(2 - 4 = -1) when int value assigned to pointer in

问题:

I am unable to get that why in this program 2 - 4 gives -1, it has assigned int values to pointers rather than addresses, I know but while I compiled it compiler gave some warnings but compiled the program and it executed but...

Program

#include<stdio.h>

int main(void) {

    int *p, *q;

    int arr[] = {1,2,3,4};

    // I know p and q are pointers and address should be assigned to them
    // but look at output, why it evaluates (p-q) to -1 while p as 2 and q as 4

    p = arr[1];
    q = arr[3];

    printf("P-Q: %d, P: %d, Q: %d", (p - q), p, q);

    return 0;
}

It gives

P-Q: -1, P: 2, Q: 4

回答1:

Strictly speaking, what happens depends entirely on your compiler and platform... but let's assume we're using a typical compiler and ignoring the warnings.

Let's simplify your question further:

p = 2;
q = 4;

printf("P-Q: %d, P: %d, Q: %d", (p - q), p, q);

which produces the same wacky result:

P-Q: -1, P: 2, Q: 4

As @gsamaras pointed out, we're trying to subtract two pointers. Let's try and see how this might result in -1:

p - q = (2 - 4) / sizeof(int)
      = (-2)    / 4
      = -1

I suggest trying a couple of your own p and q values to see what happens.

---

Examples with different p and q:

p - q = ??
==========
0 - 0 =  0
0 - 1 = -1
0 - 2 = -1
0 - 3 = -1
0 - 4 = -1
1 - 0 =  0
1 - 1 =  0
1 - 2 = -1
1 - 3 = -1
1 - 4 = -1
2 - 0 =  0
2 - 1 =  0
2 - 2 =  0
2 - 3 = -1
2 - 4 = -1
3 - 0 =  0
3 - 1 =  0
3 - 2 =  0
3 - 3 =  0
3 - 4 = -1
4 - 0 =  1
4 - 1 =  0
4 - 2 =  0
4 - 3 =  0
4 - 4 =  0

Generated using gcc -fpermissive on:

#include <stdio.h>

int main() {
    printf("p - q = ??\n");
    printf("==========\n");

    for (int i = 0; i < 5; ++i) {
        for (int j = 0; j < 5; ++j) {
            int* p = i;
            int* q = j;

            printf("%d - %d = %2d\n", p, q, (p - q));
        }
    }

    return 0;
}

回答2:

The duplicate question mentions that:

pointer subtraction yields the number of array elements between two pointers of the same type

Read more about it in Pointer subtraction confusion.

However, your code is wrong and ill-formed, since it invokes Undefined Behavior. Please compile with warnings enabled, and you will get:

main.c: In function ‘main’:
main.c:12:7: warning: assignment makes pointer from integer without a cast [-Wint-conversion]
     p = arr[1];
       ^
main.c:13:7: warning: assignment makes pointer from integer without a cast [-Wint-conversion]
     q = arr[3];
       ^
main.c:15:12: warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘long int’ [-Wformat=]
     printf("P-Q: %d, P: %d, Q: %d", (p - q), p, q);
            ^
main.c:15:12: warning: format ‘%d’ expects argument of type ‘int’, but argument 3 has type ‘int *’ [-Wformat=]
main.c:15:12: warning: format ‘%d’ expects argument of type ‘int’, but argument 4 has type ‘int *’ [-Wformat=]

The errors will occur nevertheless. For the warnings, I just used the -Wall flag.

---

In order for your code to make sense, you could just declare p and q as simple ints and and not as pointers.

Or, you could do this:

p = &arr[1];
q = &arr[3];

printf("P-Q: %td, P: %p, Q: %p", (p - q), (void *)p, (void *)q);

and get something like this:

P-Q: -2, P: 0x7ffdd37594d4, Q: 0x7ffdd37594dc

Note that I used %td for printing the result of the subtraction of pointers.