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How can I std :: sqrt (boost :: lambda :: placeholder1_type)?

How std::sqrt()

on boost::lambda::placeholder1_type

?

The following program compiles and executes fine. Anyway, it manages to convert boost::lambda::_1

to double and multiply it by 5.0.

#include <boost/lambda/lambda.hpp>
#include <iostream>
#include <cmath>

int main() {

  std::array<double, 6> a = {1, 4, 9, 16, 25, 36};

  std::for_each(a.begin(), a.end(),
      std::cout << boost::lambda::_1 * 5.0 << " "
  );
}

However, if I replace the line std::cout

with

std::cout << std::sqrt(boost::lambda::_1) << " "

the compiler (g ++) says

no known conversion for argument 1 from 'Boost :: lambda :: placeholder1_type {aka const boost :: lambda :: lambda_functor>} for "Double

So how can I take the square root of boost::lambda::_1

in this std :: for_each loop?

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3 answers


You need to postpone the call sqrt

. To achieve this, you must use bind an expression .

Note. To select the correct overload, sqrt

you need to use a throw.

#include <boost/lambda/lambda.hpp>
#include <boost/lambda/bind.hpp>
#include <iostream>
#include <cmath>

int main()
{
    std::array<double, 6> a = {1, 4, 9, 16, 25, 36};

    std::for_each(a.begin(), a.end(),
        std::cout << boost::lambda::bind(static_cast<double(*)(double)>(std::sqrt), boost::lambda::_1) << " "
    );
}


Output:

1 2 3 4 5 6

Live Sample

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Do it the C ++ 11 way:



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

int main() {

  std::array<double, 6> a = {1, 4, 9, 16, 25, 36};

  std::for_each(a.begin(), a.end(),
      [] (const auto&v ) { std::cout << sqrt(v) * 5.0 << " "; }
  );
}
+2


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Ah, the good old days of Boost.Lambda

int main() {

    std::array<double, 6> a = {1, 4, 9, 16, 25, 36};

    std::for_each(a.begin(), a.end(),
        std::cout << boost::lambda::bind(static_cast<double(*)(double)>(std::sqrt), boost::lambda::_1 * 5.0) << " "
    );
    return 0;
}
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