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Home/ Questions/Q 7412457
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Editorial Team
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Editorial Team
Asked: May 29, 20262026-05-29T06:32:34+00:00 2026-05-29T06:32:34+00:00

Consider the following code: char* p = new char[2]; long* pi = (long*) p;

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Consider the following code:

char* p = new char[2];
long* pi = (long*) p;
assert(p == pi);         // OK

char* p1 = &p[1];
long* pi1 = (long*) p1;
assert(p1 == pi1);       // OK

int d = p1 - p;
int d1 = pi1 - pi;
assert(d == d1);         // No :(

After this runs, I get d == 1 and d1 == 0, although p1 == pi1 and p == pi (I checked this in the debugger). Is this undefined behavior?

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  1. Editorial Team
    Editorial Team
    2026-05-29T06:32:35+00:00Added an answer on May 29, 2026 at 6:32 am

    As others have pointed, this is undefined behavior. However, there is a very simple explanation for what you are seeing.

    The difference between pointers is the number of elements, not the number of bytes between them.

    pi and pi1 both point to longs, but the address pointed to by pi1 is only one byte further than pi. Presuming longs are 4 bytes long, the difference in the addresses, 1, divided by the size of the element, 4, is 0.

    Another way of thinking of this is you could imagine the compiler would generate code equivalent to this for calculating d1:

    int d1 = ((BYTE*)pi1 - (BYTE*)pi)/sizeof(long).
    
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