Visual Studio C++ compiler weird behaviour

Question

I'm just curious to know why this small piece of code compiles correctly (and without warnings) in Visual Studio. Maybe the result is the same with GCC and Clang, but unfortunately I can't test them now.

struct T {
    int t;
    T() : t(0) {}
};

int main() {
    T(i_do_not_exist);
    return 0;
}

Quatrain

extremly interesting! the most WTF compiler bug I have ever seen! – Bilicki

also int(i_do_not_exist) works fine, but with undefined variable warning – Quatrain

It also compiles with g++: coliru.stacked-crooked.com/a/d8dd5a50af7c888d – Beige

See a similar question: Is Type(::x); valid? – Dirt

@DavidHaim this is not a bug – Magdalenemagdalenian

duplicate of #24622972 – Magdalenemagdalenian

Answer 1

T(i_do_not_exist); is an object declaration with the same meaning as T i_do_not_exist;.

N4567 § 6.8[stmt.ambig]p1

There is an ambiguity in the grammar involving expression-statements and declarations: An expression-statement with a function-style explicit type conversion (5.2.3) as its leftmost subexpression can be indistinguishable from a declaration where the first declarator starts with a (. In those cases the statement is a declaration.

§ 8.3[dcl.meaning]p6

In a declaration T D where D has the form

( D1 )

the type of the contained declarator-id is the same as that of the contained declarator-id in the declaration

T D1

Parentheses do not alter the type of the embedded declarator-id, but they can alter the binding of complex declarators.

Galina

I sometimes hate C++... like just now. – Trinitytrinket

That's why one should use list initialization T{i_do_not_exist} which gives a compiler error. – Laquitalar

Can someone clarify: is this an instance of a "most vexing parse" problem or something unrelated? – Monophysite

@Monophysite how about we call it "slightly vexing parse"? It could be "vexing" because it's ambiguous, but it is less vexing than the traditional "most vexing" T func( U(x), U(y) ); , and perhaps also less vexing than the "somewhat vexing parse" T func(); – Magdalenemagdalenian

Answer 2

Because it defines a variable of type T:

http://coliru.stacked-crooked.com/a/d420870b1a6490d7

#include <iostream>

struct T {
    int t;
    T() : t(0) {}
};

int main() {
    T(i_do_not_exist);
    i_do_not_exist.t = 120;
    std::cout << i_do_not_exist.t;
    return 0;
}

The above example looks silly, but this syntax is allowed for a reason.

A better example is:

int func1();
namespace A
{
   void func1(int);
   struct X {
       friend int (::func1)();
   };
}

Probably other examples could be found.

Pisa