有人可以用 C 解释一下这段代码吗？

Question

The output is :

25
28ff1c
28ff1c
25

I don't understand why. The first value and address are understandable, but why is the second output the same? Doesn't the address of g get changed in the function?

float add(int *x, int y)
{
    static float s = 100.f;
    s=s+y;
    x=x+y;
    return s;
}



int main()
{

    int g=25;
    printf("%d\n",g);
    printf("%x\n",&g);

    add(&g,35.2);

    printf("%x\n",&g);
    printf("%d\n",g);

return 0;
} 

Answer 1

You cannot change the address of a variable with a single pointer (ie int *x). As suggested in comments this may not have been your intention so sorry if you know this.

To change the address you must use a double pointer, you must deference it once to get the address and to access the value you must deference it twice.

void func(int **x)
{
    int value = **x;
    uintptr_t address = *x;

    *x = address + sizeof(int);
}

Answer 2

...but why is the second output the same?

Because your original code:

float add(int *x, int y)
{
    static float s = 100.f;
    s=s+y;
    x=x+y; // changing the pointer
    return s;
}

Does not change the value pointed to by the address of x.
Change the code as shown to allow the value to be updated:

float add(int *x, int y)
{
    static float s = 100.f;
    s=s+y;
    *x=*x+y; //changing the value pointed to by the pointer
             //(i.e., the value exposed via the de-referenced pointer is 
             //modified to a new value.)
    return s;
}

Given:

 int g=25;
 ...
 add(&g,35.2);

Results after change:

25
81feb8
81feb8
60

Note: Because the type of the 2nd argument: float add(int *x, int y)
is int , 35.2 is truncated to 30 , resulting the in the sum being
60 instead of 60.2