Type of ios manipulators

[Lionel B]

The following code compiles ok and runs as expected:

#include <iostream>
#include <iomanip>

int main()
{
typedef std:stream& (*manip_t)(std:stream&);

manip_t x = std::endl;

if (x==static_cast<manip_t>(std::endl))
std::cout << "endl detected" << std::endl;

return 0;
}

However, without the cast I get a compiler error:

main.cpp:10: error: assuming cast to type ‘std::basic_ostream<char, std::char_traits<char> >& (*)(std::basic_ostream<char, std::char_traits<char> >&)’ from overloaded function

which I interpret as:

main.cpp:10: error: assuming cast to type ‘std:stream& (*)(std:stream&)’ from overloaded function

ie.

main.cpp:10: error: assuming cast to type ‘manip_t’ from overloaded function

But my standard library ref defines:

template<typename CharT, typename Traits> basic_ostream<CharT, Traits>& endl(basic_ostream< CharT, Traits >&)

which also boils down to my manip_t... so why does my compiler
seem to think an explicit cast is necessary?

 

[Alf P. Steinbach]

* Lionel B:

The following code compiles ok and runs as expected:

#include <iostream>
#include <iomanip>

int main()
{
typedef std:stream& (*manip_t)(std:stream&);

manip_t x = std::endl;

if (x==static_cast<manip_t>(std::endl))
std::cout << "endl detected" << std::endl;

return 0;
}

However, without the cast I get a compiler error:

main.cpp:10: error: assuming cast to type ‘std::basic_ostream<char, std::char_traits<char> >& (*)(std::basic_ostream<char, std::char_traits<char> >&)’ from overloaded function

which I interpret as:

main.cpp:10: error: assuming cast to type ‘std:stream& (*)(std:stream&)’ from overloaded function

ie.

main.cpp:10: error: assuming cast to type ‘manip_t’ from overloaded function

But my standard library ref defines:

template<typename CharT, typename Traits> basic_ostream<CharT, Traits>& endl(basic_ostream< CharT, Traits >&)

which also boils down to my manip_t... so why does my compiler
seem to think an explicit cast is necessary?

A cast resolves the ambiguity of which overload you mean.

 

[Lionel B]

* Lionel B:

A cast resolves the ambiguity of which overload you mean.

Ah, ok. Comeau online gives a clearer (to me) error message:

"ComeauTest.c", line 10: error: cannot determine which instance of
function
template "std::endl" is intended
if (x == std::endl)
^
Thanks,

 

[James Kanze]

The following code compiles ok and runs as expected:

#include <iostream>
#include <iomanip>

int main()
{
typedef std:stream& (*manip_t)(std:stream&);

manip_t x = std::endl;

if (x==static_cast<manip_t>(std::endl))
std::cout << "endl detected" << std::endl;
return 0;
}

However, without the cast I get a compiler error:

main.cpp:10: error: assuming cast to type ?std::basic_ostream<char, std::char_traits<char> >& (*)(std::basic_ostream<char, std::char_traits<char> >&)? from overloaded function

which I interpret as:

main.cpp:10: error: assuming cast to type ?std:stream& (*)(std:stream&)? from overloaded function

main.cpp:10: error: assuming cast to type ?manip_t? from overloaded function

But my standard library ref defines:

template<typename CharT, typename Traits> basic_ostream<CharT, Traits>& endl(basic_ostream< CharT, Traits >&)

which also boils down to my manip_t...

If that were true, then your code wouldn't be legal. I'll bet
your manip_t is a type, and not a template.

so why does my compiler seem to think an explicit cast is
necessary?

Because std::endl is a template, and it has to know which one to
instantiation.