why is `std::istreambuf_iterator<char>` regard as function declaration when constrcuting a string?

Question

Many coders may be confused this code:

int main() {
  std::ifstream ifs("filename", std::ios::binary);
  std::string content((std::istreambuf_iterator<char>(ifs)), std::istreambuf_iterator<char>());
  //                  ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^
  ...
}

I would like to understand this code clearly, so the question is:

1. Why is std::istreambuf_iterator<char>(ifs) enclosed in ( , ) ?

2. How to determinate this code if remove the brackets here? and

3. How to determinate the code if we don't remove the brackets?

Thanks for your help.

Answer 1

In declarations declarators may be enclosed in parentheses.

Consider the following function declaration

void f( int ( x ), int ( y ) );

This declaration declares a function with two parameters of the type int.

In a declaration names of identifiers may be omitted.

So you may write

void f( int (), int () );

This declaration in turn declares a function that accepts two parameters of the function type int() .

So this construction

std::string content(std::istreambuf_iterator<char>(ifs), std::istreambuf_iterator<char>());

can be considered by the compiler as a declaration of a function that has the return type std::string and with the first parameter of the type std::istreambuf_iterator<char> and the identifier ifs and the second parameter with the function type istreambuf_iterator<char>() .

To distinguish a declaration from an expression there are used either parentheses like

std::string content( ( std::istreambuf_iterator<char>(ifs) ), std::istreambuf_iterator<char>());

or braces

std::string content(std::istreambuf_iterator<char>{ ifs }, std::istreambuf_iterator<char>());

Here is a demonstration program that shows how can look declarators in parameter declarations.

#include <iostream>

void f( int ( x ), int ( y ) )
{
    std::cout << "x = " << x << ", y = " << y << '\n';
}

void f( int(), int() );

void f( int g(), int h() )
{
    std::cout << g() + h() << '\n';
}    

int g() { return 100; }

int main()
{
    f( 10, 20 );    

    f( g, g );
}

The program output is

x = 10, y = 20
200

As for the question appeared in a comment about declaration of a parameter of a function type then for example the C function qsort uses such a parameter that specifiers the relation between two elements of an array.

void qsort(void *base, size_t nmemb, size_t size,
int compar(const void *, const void *));

Another example is of using a function type as a template argument for a class template std::function .