Returning Variables in Functions Python Not Working Right
I have been trying to return a variable in a function in a variable and use it outside of it:
test = 0 def testing(): test = 1 return test testing() print(test)
But when I run it, the result is 0. How could I fix this problem?
You are messing up a bit the scopes and/or assignment. Try this:
def testing(): test = 1 return test test = testing() print(test)
testing is different to the
test inside the module. You have to assign it on module-level to get the expected result.
Return doesn't work (python), I suspect you are trying to print(y) outside the function. The variable y is local in scope, that is only defined within fpol(). So you can print it there. Returning Variables in Functions Python Not Working Right (5 answers) Closed last year . It seems like I can't pass a value from a function to another even though I have put a return statement in the 1st function.
Because you declare
test in the function, it is not a global variable, thus, you can not access the variable
test you created in the function outside of it as they are different scopes
If you want to
return test to a variable, you have to do
result = testing() print(result)
Or, you can also add a
test = 0 def testing(): global test test = 1 return test testing() print(test)
By the way, when doing a conditional statement, you don't need the brackets around the
Learning with Python 3, But if the function should return a value, make sure all execution paths do return the value. To find the second problem above, consider what happens if you call this function with an It is preferred over this one, which also works correctly:� You need to go back and work through a tutorial about the basics of variable scope, e.g. the difference between local and global variables, as well as how returning values from a function works. You seem to be under the misapprehenesion that returning from a function makes the local variables you returned leak into the caller's (in this case
Inside the function
testing(), you're creating a new variable
test, not referring to the one that already exists. If you want to do that, you should use a
global statement in the top, as in:
def testing(): global test ...etc...
Python "return (variable)" in a function does not output, Basically “return” isn't working. The return function works when I run them through the terminal. As shown below: [code]>>> def greatest(a): All Python functions return the value None unless there is an explicit return statement with a value other than None. Consider the following common mistake made by beginning Python programmers. As you step through this example, pay very close attention to the return value in the local variables listing.
test variable inside the function does not have a global scope. So, if you want to store the return value in a variable and output it after that, you can do something like this:
result = testing() print(result)
8. Global & Return — Python Tips 0.1 documentation, The function above takes two values as input and then output their addition. to solve this problem by returning a tuple , list or dict with the required values after� Python functions can return multiple variables. These variables can be stored in variables directly. A function is not required to return a variable, it can return zero, one, two or more variables. This is a unique property of Python, other programming languages such as C++ or Java do not support this by default.
return value must be assigned to something at the call site.
test = testing()
Functions in Python have their own scope. It is created on entering (calling) the function, and destroyed when leaving it. Assignment to a name inside a scope makes that name local to this scope - causing it to be destroyed along with the scope.
# start outer scope test = 0 # create name outer:test def testing(): # start inner scope test = 1 # create name outer.testing:test return test # end inner scope # destroy name outer.testing:test testing() # new temporary inner scope outer.testing print(test) # use name outer:test # end outer scope
Notably, names in an inner scope may "shadow" names from an outer scope. While the name
test exists in both
testing and the outer scope, it does not refer to the same thing. This has two important implications:
- The assignment to the inner
testdoes not affect the outer
- At the end of
testing, the inner
testis destroyed and only the outer
This is why calling
testing() does not have the desired effect: it never modifies the outer
test passed to
return statement defines the value returned by calling a function. It does not return the name, only the value pointed to.
def testing(): test = 1 # test refers to the value 1 return test # return test => value 1
The value returned by a function is like any other value - be it from a literal, lookup, or other. Most importantly, the value does not persist unless you assign it to a name or use it directly.
testing() # call test, discard its value test = testing() # call test, store its value as `test` print(testing()) # call test, use its value in `print`
So in order to return something from a function for later use, you must store the result to a name. You can then use that name in a later statement. A minimal example for your case looks like this:
# we already can define testing here # it is overwritten later on, then def testing(): # all names we use inside of testing are gone at the end # if we just want a value, we can skip temporary names return 1 # store the return value of testing() for later use test = testing() print(test)
Addendum: It is possible for a function to modify its containing scope. However, names must then be explicitly declared as being from a foreign scope.
global keywords allow to modify names from outer scopes. A
nonlocal is the name in the closest matching function scope. A
global is the name at the module scope, regardless of any functions in-between.
test = 0 def increment(): global test # declare test as belonging to a specific scope test += 1 # no need to return something # we already modified the outer `test` print(test) # 0 increment() print(test) # 1
Note that modifying outer names is often the sign of an anti-pattern, moreso for
nonlocals. Beyond small scripts, it gets difficult to trace what is accessing and modifying
globals. Often, it is more appropriate to use classes or generators to hold state.
A function can always read names from its containing scope, provided it never writes to the same name. Such closures are very easy to create, and the lack of modification makes them easier to trace. Note that modifying a name anywhere in a function makes it local, unless declared
test = 0 def increment(): global test test += 1 def show_test(): # we never modify `test`, so it is fetched from the outside print(test) def show_and_increment1(): # this function is broken! print(test) # `test` is *not* the outer one, since we modify it in the function test += 1 # modifying `test` makes it local for the *entire* function def show_and_increment2(): # this function works! global test # force `test` to be global print(test) test += 1 show_test() # 0 increment() show_test() # 1 show_and_increment2() # 1 show_and_increment2() # 2 show_and_increment2() # 3 show_test() # 4 show_and_increment1() # UnboundLocalError: local variable 'test' referenced before assignment
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