我正在尝试对 C++ 中的两个指针的值进行算术运算，但它有时有效，有时失败。 努力在网上寻找解释

Question

I feel silly due to the basicness of the question but using my basic understanding of pointers my program is performing unexpectedly and I can't understand why.

#include <stdio.h>

void update(int *a,int *b) {
    
    *a = *b + *a;
    *b = (*a - *b);
}

int main() {

    int a, b;
    int *pa = &a, *pb = &b;
    scanf("%d %d", &a, &b);
    update(pa, pb);
    printf("%d\n%d", a, b);
    return 0;
}

So, pointers point to a memory address. If I would do arithmetic on the pointer itself I would change the address location. But a *(pointer variable) would be the value at the memory address. Is it wrong to assume that arithmetic such as (*a + *b) = *a works basically the same as (a + b) = a. I assumed I was correct due to *a = (*b + *a) working as I intended in my code.

So I'm quite confused as *b = (*a - *b) seemingly always changes *b = *a rather than actual subtraction. With inputs of 5 and 2 making *b = 5 which wouldn't make sense if subtraction took place. Input: 6,5 Outputs: 6 Input: 2, 1 Outputs: 2

Also, I have tested the code in hacker rank and visual studio code and I keep getting the same result for *b in both. The top search results for "arithmetic on pointer values", "how to change value at a memory address c++" and "can you do math on the value of pointers c++" either don't address my question (though they could be address my question but I just didn't make the connection.)

If someone could explain why I'm getting the output for *b that was unexpected, if doing arithmetic on *b and *a actually works the way I think it should and if there is a better or more efficient way to do what I'm trying to accomplish. I truly appreciate anyone who answers. Thanks

Answer 1

This has nothing to do with them being pointers, but with the operations being done.

*a = *b + *a;
*b = (*a - *b);

is done in sequence, so the *a value is changed before calculating *b .
I assume you were expecting something like

int a_before = *a;
*a = *b + *a;
*b = (a_before - *b);