How to pass variables over functions in python 3 - python

#Test
def test():
test1 = input("Type something: ")
test()
print(test1)
print(test1)
NameError: name 'test1' is not defined
'test1' is not defined
This is bothering me. I'm learning python and trying to write a small game for practice, and it seems that variables that are declared in a function don't carry over. Is there a way to get by this? Am I not supposed to use test() to close the function? Or should I figure out and use a class instead? Or perhaps is it just a quirk of python? Thanks for any help.

By default, all the names assigned inside a function definition are put in the local scope (the
namespace associated with the function call). If you need to assign a name that
lives at the top level of the module enclosing the function, you can do so by declaring
it in a global statement inside the function. If you need to assign a name
that lives in an enclosing def, as of Python 3.X you can do so by declaring it in a
nonlocal statement.
Read this: https://docs.python.org/3.4/tutorial/classes.html#python-scopes-and-namespaces
You need to return the value and store the returned value to print it.
def test():
test1 = input("Type something: ")
return test1
a = test()
print(a)
In general, functions should rely on arguments and return values instead of globals.

def test():
test1 = input("Type something: ")
These two lines define a function called test. When test is called (test()), the line inside will be run. The line inside assigns input to the variable test1. But test1 is scoped to the test function -- the name test1 only exists inside the function, so once the function ends, the name disappears.
Functions are called to produce output (or side-effects, but in this case, output). You do that by using the return keyword to end your function and return a variable. In this case, you could do return test1 inside your function (technically you could just to return input(...) directly too, and skip the creation of the variable entirely).
test()
This calls the function. You can do this at any time after the function is defined; you don't need to do it to "close" the function.
If you modify your function to return a value, then you'll need to do something with the return value on this line. Something like result = test(). That assigns the return value of test to the result variable.
print(test1)
This isn't working because test1 only exists inside the function namespace, and you are calling this line outside of the function namespace. If you've made the changes I suggested above, you can do print(result) instead.

Related

Function prototype in python? [duplicate]

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)

How come python functions can access variables that are not defined in the arguments?

For example if I write
a=1
def func():
return a
func()
It will return 1. Is this normal behavior? Are variables supposed to act in a global way in python?
If you create a variable with the same name inside a function, this variable will be local, and can only be used inside the function. The global variable with the same name will remain as it was, global and with the original value.
That is the way python language works !

General Question about function and returning variable - python [duplicate]

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)

variable not passed to lower level?

I understand that variables in python are always (?) available in subfunctions.
That is why this works:
def abc():
print(a)
def main():
abc()
pass
if __name__=='__main__':
a = 'a'
main()
However, why does it not work if the definition of ais moved to main()?
def abc():
print(a)
def main():
a = 'a'
abc()
pass
if __name__=='__main__':
main()
It is still defined before it is used and should be available in abc(), right?
No, variables defined in a function are scoped to the function by default. To make it available in abc() you would need to make it global:
def main():
global a
a = 'a'
abc()
As for the 'shadowing' warning, you get it when you redefine names at inner scopes. For example, this code creates a global variable a:
if __name__ == '__main__':
a = 'a'
main()
Now if you do a = 'b' inside main() you are effectively 'shadowing' a from the outer scope by creating a local variable with the same name.
In the first code sample, 'a' is defined as a global variable. Meaning that it can be accessed by any function called after it is instantiated.
In the second code sample, 'a' is defined as a local variable. Meaning that it only exists within the "main" function. If you wanted to pass it into the "abc" function, you would have to explicitly pass that variable. That would look like:
def abc(a):
print(a)
def main():
a = 'a'
abc(a)
pass
if __name__=='__main__':
main()
Variables at the module level of your function are available to read anywhere else in the module. When you do
if __name__=='__main__':
a = 'a'
you are defining a as a module-level variable, hence why you can read it inside abc(). This is similar to why you can see your imports within other functions.
When you define a inside abc(), it's no longer at the module-level and no longer is implicitly readable elsewhere in the module.
As an aside, global variables are usually a bad idea, it's better to pass them around explicitly. See http://wiki.c2.com/?GlobalVariablesAreBad for a good summary of why.

Access global variable inside a function

How to pass a variable as a parameter to a function?
This is my code
def hello (arg):
global my_name
my_name = "tarek"
print("my name is " + arg)
hello(my_name)
This should print
my name is tarek
but it doesn't, giving me a NameError, although I declared the variable my_name as global.
Well it works well if I declared the variable outside the function, but why is it not working when declaring the variable inside the function?
You have to define your global variable outside of your function, then you don't need to define it again in the function. Following code will print what you want:
my_name = "tarek"
def hello(arg):
print("my name is " + arg)
hello(my_name)
There is no variable my_name until the function is called so there is nothing to pass into the function, giving the error.
Define the variable outside the function, then declare it global inside the function, if you must.
If you want my_name to be global you don't need to pass it into the function. Just declare and assign it outside the function, then declare it global inside.
Passing parameters is usually done because you want it to be local to that function, then we usually return it back after its been worked on. Compartmentalising you code helps when debugging.
Nothing within hello() can make my_name exist by the time hello()is called.
Your variable declaration is scoped to inside the method (a closure), and that closure is not evaluated until you invoke it, at which point you get an error because the variable does not exist in that context.
If I understand correctly, it would seem you want to do this
def hello (arg):
global my_name
my_name = arg
print("my name is " + arg)
hello("tarek")

Categories

Resources