为什么我在尝试将 'char (*c)[6]' 传递给需要 'const char (*c)[6]' 的函数时收到警告？

Question

I know about the char ** vs const char ** thing (like described in the c faq ) but I can't see any scenario where doing so with a pointer to arrays would lead to some content inside the arrays themselves being actually modified.

My code:

void fun(const char (*p)[6])
{
    printf("%s", p[0]);
}

int main(int argc, char *argv[])
{
    char a[6] = "hello";
    char (*c)[6];

    c = &a;

    fun(c);
}

gives the below output when compiled with gcc:

test.c:17:9: warning: passing argument 1 of 'fun' from incompatible pointer type
test.c:5:10: note: expected 'const char (*)[6]' but argument is of type 'char (*)[6]'

The question here is somehow related but has no answer so far. Is it just the compiler being paranoïd and the only way to get rid of the warning is to explicitly cast ? Or is there really a chance something can go wrong ?

Answer 1

Const-conversion is covered by section 6.5.16.1 (1) of the standard:

• both operands are pointers to qualified or unqualified versions of compatible types, and the type pointed to by the left has all the qualifiers of the type pointed to by the right;

In this case it looks like T is char [6] and the rest of the requirement clearly holds, as can be seen by modifying your example:

int main(int argc, char *argv[])
{
    typedef char c6[6];
    c6 a = "hello";
    const c6 *p = &a;
}

However this is actually not the case! This intersects with 6.7.3 (8):

If the specification of an array type includes any type qualifiers, the element type is so qualified, not the array type.

So const c6 * actually names the type const char (*)[6] ; that is, pointer to array[6] of const char , not pointer to const array[6] of char .

Then the LHS points to the type const char[6] , the RHS points to the type char[6] , which are not compatible types, and the requirements for simple assignment do not hold.

Answer 2

It is just a quirk of C language specification. For another example, the char ** to const char *const * conversion is also safe from the const-correctness point of view, yet it is prohibited in C.

This quirk of const-correctness rules was "fixed" in C++ language, but C continues to stick to its original specification in this regard.

Answer 3

Actually the reasons are quite similar (char ** vs. pointer of arrays).

For what you are trying to do, the following would suffice (and it works):

void fun(const char *p)
{
    printf("%s", p);
}

int main(int argc, char *argv[])
{
    char a[6] = "hello";
    char *c;

    c = a;

    fun(c);
}

With what you are trying to do, it would be possible to modify the values as follows that defeats the purpose (just an example):

void morefun(const char *p[6])
{
    char d;
    char *p1 = &d;
    p[1] = p1;
    *p1 = 'X';
    printf("\nThe char is %c\n", *p[1]);
}

int main(int argc, char *argv[])
{
    const char *d[6];

    morefun(d);
}