Thread: If a function prototype parameter field is left empty does it assume an int?

Originally Posted by brewbuck

Well, that makes sense, since you obviously have to be able to specify the actual arguments when defining the function.

But why does it change the type promotion rules so that the call with a char is not valid? Weird.

Because you're running the type promotion rules backwards, I think. Seeing () means the compiler will run the usual type promotions, as though it were a ,... type function--integral types get promoted to int, floating point types to double. Now that's that done, you can't go backwards and say "hey that's a char" -- there's no demotion back to a char from an int.

Originally Posted by brewbuck

But I thought that char was promoted to int for argument passing, then converted back to char after invoking the function.

Maybe that's not specified and I'm just thinking of in-the-field behavior.

That might be the in-the-field behavior, but the compiler has to act as though:

If the expression that denotes the called function has a type that does include a prototype,
the arguments are implicitly converted, as if by assignment, to the types of the
corresponding parameters, taking the type of each parameter to be the unqualified version
of its declared type. The ellipsis notation in a function prototype declarator causes
argument type conversion to stop after the last declared parameter. The default argument
promotions are performed on trailing arguments.