c++ - 如何在C++中有条件地编写依赖于lambda返回类型的代码？

Question

我有为每个迭代的元素调用回调的函数:

#include <type_traits>
#include <iostream>
using namespace std;

void iterate( const auto & on_element ) {
    for ( int i = 0; i < 10; ++ i ) {
        //here I want to stop iteration if function returns bool true:
        if constexpr ( is_same< result_of( on_element )::type, bool >::value )
            if ( ! on_element( i ) )
                return;
        //but when callback doesn't return bool - simply iterate the whole range:
        else
            on_element( i );
    }
}

int main() {
    const auto & stop = [&]( const int & element ) {
        cout << "stop" << endl;
        return false;
    };
    iterate( stop );

    const auto & whole = [&]( const int & element ) {
        cout << "    " << element << endl;
    };
    cout << "Whole:" << endl;
    iterate( whole );
}

我用 g++-10 -std=gnu++2a -fconcepts main.cpp -o main编译它并得到以下错误:

main.cpp: In function ‘void iterate(const auto:1&)’:
main.cpp:7:69: error: wrong number of template arguments (1, should be 2)
    7 |         if constexpr ( is_same< result_of( on_element )::type, bool >::value )
      |                                                                     ^
In file included from main.cpp:1:
/usr/include/c++/10/type_traits:582:12: note: provided for ‘template<class, class> struct std::is_same’
  582 |     struct is_same;
      |            ^~~~~~~
main.cpp:7:78: error: qualified-id in declaration before ‘)’ token
    7 |         if constexpr ( is_same< result_of( on_element )::type, bool >::value )
      |                                                                              ^
main.cpp: In instantiation of ‘void iterate(const auto:1&) [with auto:1 = main()::<lambda(const int&)>]’:
main.cpp:21:20:   required from here
main.cpp:8:30: error: could not convert ‘(& on_element)->main()::<lambda(const int&)>(i)’ from ‘void’ to ‘bool’
    8 |             if ( ! on_element( i ) )
      |                    ~~~~~~~~~~^~~~~
      |                              |
      |                              void
main.cpp:8:30: error: in argument to unary !

在我设法对函数返回值进行推理之前，但是在这种情况下无法达到目的。是由于可调用的lambda还是由于使用了 auto限定词？

Answer 1

on_node在您的示例中不是名称，并且std::result_of不具有该语法(不建议使用std::invoke_result)。
我想你是说

std::is_same_v<std::invoke_result_t<decltype(on_element), int>, bool>