Why doesn't this very simple function work? I get NameError: name 'x' is not defined
def myfunc2():
x=5
return x
myfunc2()
print(x)
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do `x = myfunc2()` instead of just `myfunc2()` – Hippolippo Jan 23 '19 at 19:17
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3The function does work. It just doesn't do what you expect it to. You need to learn about *scope* and how return works – juanpa.arrivillaga Jan 23 '19 at 19:19
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1Possible duplicate of [What does python's return statement actually return?](https://stackoverflow.com/questions/45275046/what-does-pythons-return-statement-actually-return) – Daniel Pryden Jan 23 '19 at 19:21
2 Answers
1
You've declared and defined x inside of myfunc2 but not outside of it, so x is not defined outside of myfunc2.
If you'd like to access the value of x outside of myfunc2, you can do something like:
a = myfunc2()
print(a) # 5
I'd suggest reading up on variable scope in Python.
Henry Woody
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3I think to further clarify to a beginner this issue, one should change the names of variables. Otherwise, there's 2 different x's and the difference between the apparently 2 same things is confusing. Just my 2c. – Travis Griggs Jan 23 '19 at 19:22
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The x in myfunc2 is declared as a local. In order for this script to work, you can declare x as global:
def myfunc2():
global x
x = 5
return x
myfunc2()
print(x)
>>>5
Hope this helps.
