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Home/ Questions/Q 757325
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T15:20:57+00:00 2026-05-14T15:20:57+00:00

Although it runs correctly, the following results in the aforementioned compiler warning: return ((item

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Although it runs correctly, the following results in the aforementioned compiler warning:

return ((item - (my->items))/(my->itemSize));

‘item’ is a ‘void *’; ‘my->items’ is a ‘void *’; ‘my->itemSize’ is an ‘int’

Casting ‘item’ and ‘my->items’ as an ‘int *’ caused the program to run improperly. What is the best way to remove the warning?

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  1. Editorial Team
    Editorial Team
    2026-05-14T15:20:57+00:00Added an answer on May 14, 2026 at 3:20 pm

    Additions and subtractions with pointers work with the size of the pointed type:

    int* foo = 0x1000;
    foo++;
    // foo is now 0x1004 because sizeof(int) is 4
    

    Semantically speaking, the size of void should be zero, since it doesn’t represent anything. For this reason, pointer arithmetic on void pointers should be illegal.

    However, for several reasons, sizeof(void) returns 1, and arithmetic works as if it was a char pointer. Since it’s semantically incorrect, you do, however, get a warning.

    To suppress the warning, use char pointers.

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