Question

Consider

#include <cstdio>

int main() {
    char a[10];
    char *begin = &a[0];
    char *end = &a[9];
    int i = end - begin;
    printf("%i\n", i);
    getchar();
}

---

#include <cstdio>

int main() {
    int a[10];
    int *begin = &a[0];
    int *end = &a[9];
    int i = end - begin;
    printf("%i\n", i);
    getchar();
}

---

#include <cstdio>

int main() {
    double a[10];
    double *begin = &a[0];
    double *end = &a[9];
    int i = end - begin;
    printf("%i\n", i);
    getchar();
}

All the above three examples also print 9

May I know, how I should interpret the meaning of 9. What does it mean?

Answer 1

the compiler will automatically calculate pointer arithmetic based on type of pointer, which is why you cant perform operation using void* (no type information) or mixed pointer type (ambiguous type).

In MSVC2008 (and in most other compiler i believe), the syntax is interpreted as calculate the amount of element difference between two pointer.

int i = end - begin;
00411479  mov         eax,dword ptr [end] 
0041147C  sub         eax,dword ptr [begin] 
0041147F  sar         eax,3 
00411482  mov         dword ptr [i],eax 

Since the subtraction is followed by a right-shifting, the result will be round-down, hence guarantee N element can fit into memory space between two pointer (and there might be unused gap). This is proven in code below which yield result of 9 too.

int main()
{
    double a[10];
    double *begin = &a[0];
    char *endc = (char*)&a[9];
    endc += 7;
    double *end = (double*)endc;
    int i = end - begin;
    printf("%i\n", i);
    getchar();

    return 0;
}