How does this declaration invoke the Most Vexing Parse?

Question

Consider the following program:

#include 

struct A {};

int main(int argc, char** argv) {
    A a(std::fstream(argv[1]));
}

Answer 1

I think it's being parsed as

A a(std::fstream argv[1]);

i.e. a function that takes in an array of 1 std::fstream, where the extra parentheses are redundant.

Of course, in reality, that array parameter decays to a pointer, so what you end up with semantically is:

A a(std::fstream* argv);

Answer 2

The parens are superfluous. All of the following are the declarations:

T a;
T (a);
T ((a));
T (((a))));

so are these:

T a[1];
T (a[1]);
T ((a[1]));
T (((a[1]))));

So is this:

 A a(std::fstream(argv[1]));

which is same as:

 A a(std::fstream argv[1]);

which is same as:

 A a(std::fstream *argv);

Hope that helps.