Re: [Boost-users] apply std::pow to vector or matrix

7 Jul 2010

      On Wed, Jul 7, 2010 at 8:06 AM, Robert Jones <robertgbjones@gmail.com>wrote:
...
On Wed, Jul 7, 2010 at 3:30 AM, Max S. Kaznady <max.kaznady@gmail.com>wrote:
...
Hi Robert,
Yep, the following works great. Not sure why the same code doesn't
work when I have it in my larger project, but I guess that's for me to
figure out.
There is still the problem with the std::for_each, see below:
#include <iostream>
#include <vector>
#include <algorithm>
#include <math.h>
#include "boost/bind.hpp"
#include <boost/numeric/ublas/vector.hpp>
#include <boost/numeric/ublas/io.hpp>
int main( )
{
 boost::numeric::ublas::vector<double> v(10);
 double exponent = 2.1;
 std::fill(v.begin(), v.end(), 8.0);
 // The following works just fine:
  std::transform( v.begin(), v.end(), v.begin(), boost::bind( pow, _1,
exponent ) );
  // But the for_each doesn't apply the power function, just leaves
the vector unaltered
 std::for_each( v.begin(), v.end(), boost::bind( pow, _1, exponent ) );
 // Print the result
 std::cout<< v << std::endl;
return 0;
}
Compiled with: g++ -O2 -Wfatal-errors -Wall -g -ansi
-I/usr/local/boost/include -o main main.cpp
Hi Max
I haven't taken the time to try this, but does this do the trick?
std::for_each(v.begin(), v.end(), boost::bind(pow, boost::ref(_1),
exponent));
Cheers
- Rob.
Forget that, it was complete gibberish. Your example doesn't do as you
expect
because std::pow does not alter its argument, but rather returns a result.
You have
to do something with the result of std::pow, hence my original use of
transform.

To understand this look at this example.

#include <vector>
#include <algorithm>
#include <math.h>
#include "boost/bind.hpp"

void my_pow( double & val, double exponent )
{
    val = pow( val, exponent );
}

int main( )
{
    std::vector<double> v;
    double exponent = 2.1;

    std::for_each( v.begin(), v.end(), boost::bind( my_pow, _1, exponent )
);
}

HTH

- Rob.