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Home/ Questions/Q 6950385
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T14:05:09+00:00 2026-05-27T14:05:09+00:00

I came across the following code: int main() { char *A=(char *)malloc(20); char *B=(char

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I came across the following code:

int main()
{
    char *A=(char *)malloc(20);
    char *B=(char *)malloc(10);
    char *C=(char *)malloc(10);
    printf("\n%d",A);
    printf("\t%d",B);
    printf("\t%d\n",C);
    return 0;
}  
//output--   152928264     152928288    152928304

I want to know how the allocation and padding is done by malloc(). Looking at the output I can see that the starting address is a multiple of 8. Arethere any other rules?

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  1. Editorial Team
    Editorial Team
    2026-05-27T14:05:09+00:00Added an answer on May 27, 2026 at 2:05 pm

    Accdording to this documentation page,

    the address of a block returned by malloc or realloc in the GNU system is always a multiple of eight (or sixteen on 64-bit systems).

    In general, malloc implementations are system-specific. All of them keep some memory for their own bookkeeping (e.g. the actual length of the allocated block) in order to be able to release that memory correctly when you call free. If you need to align to a specific boundary, use other functions, such as posix_memalign.

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