memcpy with destination pointer to const data

Question

I always thought that an statement like const int *a means a is an int pointer to const data and as such one should not be able to modify the value it points to. Indeed if you do const int a [] = {1,2,3} and then issue a[0] = 10 you'll get compiler errors.

To my surprise, however, the following compiles without any warning and runs just fine.

#include <stdio.h>
#include <string.h> 

int main (){

    const int a [] = {1, 1, 1};
    const int b [] = {2, 2, 2};

    memcpy((void*) &a[0], (const void*)&b[0], 3*sizeof(int));

    int i;
    for (i=0; i<3; i++) printf("%d\n",a[i]);

    return 0;
} 

Why is this allowed? Is this due to the cast? When I do memcpy(&a[0], (const void*)&b[0], 3*sizeof(int)); compiler promptly generates the following warning:

cpy.c: In function ‘main’:
cpy.c:9:3: warning: passing argument 1 of ‘memcpy’ discards ‘const’ qualifier from pointer target type [enabled by default]
/usr/include/string.h:44:14: note: expected ‘void * __restrict__’ but argument is of type ‘const int *’

Answer 1

You told the compiler to disregard the initial declaration when you performed the cast. It listened. That doesn't mean that your program is correct however. Modifying what was originally declared to be const results in undefined behavior (for example, the compiler is free to store that data in read only memory).

C doesn't hold your hand. If you chose to do something dangerous then it will let you.