Construction of temporary in function call is interpreted as declaration

Question

Lately I ran into a problem which somehow (but only somehow) makes sense to me. It is based on interpreting the construction of a temporary as declaration of the single (!)

Answer 1

The rule is that if a declaration has the syntax of a function declaration then it is one; otherwise it is a variable declaration. Surprising instances of this are sometimes called most-vexing-parse.

Bar bar0(Foo0(x), Foo1(x), Foo2(x)); // Does not work: conflicting declaration ‘Foo1 x’ previous declaration as ‘Foo0 x’; conflicting declaration ‘Foo2 x’ previous declaration as ‘Foo0 x’

This is a function declaration: bar0 is the name, Bar is the return type, and the parameter types are Foo0, Foo1 and Foo2. The parameter names are all x which is illegal - the names of function parameters must be different. If you change x x x to x y z the error goes away).

Bar bar1(Foo0{x}, Foo1(x), Foo2(x)); // Works WTF
Bar bar2(Foo0(x), Foo1{x}, Foo2(x)); // Works WTF
Bar bar4(Foo0{x}, Foo1{x}, Foo2{x}); // Works totally makes sens to me

These lines and create objects bar1, bar2, and bar4 of type Bar. They cannot be parsed as function declarations because the { } notation is not valid syntax in a function declaration.

Therefore, Foo0{x} etc. are expressions which provide the arguments to Bar's constructor. Foo0{x} and Foo0(x) are equivalent ways of declaring a temporary of type Foo0 with initializer x.

Bar bar3(Foo0(x), Foo1(x), Foo2{x}); // Does not work: conflicting declaration ‘Foo1 x’ previous declaration as ‘Foo0 x’

I think this is a compiler bug; the part Foo2{x} means that this line cannot be a function declaration; and it looks like a valid declaration of a variable bar3.

x.doStuff(); //Dose not work. This makes sens to me. But in the context its curious

x is an int ; it does not have any methods.