Why this is not a vexing parse?

Basically this is a 开发者_Go百科follow up of this question about most vexing parse. I can understand that this is due to the ambiguity between the function declaration and variable definition.

But in Comeau online, I just tired the following.

class T{

public:

    T(int i){
    }

    int fun1(){
        return 1;
    }

};

int main()
{
    T myT(10); // I thought it'd be a function declaration that takes an int and returns a type T
    myT.fun1(); // and a compiler error out here.
} 

But it compiles fine and there were no errors. I looked into the standard docs but couldn't come to a reasoning.

So, what am I missing here?

---

Because 10 is not a type. :)

This would be a Most Vexing Parse:

T myT(T());
// T() gets interpreted as function pointer argument to a function returning T();
// This is equivalent to:
T myT(T (*fn)());

Another variety of the Most Vexing Parse is this one:

unsigned char c = 42;
T myT(int(c));
// int(c) gets interpreted as an int argument called c.
// This is equivalent to:
T myT(int c);

---

The 10 cannot be a parameter type name, so this must be a variable declaration.

The compiler must choose a function declaration when it can do that, but in many cases like this it cannot and there is no ambiguity.

---

It's not a vexing parse because you used an integer literal rather than, say:

T myT(T());

As in this complete example:

#include <iostream>

struct T { int f() { return 1; } };

int main(int argc, char** argv) {
    T t(T());
    std::cout << t.f() << '\n';
    return 0;
}

Which is ambiguous because it could mean:

• myT is a T initialised with a default-constructed T; or
• myT is a function returning a T and taking one argument of type T(), which denotes a zero-argument function whose return type is also T.

The latter interpretation is the default one, which is why a compiler error results from attempting to use the newly declared function as though it were the object you expected it to be.

See the Wikipedia article about it.