Question

Consider the following code:

struct Foo
{
};

struct Bar
{
  explicit Bar(const Foo&)
  {
  }
};

int main()
{
  Foo foo;

  Bar bar(foo); // Okay.
  Bar(foo);     // Will not compile.
  (Bar(foo));   // Okay. Unnamed temporary requires parenthesis.
}

Why are the parenthesis around the temporary version required? What ambiguity do they solve?

My hunch is: I think the compiler sees Bar(foo) as a declaration for a function, but I’m not sure why that would be the case since foo (the instance) is not a type. The parenthesis, therefore, force the above to be treated as an expression, not as a forward declaration.

Answer 1

Congratulations on discovering the most vexing parse.

Scott Meyers describes it as follows:

In general, the [C++] language (thanks, unfortunately, to its history) will try to interpret any declaration made like this as the declaration of a function.

In your case, introducing parentheses disambiguates the parse, forcing it to become a local.