warning: format ‘%p’ expects argument of type ‘void *’, but argument 3 has type ‘char **’

Question

I\'m trying to print out the value of argv at this point in my code, but when I try to compile it, I get warning: format ‘%p’ expects argument of type ‘vo

Answer 1

Well, not any pointer (type).

According to C11, chapter §7.21.6.1, for the %p conversion specifier,

p

The argument shall be a pointer to void. The value of the pointer is converted to a sequence of printing characters, in an implementation-defined manner.

So, the argument must be of type void * (or, a char*, given they have the same alignment requirements). Any other type of pointers, must be converted through an explicit cast, as there is no default argument promotion for pointers.

Something like

printf("%s%p", "argv = ", (void *)argv);

should do.