如何在不过于冗长的情况下获得可变参数模板函数的正确解析

Question

我的问题是关于我在定义和使用可变参数函数时遇到的编译错误。我将在下面给出一个最小的例子。

首先我定义了一个函数来打印一个对象：（所有代码都是here）

template <typename T>
void print(T t){
    std::cout << t<<std::endl;
}

接下来，我定义了一个函数，允许您转换对象然后打印它：

template <typename T>
void print(T t,std::function<void(T& t)> transformer){
    transformer(t);
    std::cout << t << std::endl;
}

接下来，我想定义一个函数，让您可以调用上述两个函数中的任何一个，但会导致打印标签和结果。我可以使用可变函数模板实现这一点：

template <typename... Params>
void printLabeled(std::string label,Params... params){
    std::cout << label << ": ";
    print(params...);
}

现在我可以尝试使用代码了。我想做的主要事情是调用 printLabeled("answer",1,{[](int& num){num+=1;}});，但这不起作用。我必须做一些更冗长的事情。有没有比 printLabeled("answer",1,{[](int& num){num+=1;}}); 更冗长的函数调用方法？同样，您可以运行代码 here。下面是有效和无效的函数调用示例。

int main() {
    //prints the number '1', as expected
    print(1);

    //prints the number the label "answer" and then the number one
    printLabeled("answer",1);

    //I intended this to increment 1 to 2 and then print 2,
    //but I guess the lambda does not get converted to an std::function,
    //so this doesn't work
    //print(1,[](int& num){num+=1;});

    //This does was the above line was intended to do
    print(1,{[](int& num){num+=1;}});

    //I intended this to be the labeled version of the above line,
    //but for some reason, this doesn't work anymore
    // printLabeled("answer",1,{[](int& num){num+=1;}});

    //This labeled version does work finally,
    //but its pretty verbose
    printLabeled("answer",1,std::function<void(int&)>([](int& num){num+=1;}));
    return 0;
}

Answer 1

#include <iostream>
#include <string>

template <typename T, typename F>
void print(T t, F transformer){
    transformer(t);
    std::cout << t << std::endl;
}

template <typename T>
void print(T t){
    std::cout << t<<std::endl;
}

template <typename... Params>
void printLabeled(std::string label, Params&&... params){
    std::cout << label << ": ";
    print(std::forward<Params>(params)...);
}

int main() {
    //prints the number '1', as expected
    print(1);

    //prints the number the label "answer" and then the number one
    printLabeled("answer",1);

    //This does was the above line was intended to do
    print(1, [](int& num){num+=1;});

    //I intended this to be the labeled version of the above line,
    //but for some reason, this doesn't work anymore
    printLabeled("answer", 1, [](int& num){num+=1;});

    return 0;
}

https://ideone.com/JTYUpa

如何在不过于冗长的情况下获得可变参数模板函数的正确解析

How to get correct resolution of variadic template function without being too verbose

c++

variadic-templates

c++11

template-argument-deduction