Question

With my compiler (Apple llvm-gg-4.2) this code:

void fun1(const char *s)
{
    char* t = s+1;
}
void fun2(char *s)
{
    char* t = s+1;
}

int main(void)
{
    char* a;
    fun1(a);
    fun2(a);
}

gives this warning:

junk.c:3: warning: initialization discards qualifiers from pointer target type

on fun1 but not on fun2. Why?

Answer 1

fun1 is taking const char* and is being assigned to char*
Whereas fun2 is taking a char* and being assigned to char* which is fine.

If you are assigning a constant pointer to a non-const pointer, this means you can modify the const pointer by using the const pointer

In this case, inside fun1 if you do t[0] = 'a' its not legal because you are modifying const memory, which is why compiler warns you