g++ warning when using optional 'struct' keyword

Question

If I write this program:

#include <iostream>

namespace foo {
    struct bar {
        int x;
    };
}

int main (void) {
    struct foo::bar *a = new struct foo::bar;
    delete a;
    return 0;
}

and compile it with:

g++ main.cxx -Wall -Wextra

It gives me this warning:

main.cxx: In function ‘int main()’:
main.cxx:10:39: warning: declaration ‘struct foo::bar’ does not declare anything [enabled by default]

However, if I take out the struct keyword after the new keyword:

#include <iostream>

namespace foo {
    struct bar {
        int x;
    };
}

int main (void) {
    struct foo::bar *a = new foo::bar;
    delete a;
    return 0;
}

and compile it the same way, g++ outputs no warnings. Why does g++ output that warning if I use the struct keyword?

Interestingly neither clang++ nor comeau complain about the syntax, and a cursory look into the standard (C++03) didn't help understand the problem either. — David Rodríguez - dribeas
– David Rodríguez - dribeas, Commented Jun 1, 2011 at 21:05
As it is only a warning, I hesitate to say bug - but a GCC bug? The code looks OK to me. Also, it is namespace associated - doing the same thing without a namespace does not produce a warning. — user2100815
– user2100815, Commented Jun 1, 2011 at 21:07

Thomas Matthews · Accepted Answer · 2011-06-01 20:54:30Z

5

In C++, the struct keyword defines a type, and the new type no longer needs the struct keyword. This is one of the many differences between C and C++.

answered Jun 1, 2011 at 20:54

Thomas Matthews

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4 Comments

tyree731 Over a year ago

This explain why the latter code works, but not why the first code produced a warning.

Adetque Over a year ago

I understand that it's unnecessary to use the keyword in C++; I say that it's optional in the thread title. Thanks for responding though.

J T Over a year ago

Is it optional? What does the standard have to say about that?

Adetque Over a year ago

According to MSDN, the struct keyword is "is unnecessary once the type has been defined." This is what I meant, so I apologize if I was being ambiguous.

ildjarn · Accepted Answer · 2011-06-01 23:36:23Z

3

Examining the error:

main.cxx:10:39: warning: declaration ‘struct foo::bar’ does not declare anything [enabled by default]

g++ thinks you're declaring a new struct with the name foo::bar instead of allocating memory of type struct foo::bar. My guess would be because g++ assumes any usage of struct that doesn't declare an lvalue is for the purpose of declaring a type.

edited Jun 1, 2011 at 23:36

ildjarn

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answered Jun 1, 2011 at 20:59

tyree731

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4 Comments

David Rodríguez - dribeas Over a year ago

I understand what you say, but if the compiler is considering that a new type is being declared, then it should bail out with an error, not produce a simple warning, as the standard explicitly states that you cannot declare new types in the new expression

user2100815 Over a year ago

@Adetque It is a good idea to wait at least 24 hours before accepting an answer, so others can have chance to reply. In this case, I think the answer you have accepted is wrong.

ildjarn Over a year ago

@David : Maybe we should just be happy that g++ is punishing users who treat C++ as C-with-classes. ;-P

Adetque Over a year ago

@Neil Butterworth: Ok, I will be sure to do that. @ildjarn: I try to avoid treating it like that. It's just a habit I have from C.

user2100815 · Accepted Answer · 2011-06-01 21:28:06Z

2

Just to post the minimal code that does and does not illustrate the problem:

namespace N {
struct A {};
}

struct B {};

int main() {
    struct N::A * a  = new struct N::A; // problem
    struct B * b  = new struct B;       // ok
}

Personally, I think this is a small GCC bug, but I am not wise in the myriad ways of namespaces.

answered Jun 1, 2011 at 21:28

user2100815

Comments

Loki Astari · Accepted Answer · 2011-06-02 06:01:00Z

1

Note:

   struct foo::bar *a = new struct foo::bar;
// ^^^^^^ (1)
                        //  ^^^^^^ (2)

Not needed in C++
- Putting struct here is a hangover from C and is no longer needed in C++
  As all the types are all in the same namespace.
  While in C structs had there own namespace separate from other types.
  It is allowed to retain backward compatibility with C.
struct here looks like it is forward declaring something.

In C++ the above is written as:

   foo::bar* a = new foo::bar;

// or preferably

   foo::bar* a = new foo::bar(); // This makes a difference if you don't define
                                 // a default constructor.

From David (Below)

Important enough I think it needs to be in the answer rather than the comments:

namespace foo
{
    struct bar {};
    void bar() {}
}

int main()
{
    struct foo::bar* p = new struct foo::bar;
    // Needed here because the compiler can't tell the difference between
    // a struct and a function otherwise
}

edited Jun 2, 2011 at 6:01

answered Jun 1, 2011 at 21:06

Loki Astari

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1 Comment

David Rodríguez - dribeas Over a year ago

1. Not needed in C++, or needed, depending on the rest of the code: namespace foo { struct bar {}; void bar() {} } int main() { struct foo:bar* p = new struct foo::bar; }. In that particular convoluted example, the use of struct is needed in C++ to tell the compiler to search for bar in namespace foo but only in the set of user defined types --and avoid other types of identifiers, like functions or variables.

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