What is meant by this?

[maluk]

I found a declaration like this:

void h(int());

What does this mean? Is that argument a pointer to function or what? If it was intended as a pointer to function then should it be declared like this:

void h(int (*)());

Rome did not create a great empire by having meetings, they did it by
killing all those who opposed them.

- janvier -

 

[ArkM]

Syntactically it's a function prototype (incorrect, of course). May be, it's:

void h(int[]); // square brackets

?..

 

[maluk]

Nope.

This one I found on the C++ Standards (ANSI ISO/IEC 14882) document. Here's the link here:

http://www.kuzbass.ru:8086/docs/isocpp/over.html

It is in the third item of section 3. Here are the complete examples:

void h(int());
void h(int (*)());              //  redeclaration of  h(int())
void h(int x()) { }             //  definition of  h(int())
void h(int (*x)()) { }          //  ill-formed: redefinition of  h(int())

Rome did not create a great empire by having meetings, they did it by
killing all those who opposed them.

- janvier -

 

[maluk]

hello? anyone?

Rome did not create a great empire by having meetings, they did it by
killing all those who opposed them.

- janvier -

 

[cpjust]

I thought you already found your answer?
int() is a declaration for a function pointer taking no arguments and returning an int.

 

[maluk]

So int() == int (*)()?

I thought the correct declaration for a pointer to a function is int (*)() rather than int().

Can anyone explain why int () is same as int (*)()? I am confused because int () for me looks like an int declaration that is very different from int(*)().

Rome did not create a great empire by having meetings, they did it by
killing all those who opposed them.

- janvier -

 

[ArkM]

- Parameter declarations that differ only in that one is a function type and the other is a pointer to the same function type are equivalent. That is, the function type is adjusted to become a pointer to function type (8.3.5).
Example:
void h(int());
void h(int (*)());
...

As usually, we can miss names on unused parameters in parameter list.
So it's true equivalent. But many of old C++ compilers do not recognize this construct...
How funny!

 

[cpjust]

Can anyone explain why int () is same as int (*)()?

I don't know why, but unfortunately, that's what the standard says. There's a lot of things in the standard that make me ask "Why the hell did they do that??" ;-)

 

[maluk]

Haha..I guess I should just accept the fact that int() is equivalent to int (*)().

Thanks guys!

Rome did not create a great empire by having meetings, they did it by
killing all those who opposed them.

- janvier -

 

[chipperMDW]

Actually, they aren't the same.

int() is the type of a function.

int(*)() is the type of a pointer to a function.

When used in a prototype, however, int() is treated as exactly equivalent to int(*)().

It's almost like how int[] gets treated the same as int*.