Overloading function when passing lambda as parameter

问题

I'm trying to implement template function when return parameter is either void or T. I tried different variations of the code above using sfinae but still not sure if that is in general possible in case lamdba is function parameter. The following code does not compile :

#include <functional>

template <typename T>
T Apply(const std::function<T()>& func)
{
    return func();
}

template <>
void Apply(const std::function<void()>& func)
{
    func();
}

int main(int argc, char *argv[])
{
    int i1 = Apply([]() { return 10; });
    bool b1 = Apply([]() { return true; });
    Apply([]() { return; });

    return 0;
}

Error :

error C2672: 'Apply': no matching overloaded function found
error C2784: 'T Apply(const std::function<T(void)> &)': could not deduce     template argument for 'const std::function<T(void)> &' from 'main::<lambda_536cc9cae26ef6d0d1fbeb7b66a2e26b>'

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回答1:

This is because template deduction needs a perfect-ish match for each of the function arguments in order to successfully deduce the template parameters.

You'd need to templetize the function object itself:

#include <type_traits>

template <class Function, class Return = std::result_of_t<Function()>>
Return Apply(Function func)
{
    return func();
}

Usage:

#include <iostream>

int main()
{
    std::cout << Apply([]() { return 42; }) << "\n";
}

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回答2:

Unfortunately you can't do that because the implicit conversion (from lambda closure type to std::function) is not considered in template argument deduction; the code fails because T can't be deduced.

You can use the lambda closure type as the parameter type directly, and declare the return type as auto to be deduced automatically. e.g.

template <typename T>
auto Apply(T func)
{
    return func();
}

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来源：https://stackoverflow.com/questions/48424942/overloading-function-when-passing-lambda-as-parameter

标签

c++

templates

lambda

sfinae