'Get returned value of function in function which called it
I am newbie in python, and I build two functions in which, I am calling second function with 1 parameters in first function and I am trying to access second function's returned data in first function.
def second_function(first_param):
final = first_param + 50
return final
def first_function():
second_function(50)
# trying to access second_function's returned data HERE
print(second_function)
But it is not showing any returned data.
Any help would be much Appreciated. Thank You in Advance.
Solution 1:[1]
The problem here is that you are using print(second_function), so that will simply output the name of the function. Now, if you want to output the result of the function, you should do:
def second_function(first_param):
final = first_param + 50
return final
def first_function():
output = second_function(50)
print(output)
Solution 2:[2]
you could first put the returned value in a variable like this
def second_function(first_param):
final = first_param + 50
return final
def first_function():
value = second_function(60)
print(value )
or print the returned value with out using any variable
def second_function(first_param):
final = first_param + 50
return final
def first_function():
print(second_function(50))
Solution 3:[3]
That's because second_function is an object in its own right. Try either of the following:
def first_function():
out = second_function(50)
# trying to access second_function's returned data HERE
print(out)
def first_function_alternate():
print(second_function(50))
What's happening when you do print(second_function) is that the computer is trying to print the value of the function itself, not what it returns. We can store this value to a variable (my first answer) or simply generate it on-the-fly (my second answer).
Solution 4:[4]
In Python, the returned data from a function will be assigned to a variable. So you would use:
my_value = second_function(60)
and the returned value would be stored in the variable my_value
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Alberto Agudo Dominguez |
| Solution 2 | |
| Solution 3 | Jakob Lovern |
| Solution 4 | Coddo |
