问题:

Why this shows warning:

#include<stdio.h>
foo (const char **p)
{ 

}

int main(int argc , char **argv)
{
    foo(argv);
}

But following does not show any warning

char * cp;
const char *ccp;
ccp = cp;

The first code snippet shows warning passing arg 1 of foo from incompatible pointer type. But the second snippet does not show any warning. Both are const pointers

回答1:

See the C FAQ list

You can cast in order to remove warnings:

foo((const char **)argv);

But as FAQ says: the need for such a cast may indicate a deeper problem which the cast doesn't really fix.

回答2:

Depending on your compilation flags, you might need an explicit cast when assigning cp's content to ccp.

回答3:

In the first version you are casting between two different types of pointer not simply adding a const to the pointer.

char ** is a pointer to a (pointer to a char)
const char ** is a pointer to a (pointer to a const char)

As you can see these pointer point to different types similar to the more obviously questionable:

int *i;
double *d;
d = i;

In your second example you see that you can cast from a pointer to a const pointer so if you were to apply this to your situation you would need to have a const pointer to (a pointer to a char).