为什么--x; work和x-给出递归代码中的seg错误

Question

This code gives me a seg fault, but when I change the x-- to --x it prints correctly.

Are not they the same????

int main()
{    
    myFunc(5); 
    return 0;
}

void myFunc (int x) {  
    if (x > 0) {
        myFunc(x--);
        printf("%d, ", x);
    }
    else
        return;
}

Answer 1

No they are not the same.

The difference between x-- and --x is whether the returned value is before or after the decrement.

In myFunc(x--) , x-- returns the old value. So myFunc() gets called repeatability with the same value -> infinite recursion.

In myFunc(--x) , --x returns the new value. So myFunc() gets called with a decreasing number each time -> no infinite recursion.

---

It will be easier to see this if you moved your printf to the start of the function call:

void myFunc (int x) {  
    printf("%d, ", x);

    if (x > 0) {
        myFunc(x--);

    }
    else
        return;
}

Output: (when called with 10)

10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, ...

Answer 2

The post-decrement operation ( x-- ) takes place after the argument is evaluated, so myFunc is repeatedly called with the same value. The pre-decrement ( --x ) operation takes place before the argument is evaluated, which works as expected.

Answer 3

Because the modification of -- is either post or pre. When you call x-- it is post, and therefore runs after myFunc is called, in this case doing nothing. --x will be called before myFunc and therefore will affect x.

Answer 4

x-- decrements the variable after the use. This is the reason why you pass to your function always the same value, that should be decreased, but doesn't, as well as the function repeats itself.