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Home/ Questions/Q 328427
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Editorial Team
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Editorial Team
Asked: May 12, 20262026-05-12T09:29:39+00:00 2026-05-12T09:29:39+00:00

So I have this code: #include boost_bind.h #include <math.h> #include <vector> #include <algorithm> double

  • 0

So I have this code:

#include "boost_bind.h"
#include <math.h>
#include <vector>
#include <algorithm>

double foo(double num, double (*func)(double)) {
  return 65.4;
}

int main(int argc, char** argv) {
  std::vector<double> vec;
  vec.push_back(5.0);
  vec.push_back(6.0);
  std::transform(vec.begin(), vec.end(), vec.begin(), boost::bind(foo, _1, log));
}

And receive this error:

        return unwrapper<F>::unwrap(f, 0)(a[base_type::a1_], a[base_type::a2_]);
.............................................................^
%CXX-E-INCOMPATIBLEPRM, argument of type "double (* __ptr64 )(double) C" is
          incompatible with parameter of type "double (* __ptr64 )(double)"
          detected during:
            instantiation of ...5 pages of boost

So this error is because ‘log’ is extern “C”‘d in math.h

I was wondering how to declare my function pointer argument in foo() so it handles extern “C”‘d functions.

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  1. Editorial Team
    Editorial Team
    2026-05-12T09:29:39+00:00Added an answer on May 12, 2026 at 9:29 am

    You can try including cmath instead, and using static_cast<double(*)(double)>(std::log) (cast necessary to resolve to the double overload).

    Otherwise, you will limit your function to extern C functions. This would work like

    extern "C" typedef double (*ExtCFuncPtr)(double);
    
    double foo(double num, ExtCFuncPtr func) {
      return 65.4;
    }
    

    Another way is to make foo a functor

    struct foo {
      typedef double result_type;
      template<typename FuncPtr>
      double operator()(double num, FuncPtr f) const {
        return 65.4;
      }
    };
    

    Then you can pass foo() to boost::bind, and because it’s templated, it will accept any linkage. It will also work with function objects, not only with function pointers.

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