Question

I have a function whose prototype is as follows:

void foo(const char * data);

Elsewhere in my code, I have a global variable declared as follows

volatile char var[100];

Whenever I try to do this:

foo(var);

The compiler throws up the following error message:

Argument of type “volatile char *” is incompatible with parameter of type “const char *”

Why is that the case? As I understand it, the variable in my function is not allowed to change the pointer or its contents. I understand that because my global variable is volatile, it could potentially change at any time, but seeing as it is perfectly legal to have a volatile const variable, I don’t see why I am getting this compiler error.

Thanks

–Amr

Answer 1

It’s because implicit conversions can add qualifiers to the target of pointer types, but not remove them. So if you want your function to be able to accept volatile and/or const qualified pointers, you must declare it with both:

void foo(const volatile char * data);