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c++templateslambdac++14std-function

templated function argument in C++14

This code does not compile, not even under C++14, because of problems with template type deduction. What is the least inelegant workaround?

#include <vector>
#include <functional>
#include <iostream>

template <class T>
std::vector<T> merge_sorted(
    const std::vector<T>& a, const std::vector<T>& b,
    std::function<bool(const T, const T)> a_before_b)
{
    std::vector<T> ret;
    auto ia=a.begin();
    auto ib=b.begin();
    for (;;ia!=a.end() || ib!=b.end())
        ret.push_back( a_before_b(*ia,*ib) ? *(ia++) : *(ib++) );
    return ret;
}

int main()
{
    std::vector<double> A { 1.1, 1.3, 1.8 };
    std::vector<double> B { 2.1, 2.2, 2.4, 2.7 };
    auto f = [](const double a, const double b) -> bool {
        return (a-(long)(a))<=(b-(long(b))); };
    std::vector<double> C = merge_sorted(A, B, f);
    for (double c: C)
        std::cout << c << std::endl;
    // expected outout: 1.1 2.1 2.2 1.3 2.4 2.7 1.8
}

Here the error message from g++ -std=c++14 main.cpp:

main.cpp: In function ‘int main()’:
main.cpp:23:49: error: no matching function for call to ‘merge_sorted(std::vector<double>&, std::vector<double>&, main()::<lambda(double, double)>&)’
     std::vector<double> C = merge_sorted(A, B, f);
                                                 ^
main.cpp:6:16: note: candidate: template<class T> std::vector<T> merge_sorted(const std::vector<T>&, const std::vector<T>&, std::function<bool(T, T)>)
 std::vector<T> merge_sorted(
                ^~~~~~~~~~~~
main.cpp:6:16: note:   template argument deduction/substitution failed:
main.cpp:23:49: note:   ‘main()::<lambda(double, double)>’ is not derived from ‘std::function<bool(T, T)>’
     std::vector<double> C = merge_sorted(A, B, f);

==

Later edit, just for the record: Here comes a version of the code that compiles (thanks to received answers) and that executes correctly (several corrections of the above untested code):

#include <vector>
#include <functional>
#include <iostream>

template <class T, class Pred>
std::vector<T> merge_sorted(const std::vector<T>& a, const std::vector<T>& b, Pred a_before_b)
{
    std::vector<T> ret;
    auto ia=a.begin();
    auto ib=b.begin();
    for (;ia!=a.end() && ib!=b.end();)
        ret.push_back( a_before_b(*ia,*ib) ? *(ia++) : *(ib++) );
    for (;ia!=a.end();)
        ret.push_back( *(ia++) );
    for (;ib!=b.end();)
        ret.push_back( *(ib++) );
    return ret;
}

int main()
{
    std::vector<double> A { 1.1, 1.3, 1.8 };
    std::vector<double> B { 2.1, 2.2, 2.4, 2.7 };
    auto f = [](const double a, const double b) -> bool {
        return (a-(long)(a))<=(b-(long(b))); };
    std::vector<double> C = merge_sorted(A, B, f);
    for (double c: C)
        std::cout << c << std::endl;
    // expected outout: 1.1 2.1 2.2 1.3 2.4 2.7 1.8
}
Solution

  • The problem here is that f is not a std::function. It is some unnamed class type but it is not a std::function. When the compiler does template argument deduction it does not do any conversions, it works with the parameters as is to deduce their type. That means where it expects to see a std::function<bool(const T, const T)> it sees main()::<lambda(double, double)> as that is the type of the lambda and since those types do not match the deduction fails. In order to get deduction to succeed you need to get them to match.

    Without changing the function signature you have to cast f to a std::function in order to get it to work. That would look like

    std::vector<double> C = merge_sorted(A, B, static_cast<std::function<bool(const double,const double)>>(f));

    If you do not mind changing the function signature then we can use

    template <class T, class Func>
std::vector<T> merge_sorted(
    const std::vector<T>& a, const std::vector<T>& b,
    Func a_before_b)

    And now it doesn't matter if you pass a std::function or a lambda or a functor.