Python导入其他文件函数时避免原文件输出及解决变量未定义问题

Let's break down what's happening here and fix it without needing new files.

Why you're seeing f1's output when running f2 #

When you use from f1 import *, Python executes every line of code in f1 to load its contents into f2's scope. That means the y = fS(a,b), print(y), w = fM(a,b,y), and print(w) lines in f1 run during the import, which is why you see f1's 7 and 84 alongside f2's results.

Why moving a and b into if __name__ == '__main__': caused an error #

The if __name__ == '__main__': block only runs when you execute f1 directly (e.g., python f1.py). When f2 imports f1, this block doesn't run—so if you put a = 3 and b = 4 inside it, those variables never get defined in f1's global scope. The line y = fS(a,b) (which was still in f1's global scope) tries to access a and b that don't exist, hence the "name not defined" error.

The fix: Move execution code (not function definitions) into the guard block #

You only need to wrap the code that runs when f1 is executed directly into the if __name__ == '__main__': block. Keep your function definitions in the global scope so they can be imported by f2. Here's the corrected f1:

# Keep function definitions global so they're importable
def fS(a,b):
    x = a+b
    return x

def fM(a,b,y):
    z = a*b*y
    return z

# Only run this code when f1 is executed directly, not when imported
if __name__ == '__main__':
    a = 3
    b = 4
    y = fS(a,b)
    print(y)
    w = fM(a,b,y)
    print(w)

How this works: #

• When you run python f1.py, __name__ equals '__main__', so the block runs and you get the original output (7 84).
• When f2 imports f1, only the function definitions are loaded—none of the code inside the guard block runs, so f1's print statements don't execute.
• Your f2 code stays exactly the same: when it calls fS(c,d) and fM(c,d,p), it uses the arguments you pass (not any global variables from f1), so you'll only see f2's output (12 420).

内容的提问来源于stack exchange，提问作者Chr_8580