int a=int(); what happens in C++98?

Question

Please read the question entirely before you think to mark it as duplicate. The statement like

int i=int();

most programmers will say that there is value initialization here & i will be value initialized. (0 as output). But it also prints 0 as output on C++98 compiler. Following program that I tested on C++98 implementation and gives me 0 as output.

#include <iostream>
int main()
{
     int i=int();
     std::cout<<i;
}

Don't say that i is value initialized in above C++98 program ,because value initialization introduced in C++03. So How i is initialized here? Is it really constructor call? int() looks like constructor call. Primitive types have also default constructors in C++ as said by Bjarne stroustrup in his book C++ programming language & TC++PL.

The C++ programming language Bjarne stroustrup:

10.4.2 Built in types also have default constructors

also read section 6.2.8 of same book.

The following links also says that built in types have default constructors in C++.

1) http://www.informit.com/guides/content.aspx?g=cplusplus&seqNum=15

2) http://www.geeksforgeeks.org/c-default-constructor-built-in-types/

So can I really say that it is a constructor call of the integer type?