name__ returns module name).
"just reorganize my code so that I don't have this problem." Correct. Easy to do. Always works.
You can always provide the function prior to it's reference.
"However, there are cases when this is probably unavoidable, for instance when implementing some forms of recursion"
Can't see how that's even remotely possible. Please provide an example of a place where you cannot define the function prior to it's use.
Now wait a minute. When your module reaches the print statement in your example, before cmp_configs has been defined, what exactly is it that you expect it to do?
If your posting of a question using print is really trying to represent something like this:
fn = lambda mylist:"\n".join([str(bla)
for bla in sorted(mylist, cmp = cmp_configs)])
then there is no requirement to define cmp_configs before executing this statement, just define it later in the code and all will be well.
Now if you are trying to reference cmp_configs as a default value of an argument to the lambda, then this is a different story:
fn = lambda mylist,cmp_configs=cmp_configs : \
"\n".join([str(bla) for bla in sorted(mylist, cmp = cmp_configs)])
Now you need a cmp_configs variable defined before you reach this line.
[EDIT - this next part turns out not to be correct, since the default argument value will get assigned when the function is compiled, and that value will be used even if you change the value of cmp_configs later.]
Fortunately, Python being so type-accommodating as it is, does not care what you define as cmp_configs, so you could just preface with this statement:
cmp_configs = None
And the compiler will be happy. Just be sure to declare the real cmp_configs before you ever invoke fn.
Python technically has support for forward declaration.
if you define a function/class then set the body to pass, it will have an empty entry in the global table.
you can then "redefine" the function/class later on to implement the function/class.
unlike c/c++ forward declaration though, this does not work from outside the scope (i.e. another file) as they have their own "global" namespace
example:
def foo(): pass
foo()
def foo(): print("FOOOOO")
foo()
foo is declared both times
however the first time foo is called it does not do anything as the body is just pass
but the second time foo is called. it executes the new body of print("FOOOOO")
but again. note that this does not fix circular dependancies. this is because files have their own name and have their own definitions of functions
example 2:
class bar: pass
print(bar)
this prints <class '__main__.bar'> but if it was declared in another file it would be <class 'otherfile.foo'>
i know this post is old, but i though that this answer would be useful to anyone who keeps finding this post after the many years it has been posted for
One way is to create a handler function. Define the handler early on, and put the handler below all the methods you need to call.
Then when you invoke the handler method to call your functions, they will always be available.
The handler could take an argument nameOfMethodToCall. Then uses a bunch of if statements to call the right method.
This would solve your issue.
def foo():
print("foo")
#take input
nextAction=input('What would you like to do next?:')
return nextAction
def bar():
print("bar")
nextAction=input('What would you like to do next?:')
return nextAction
def handler(action):
if(action=="foo"):
nextAction = foo()
elif(action=="bar"):
nextAction = bar()
else:
print("You entered invalid input, defaulting to bar")
nextAction = "bar"
return nextAction
nextAction=input('What would you like to do next?:')
while 1:
nextAction = handler(nextAction)
$ cat declare_funcs.py
#!/usr/bin/python3
def declared_after():
print("good declared after")
declared_after()
$ python3 declare_funcs.py
good declared after
Change call place:
$ cat declare_funcs.py
#!/usr/bin/python3
declared_after()
def declared_after():
print("good declared after")
$ python3 declare_funcs.py
Traceback (most recent call last):
File "declare_funcs.py", line 4, in <module>
declared_after()
NameError: name 'declared_after' is not defined
Is there way to declare only header of function like it was in C/C++?
For example:
#!/usr/bin/python3
def declared_after() # declaration about defined function
declared_after()
def declared_after():
print("good declared after")
I found this Declare function at end of file in Python
Any way there appear another function in the beginning like wrapper, and this wrapper must be called after declaration of wrapped function, this is not an exit. Is there more elegant true-python way?
You can't forward-declare functions in Python. It doesn't make a lot of sense to do so, because Python is dynamically typed. You could do something silly like this, and what would expect it to do?
foo = 3
foo()
def foo():
print "bar"
Obviously, you are trying to __call__ the int object for 3. It's absolutely silly.
You ask if you can forward-declare like in C/C++. Well, you typically don't run C through an interpreter. However, although Python is compiled to bytecode, the python3 program is an interpreter.
Forward declaration in a compiled language makes sense because you are simply establishing a symbol and its type, and the compiler can run through the code several times to make sense of it. When you use an interpreter, however, you typically can't have that luxury, because you would have to run through the rest of the code to find the meaning of that forward declaration, and run through it again after having done that.
You can, of course, do something like this:
foo = lambda: None
foo()
def foo():
print "bar"
But you instantiated foo nonetheless. Everything has to point to an actual, existing object in Python.
This doesn't apply to def or class statements, though. These create a function or class object, but they don't execute the code inside yet. So, you have time to instantiate things inside them before their code runs.
def foo():
print bar()
# calling foo() won't work yet because you haven't defined bar()
def bar():
return "bar"
# now it will work
The difference was that you simply created function objects with the variable names foo and bar representing them respectively. You can now refer to these objects by those variable names.
With regard to the way that Python is typically interpreted (in CPython) you should make sure that you execute no code in your modules unless they are being run as the main program or unless you want them to do something when being imported (a rare, but valid case). You should do the following:
Put code meant to be executed into function and class definitions.
Unless the code only makes sense to be executed in the main program, put it in another module.
Use if __name__ == "__main__": to create a block of code which will only execute if the program is the main program.
In fact, you should do the third in all of your modules. You can simply write this at the bottom of every file which you don't want to be run as a main program:
if __name__ = "__main__":
pass
This prevents anything from happening if the module is imported.
Python doesn't work that way. The def is executed in sequence, top-to-bottom, with the remainder of the file's contents. You cannot call something before it is defined as a callable (e.g. a function), and even if you had a stand-in callable, it would not contain the code you are looking for.
This, of course, doesn't mean the code isn't compiled before execution begins—in fact, it is. But it is when the def is executed that declared_after is actually assigned the code within the def block, and not before.
Any tricks you pull to sort-of achieve your desired effect must have the effect of delaying the call to declared_after() until after it is defined, for example, by enclosing it in another def block that is itself called later.
One thing you can do is enclose everything in a main function:
def main():
declared_after()
def declared_after():
print("good declared after")
main()
However, the point still stands that the function must be defined prior to calling. This only works because main is called AFTER declared_after is defined.
As zigg wrote, Python files are executed in order they are written from top to bottom, so even if you could “declare” the variable before, the actual function body would only get there after the function was called.
The usual way to solve this is to just have a main function where all your standard execution stuff happens:
def main ():
# do stuff
declared_after();
def declared_after():
pass
main()
You can then also combine this with the __name__ == '__main__' idiom to make the function only execute when you are executing the module directly:
def main ():
# do stuff
declared_after();
def declared_after():
pass
if __name__ == '__main__':
main()
How can I get a variable that contains the currently executing function in Python? I don't want the function's name. I know I can use inspect.stack to get the current function name. I want the actual callable object. Can this be done without using inspect.stack to retrieve the function's name and then evaling the name to get the callable object?
Edit: I have a reason to do this, but it's not even a remotely good one. I'm using plac to parse command-line arguments. You use it by doing plac.call(main), which generates an ArgumentParser object from the function signature of "main". Inside "main", if there is a problem with the arguments, I want to exit with an error message that includes the help text from the ArgumentParser object, which means that I need to directly access this object by calling plac.parser_from(main).print_help(). It would be nice to be able to say instead: plac.parser_from(get_current_function()).print_help(), so that I am not relying on the function being named "main". Right now, my implementation of "get_current_function" would be:
import inspect
def get_current_function():
return eval(inspect.stack()[1][3])
But this implementation relies on the function having a name, which I suppose is not too onerous. I'm never going to do plac.call(lambda ...).
In the long run, it might be more useful to ask the author of plac to implement a print_help method to print the help text of the function that was most-recently called using plac, or something similar.
The stack frame tells us what code object we're in. If we can find a function object that refers to that code object in its __code__ attribute, we have found the function.
Fortunately, we can ask the garbage collector which objects hold a reference to our code object, and sift through those, rather than having to traverse every active object in the Python world. There are typically only a handful of references to a code object.
Now, functions can share code objects, and do in the case where you return a function from a function, i.e. a closure. When there's more than one function using a given code object, we can't tell which function it is, so we return None.
import inspect, gc
def giveupthefunc():
frame = inspect.currentframe(1)
code = frame.f_code
globs = frame.f_globals
functype = type(lambda: 0)
funcs = []
for func in gc.get_referrers(code):
if type(func) is functype:
if getattr(func, "__code__", None) is code:
if getattr(func, "__globals__", None) is globs:
funcs.append(func)
if len(funcs) > 1:
return None
return funcs[0] if funcs else None
Some test cases:
def foo():
return giveupthefunc()
zed = lambda: giveupthefunc()
bar, foo = foo, None
print bar()
print zed()
I'm not sure about the performance characteristics of this, but i think it should be fine for your use case.
I recently spent a lot of time trying to do something like this and ended up walking away from it. There's a lot of corner cases.
If you just want the lowest level of the call stack, you can just reference the name that is used in the def statement. This will be bound to the function that you want through lexical closure.
For example:
def recursive(*args, **kwargs):
me = recursive
me will now refer to the function in question regardless of the scope that the function is called from so long as it is not redefined in the scope where the definition occurs. Is there some reason why this won't work?
To get a function that is executing higher up the call stack, I couldn't think of anything that can be reliably done.
This is what you asked for, as close as I can come. Tested in python versions 2.4, 2.6, 3.0.
#!/usr/bin/python
def getfunc():
from inspect import currentframe, getframeinfo
caller = currentframe().f_back
func_name = getframeinfo(caller)[2]
caller = caller.f_back
from pprint import pprint
func = caller.f_locals.get(
func_name, caller.f_globals.get(
func_name
)
)
return func
def main():
def inner1():
def inner2():
print("Current function is %s" % getfunc())
print("Current function is %s" % getfunc())
inner2()
print("Current function is %s" % getfunc())
inner1()
#entry point: parse arguments and call main()
if __name__ == "__main__":
main()
Output:
Current function is <function main at 0x2aec09fe2ed8>
Current function is <function inner1 at 0x2aec09fe2f50>
Current function is <function inner2 at 0x2aec0a0635f0>
Here's another possibility: a decorator that implicitly passes a reference to the called function as the first argument (similar to self in bound instance methods). You have to decorate each function that you want to receive such a reference, but "explicit is better than implicit" as they say.
Of course, it has all the disadvantage of decorators: another function call slightly degrades performance, and the signature of the wrapped function is no longer visible.
import functools
def gottahavethatfunc(func):
#functools.wraps(func)
def wrapper(*args, **kwargs):
return func(func, *args, **kwargs)
return wrapper
The test case illustrates that the decorated function still gets the reference to itself even if you change the name to which the function is bound. This is because you're only changing the binding of the wrapper function. It also illustrates its use with a lambda.
#gottahavethatfunc
def quux(me):
return me
zoom = gottahavethatfunc(lambda me: me)
baz, quux = quux, None
print baz()
print zoom()
When using this decorator with an instance or class method, the method should accept the function reference as the first argument and the traditional self as the second.
class Demo(object):
#gottahavethatfunc
def method(me, self):
return me
print Demo().method()
The decorator relies on a closure to hold the reference to the wrapped function in the wrapper. Creating the closure directly might actually be cleaner, and won't have the overhead of the extra function call:
def my_func():
def my_func():
return my_func
return my_func
my_func = my_func()
Within the inner function, the name my_func always refers to that function; its value does not rely on a global name that may be changed. Then we just "lift" that function to the global namespace, replacing the reference to the outer function. Works in a class too:
class K(object):
def my_method():
def my_method(self):
return my_method
return my_method
my_method = my_method()
I just define in the beginning of each function a "keyword" which is just a reference to the actual name of the function. I just do this for any function, if it needs it or not:
def test():
this=test
if not hasattr(this,'cnt'):
this.cnt=0
else:
this.cnt+=1
print this.cnt
The call stack does not keep a reference to the function itself -
although the running frame as a reference to the code object that is the code associated to a given function.
(Functions are objects with code, and some information about their environment, such as closures, name, globals dictionary, doc string, default parameters and so on).
Therefore if you are running a regular function, you are better of using its own name on the globals dictionary to call itself, as has been pointed out.
If you are running some dynamic, or lambda code, in which you can't use the function name, the only solution is to rebuild another function object which re-uses thre currently running code object and call that new function instead.
You will loose a couple of things, like default arguments, and it may be hard to get it working with closures (although it can be done).
I have written a blog post on doing exactly that - calling anonymous functions from within themselves - I hope the code in there can help you:
http://metapython.blogspot.com/2010/11/recursive-lambda-functions.html
On a side note: avoid the use o inspect.stack -- it is too slow, as it rebuilds a lot of information each time it is called. prefer to use inspect.currentframe to deal with code frames instead.
This may sounds complicated, but the code itself is very short - I am pasting it bellow. The post above contains more information on how this works.
from inspect import currentframe
from types import FunctionType
lambda_cache = {}
def myself (*args, **kw):
caller_frame = currentframe(1)
code = caller_frame.f_code
if not code in lambda_cache:
lambda_cache[code] = FunctionType(code, caller_frame.f_globals)
return lambda_cache[code](*args, **kw)
if __name__ == "__main__":
print "Factorial of 5", (lambda n: n * myself(n - 1) if n > 1 else 1)(5)
If you really need the original function itself, the "myself" function above could be made to search on some scopes (like the calling function global dictionary) for a function object which code object would match with the one retrieved from the frame, instead of creating a new function.
sys._getframe(0).f_code returns exactly what you need: the codeobject being executed. Having a code object, you can retrieve a name with codeobject.co_name
OK after reading the question and comments again, I think this is a decent test case:
def foo(n):
""" print numbers from 0 to n """
if n: foo(n-1)
print n
g = foo # assign name 'g' to function object
foo = None # clobber name 'foo' which refers to function object
g(10) # dies with TypeError because function object tries to call NoneType
I tried solving it by using a decorator to temporarily clobber the global namespace and reassigning the function object to the original name of the function:
def selfbind(f):
""" Ensures that f's original function name is always defined as f when f is executed """
oname = f.__name__
def g(*args, **kwargs):
# Clobber global namespace
had_key = None
if globals().has_key(oname):
had_key = True
key = globals()[oname]
globals()[oname] = g
# Run function in modified environment
result = f(*args, **kwargs)
# Restore global namespace
if had_key:
globals()[oname] = key
else:
del globals()[oname]
return result
return g
#selfbind
def foo(n):
if n: foo(n-1)
print n
g = foo # assign name 'g' to function object
foo = 2 # calling 'foo' now fails since foo is an int
g(10) # print from 0..10, even though foo is now an int
print foo # prints 2 (the new value of Foo)
I'm sure I haven't thought through all the use cases. The biggest problem I see is the function object intentionally changing what its own name points to (an operation which would be overwritten by the decorator), but that should be ok as long as the recursive function doesn't redefine its own name in the middle of recursing.
Still not sure I'd ever need to do this, but thinking about was interesting.
Here a variation (Python 3.5.1) of the get_referrers() answer, which tries to distinguish between closures that are using the same code object:
import functools
import gc
import inspect
def get_func():
frame = inspect.currentframe().f_back
code = frame.f_code
return [
referer
for referer in gc.get_referrers(code)
if getattr(referer, "__code__", None) is code and
set(inspect.getclosurevars(referer).nonlocals.items()) <=
set(frame.f_locals.items())][0]
def f1(x):
def f2(y):
print(get_func())
return x + y
return f2
f_var1 = f1(1)
f_var1(3)
# <function f1.<locals>.f2 at 0x0000017235CB2C80>
# 4
f_var2 = f1(2)
f_var2(3)
# <function f1.<locals>.f2 at 0x0000017235CB2BF8>
# 5
def f3():
print(get_func())
f3()
# <function f3 at 0x0000017235CB2B70>
def wrapper(func):
functools.wraps(func)
def wrapped(*args, **kwargs):
return func(*args, **kwargs)
return wrapped
#wrapper
def f4():
print(get_func())
f4()
# <function f4 at 0x0000017235CB2A60>
f5 = lambda: get_func()
print(f5())
# <function <lambda> at 0x0000017235CB2950>
Correction of my previous answer, because the subdict check already works with "<=" called on dict_items and the additional set() calls result in problems, if there are dict-values which are dicts themself:
import gc
import inspect
def get_func():
frame = inspect.currentframe().f_back
code = frame.f_code
return [
referer
for referer in gc.get_referrers(code)
if getattr(referer, "__code__", None) is code and
inspect.getclosurevars(referer).nonlocals.items() <=
frame.f_locals.items()][0]