[SOLVED] function pointer argument

atlantis43 · 06-12-2013, 07:47 AM

Wondering if anyone can explain, in simple terms, what the argument "(char const *,...)" means in a C99 command such as:

Code:

int (*funcPtr)(char const *,...);

------especially, the *,... portion of the argument.

rtmistler · 06-12-2013, 08:01 AM

"char const" and "const char" mean the same thing in this case given the parenthesis.

The const modifier would normally apply to the variable type on it's left.

Many people write the const first as in

Code:

const char myArray[64];

and when they do so on a variable declaration line, either at the start of the line or just after a comma delimiter because there was another term, the const ends up affecting the variable type to the right of it instead.

Short answer is it means pointer to a constant character, same as

Code:

const char *

would mean the same thing.

ta0kira · 06-12-2013, 08:25 AM

The ... means it's a variadic function, i.e. it takes an arbitrary number of arguments. It looks like a function pointer to a printf type of function.

Kevin Barry

atlantis43 · 06-12-2013, 08:38 AM

Quote:

Originally Posted by ta0kira

The ... means it's a variadic function, i.e. it takes an arbitrary number of arguments. It looks like a function pointer to a printf type of function.

Kevin Barry

Yes, it is a pointer to printf(), and both responses combined successfully addressed my question.