const char* 指针算术警告

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

Answer 2

In fun1 , s is a const char * . When you do char* t = s+1; , you "remove" that const status from s . Hence the "discards qualifiers". If this were C++, you would get a compiler error instead of a warning.

Answer 3

The reason why is in fun1 you are converting const char * to a char * . This is losing the const qualifier and opening the door to modify data the function likely didn't intend to be modified.

To fix this change the fun1 body to

const char* t = s+1;