Question

The following C++ code does nothing (using GCC 4.4.3) – it does not print the text:

struct MyStruct { MyStruct() { cout << "Hello" << endl; } };

void foo() {
  MyStruct ();
}

I think this is not so obvious… Let alone the danger of forgetting to give a variable name. Is there a compiler option/warning to forbid compilation of such code or is there any hidden secret behind allowing it?

Edit: I am sorry. The version above MyStruct(); acutally prints. The version which is not printing is:

void bar() {
    MyStruct a();
}

So now I am a bit confused.

Answer 1

This is not a declaration, because in MyStruct ();, the MyStruct would be part of the decl-specifier-seq and forms a type-name therein. And then () can only be a function-declarator. This requires that there is a declarator-id specified, which in your case is not. A special syntactical form is needed to allow such a syntax in declaring a constructor. But such syntactical exceptions are not made in a declaration-statement.

So this construct cannot be a declaration. It is parsed as an expression which specifies a functional cast creating a temporary of type MyStruct.

If the compiler does not print Hello it is either non-conforming or you are not calling foo in your program.

---

Your edit also does not specify a declaration without a name. It instead specifies a declaration that does have a name. It declares a function that is called a. The compiler has no way that you meant something else by this.

MyStruct a();

It could in an effort deduce this by having a recovering rule when it later discovers errors in your code such as the following

a.f();

If you have this in your code to try and call a member function and “a” is a function, the compiler could check to see if MyStruct contains a member f such that this expression is well-formed. But what if you just forgot to place parentheses? The following would be valid for the above declared function that returns a MyStruct, assuming a suitably declared member f.

a().f();

So in effect, the compiler can’t really know what you mean.