Function not working in this specific code - interprets wrong "if" statement - Python

Question

Python function that I just wrote is keep on looping up to a hundred. On the interpreter (or rather RUN), I am entering "5", but it executes the wrong "if" statement.

def range_v2(c):
    int(c)
    if c == 5:
        for x in range(1, 50, +5):
            print(x)
    else:
        for x in range(0, 100, +5):
            print(x)

y = input()
int(y)
range_v2(y)

You don't need to set the input as an `int` twice by the way. — HavelTheGreat, Mar 06 '15 at 17:55

score 3 · Answer 1 · answered Mar 06 '15 at 17:33

You are redefining your "c" variable in the function. Your function accepts an argument called "c" and then in the first line of the function you set the value of "c" to 0, so no matter what argument you pass to the function, it will always be evaluated for 0. Remove the line "c=0" and it should work how you expect.

score 1 · Accepted Answer · answered Mar 06 '15 at 17:34

1

int(c) doesn't in-place convert c to an int if that's what you think it does. You'd have to do

c = int(c)

answered Mar 06 '15 at 17:34

Cory Kramer

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@JimmyBernljung because He came up with the answer first – Krystian Mikołajczyk Mar 07 '15 at 17:45

Function not working in this specific code - interprets wrong "if" statement - Python

2 Answers2