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Asked: 2026-05-26T08:39:45+00:00

This is an example of what my code looks like: void *a = NULL;

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This is an example of what my code looks like:

void *a = NULL;
void *b = //something;
a = (void *)(*((char *)b + 4));

The value that (b+4) is pointing to is an address that I want to store in a. When I try to compile, I get “warning: cast to pointer from integer of different size.” What does this mean, and what should I do to fix it?

EDIT: To clarify, I don’t want ‘a’ to point to an address that is 4 bytes greater than ‘b’. In my program, I know that the value stored at ((char *)b + 4) is itself another pointer, and I want to store this pointer in ‘a’.

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1 Answer

  1. Editorial Team

    This is a char*:

    (char *)b + 4

    That means that this:

    *((char *)b + 4)

    is a char. Then you proceed to cast that char to a void* and the compiler complains. You don’t have to manually cast to void* in C so just this should do:

    a = (char *)b + 4;

    Update for comments: Sounds like you’re after this:

    a = *(char **)((char *)b + 4)
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