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Editorial Team
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Asked: 2026-05-22T14:41:33+00:00

My question is very simple one. Say we have: char* ptr = (char*) malloc(sizeof(char)*SIZE);

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My question is very simple one. Say we have:

char* ptr = (char*) malloc(sizeof(char)*SIZE);
ptr+= SIZE/2;
free(ptr);

What happens when we free the pointer? Is it undefined operation?
Does it free all of SIZE buffer or only the remaining SIZE/2?
Thanks in advance for disambiguating this for me.

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1 Answer

  1. Editorial Team

    Your program will probably crash: the free() operation is actually quite simple in C, but works only on the original allocated address.

    The typical memory allocator works like this pseudo code:

    • ask to alloc 64 bytes
    • allocator allocs 70 bytes (6 bytes more)
    • the first 2 bytes is set to a “signature”, a pattern recognized by the allocator to identify the memory allocated by him
    • the next 4 bytes denote the allocated size
    • return a pointer to the start of the 7th byte

    So when you call free(ptr), the allocator goes 6 bytes before your pointer to check for the signature. If it doesn’t find the signature, it crashes 🙂

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