Variable declaration taken as a function pointer declaration

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?

B.

re: 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

re: 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) {};
> 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

re: Variable declaration taken as a function pointer declaration

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.

re: Variable declaration taken as a function pointer declaration

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