The following code compiles ok and runs as expected:

#include <iostream>
#include <iomanip>

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

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::ostream& (*)(std::ostream&)’ 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 Traitsbasic_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?

--
Lionel B

* Lionel B:

The following code compiles ok and runs as expected:
>
#include <iostream>
#include <iomanip>
>
int main()
{
typedef std::ostream& (*manip_t)(std::ostream&);
>
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::ostream& (*)(std::ostream&)’ 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 Traitsbasic_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.

--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?

On Fri, 20 Apr 2007 13:03:09 +0200, Alf P. Steinbach wrote:

* 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,

--
Lionel B

On Apr 20, 12:56 pm, Lionel B <m...@privacy.netwrote:

The following code compiles ok and runs as expected:

#include <iostream>
#include <iomanip>

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

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::ostream& (*)(std::ostream&)? 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 Traitsbasic_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.

--
James Kanze (Gabi Software) email: james.kanze@gmail.com
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