why do I need ::sin, not std::sin?

[Alexander Stippler]

Hi,

simple question. Why does this not work:

#include <algorithm>
#include <cmath>
#include <iostream>

int
main()
{
double a[2];
a[0] = 0;a[1] = M_PI/2.;
std::transform( a, a+2, a, std::sin);
std::cout << a[0] << ", " << a[1] << std::endl;
return 0;
}

for gcc and icc the type of the third argument of transform is unknown. If I
use the global ::sin it works fine. Why? I include cmath and there sin is
defined in namespace std, so why is it unknown?

regards,
alex

[red floyd]

Alexander Stippler wrote:

Hi,

simple question. Why does this not work:

#include <algorithm>
#include <cmath>
#include <iostream>

int
main()
{
double a[2];
a[0] = 0;a[1] = M_PI/2.;
std::transform( a, a+2, a, std::sin);
std::cout << a[0] << ", " << a[1] << std::endl;
return 0;
}

for gcc and icc the type of the third argument of transform is unknown. If I
use the global ::sin it works fine. Why? I include cmath and there sin is
defined in namespace std, so why is it unknown?

regards,
alex

IANAG (I am not a guru), and am probably way off base (please correct me
guys -- please no flames).

Isn't std::sin a template? Would you need to specifically instantiate
std::sin, i.e.

std::transform( a, a+2, a, std::sin<double >)

You might also want to use std::ptr_fun as a wrapper to std::sin (or ::sin).

[Rob Williscroft]

Alexander Stippler wrote in news:40******@n ews.uni-ulm.de in
comp.lang.c++:

Hi,

simple question. Why does this not work:

#include <algorithm>
#include <cmath>
#include <iostream>

int
main()
{
double a[2];
a[0] = 0;a[1] = M_PI/2.;
std::transform( a, a+2, a, std::sin);
std::cout << a[0] << ", " << a[1] << std::endl;
return 0;
}

for gcc and icc the type of the third argument of transform is
unknown. If I use the global ::sin it works fine. Why? I include cmath
and there sin is defined in namespace std, so why is it unknown?

regards,
alex

There are several overloads of std::sin, you need to pick one:

std::transform( a, a+2, a, (double (*)(double)std: :sin);

Note M_PI isn't AFAICT Standard C++, you can use:

double const M_PI = (2.0 * std::acos( 0.0 ));

Though maybe:

double const double_pi = (2.0 * std::acos( 0.0 ));

Whould be better.

HTH.

Rob.
--
http://www.victim-prime.dsl.pipex.com/

[Pete Becker]

red floyd wrote:

Isn't std::sin a template?

No, just an ordinary function, with overloads for arguments of type
float, double, and long double. (There's also a sin template for
arguments of type complex<T>, but that's not involved here)

--

Pete Becker
Dinkumware, Ltd. (http://www.dinkumware.com)

[Alexander Stippler]

red floyd wrote:

Alexander Stippler wrote:

Hi,

simple question. Why does this not work:

#include <algorithm>
#include <cmath>
#include <iostream>

int
main()
{
double a[2];
a[0] = 0;a[1] = M_PI/2.;
std::transform( a, a+2, a, std::sin);
std::cout << a[0] << ", " << a[1] << std::endl;
return 0;
}

for gcc and icc the type of the third argument of transform is unknown.
If I use the global ::sin it works fine. Why? I include cmath and there
sin is defined in namespace std, so why is it unknown?

regards,
alex

IANAG (I am not a guru), and am probably way off base (please correct me
guys -- please no flames).

Isn't std::sin a template? Would you need to specifically instantiate
std::sin, i.e.

std::transform( a, a+2, a, std::sin<double >)

You might also want to use std::ptr_fun as a wrapper to std::sin (or

std::transform( a, a+2, a, std::sin<double >)

What about this line if you have <complex> included? I took a look at the
<cmath> source code. It defines (in namespace std!) sin for float argument
as non template function, but the example above results in the same error
for float. I'm confused.

regards,
alex

[Victor Bazarov]

Rob Williscroft wrote:

Alexander Stippler wrote in news:40******@n ews.uni-ulm.de in
comp.lang.c++:

Hi,

simple question. Why does this not work:

#include <algorithm>
#include <cmath>
#include <iostream>

int
main()
{
double a[2];
a[0] = 0;a[1] = M_PI/2.;
std::transform( a, a+2, a, std::sin);
std::cout << a[0] << ", " << a[1] << std::endl;
return 0;
}

for gcc and icc the type of the third argument of transform is
unknown. If I use the global ::sin it works fine. Why? I include cmath
and there sin is defined in namespace std, so why is it unknown?

regards,
alex

There are several overloads of std::sin, you need to pick one:

std::transform( a, a+2, a, (double (*)(double)std: :sin);

Unmatched parentheses. Did you mean to write

std::transform( a, a+2, a, (double (*)(double))std ::sin);

(i.e. use the C-style cast)?

BTW, why can't it figure out the type of 'std::sin'? Is that because
it's overloaded?
[...]

V

[Pete Becker]

Rob Williscroft wrote:

Note M_PI isn't AFAICT Standard C++, you can use:

double const M_PI = (2.0 * std::acos( 0.0 ));

Though maybe:

double const double_pi = (2.0 * std::acos( 0.0 ));

Whould be better.

The latter, definitely. Otherwise you end up calling acos every time you
use the value of M_PI.

--

Pete Becker
Dinkumware, Ltd. (http://www.dinkumware.com)

[Rob Williscroft]

Victor Bazarov wrote in news:Is******** ********@dfw-read.news.verio .net
in comp.lang.c++:

There are several overloads of std::sin, you need to pick one:

std::transform( a, a+2, a, (double (*)(double)std: :sin);
Unmatched parentheses. Did you mean to write

std::transform( a, a+2, a, (double (*)(double))std ::sin);

Thanks, I should have "cut-n-paste" from my editor.
(i.e. use the C-style cast)?

BTW, why can't it figure out the type of 'std::sin'? Is that because
it's overloaded?

Yes, but its also because std::transform' s signature doesn't do
anything to help:

template<class InputIterator, class OutputIterator, class UnaryOperation>
OutputIterator
transform(
InputIterator first, InputIterator last,
OutputIterator result, UnaryOperation op
)
;

I.e. UnaryOperation is independant of InputIterator and OutputIterator.

Deducable example:

#include <cmath>

template < typename T >
T f( T arg, T (*func)( T ) )
{
return func( arg );
}

int main()
{
f( 0.0, std::sin );
}
Rob.
--
http://www.victim-prime.dsl.pipex.com/

[Victor Bazarov]

Pete Becker wrote:

Rob Williscroft wrote:

Note M_PI isn't AFAICT Standard C++, you can use:

double const M_PI = (2.0 * std::acos( 0.0 ));

Though maybe:

double const double_pi = (2.0 * std::acos( 0.0 ));

Whould be better.

The latter, definitely. Otherwise you end up calling acos every time you
use the value of M_PI.

Huh? I must be missing something. Did you mean to support the usual
const versus macro argument but instead implied that if the name of the
constant is M_PI, then it's somehow a function call instead of a const,
and if the name is 'double_pi', then it's reverse?

Rob didn't try to suggest

#define M_PI (2.0 * std::acos(0))

, did he? So, the latter or the former should be basically the same.

Victor

[Victor Bazarov]

red floyd wrote:

Alexander Stippler wrote:

Hi,

simple question. Why does this not work:

#include <algorithm>
#include <cmath>
#include <iostream>

int
main()
{
double a[2];
a[0] = 0;a[1] = M_PI/2.;
No such thing as 'M_PI' in the standard C++, BTW.
std::transform( a, a+2, a, std::sin);
std::cout << a[0] << ", " << a[1] << std::endl;
return 0;
}

for gcc and icc the type of the third argument of transform is
unknown. If I
use the global ::sin it works fine. Why? I include cmath and there sin is
defined in namespace std, so why is it unknown?

regards,
alex

IANAG (I am not a guru), and am probably way off base (please correct me
guys -- please no flames).

Isn't std::sin a template?

Not for doubles. For complex<> and for valarray<>, yes. For
double, float, long double, it's just an overloaded function.
Would you need to specifically instantiate
std::sin, i.e.

std::transform( a, a+2, a, std::sin<double >)
No.

You might also want to use std::ptr_fun as a wrapper to std::sin (or
::sin).

That might help. I am not sure I understand why, though.

For all I can see in the Standard, the 'std::transform ' template has
to be able to figure out the type of 'std::sin'. It's an overloaded
function and the lookup should be limited to the type of the first
argument.

I would like to really hear from Greg Comeau on this, too. His online
try-it-out thing refuses to compile and I cannot figure out how to make
it.

Victor