Variable declaration taken as a function pointer declaration

Bolin wrote:[color=blue]
> When compiling the following code:
>
> [code]
> #include <iostream>
>
> struct B {};
>
> struct A
> {
> A(B b1, B b2) {};[/color]
.. ^^^
Trailing semicolon is extraneous here.
[color=blue]
> void foo() { std::cout << "foo called" << std::endl; }
> };
>
> int main(int argc, char * argv[])
> {
> A a(B(), B());
> a.foo();
> return 0;
> }
> [\code]
>
> the compiler will interprete the first line of the main function as a
> function pointer declaration, and thus will fail at the next line. Does
> somebody know why this is so, and if there is an elegant way to solve
> this problem that does not involve temporary variables?[/color]

It is so because the language designers had to make a decision and they
picked the "If it looks like a declaration, it is a declaration" solution.

You can work around it by adding an extra set of parentheses around the
arguments:

A a((B()), (B()));

V

Bolin wrote:[color=blue]
> When compiling the following code:
>
> [code]
> #include <iostream>
>
> struct B {};
>
> struct A
> {
> A(B b1, B b2) {};
> void foo() { std::cout << "foo called" << std::endl; }
> };
>
> int main(int argc, char * argv[])
> {
> A a(B(), B());
> a.foo();
> return 0;
> }
> [\code]
>
> the compiler will interprete the first line of the main function as a
> function pointer declaration, and thus will fail at the next line. Does
> somebody know why this is so, and if there is an elegant way to solve
> this problem that does not involve temporary variables?[/color]

It is so because your code could either be read as a function
declaration or a variable definition and the rule to disambiguate this
is "if it could be a function declaration, it is a function
declaration". The solution is to change the line that defines a to

A a((B()), (B()));

Gavin Deane

Thanks for your answer and your solution -- the thing that surprised me
is that a bare B() could be interpreted as a function pointer.

B.

Bolin wrote:[color=blue]
> Thanks for your answer and your solution -- the thing that surprised me
> is that a bare B() could be interpreted as a function pointer.[/color]

Take a look at

http://www.gotw.ca/gotw/075.htm

particularly the logic up to examples 2d and 2e.

Gavin Deane