如何使std :: cout的可变宏？ [英] How to make a variadic macro for std::cout?

问题描述

我如何创建一个带有可变参数的宏，并使用std :: cout打印出来？

How would I make a macro that took a variable amount of arguments, and prints its out using std::cout? Sorry if this is a noob question, couldn't find anything that clarified variadic macros after searching around for the answer.

概念示例：

#include <iostream>
#define LOG(...) std::cout << ... << ... << std::endl
int main() {
    LOG("example","output","filler","text");
    return 0;
}

会输出：

exampleoutputfillertext

推荐答案

你不需要预处理器宏来做到这一点。你可以写成普通的
C ++：

You do not need preprocessor macros to do this. You can write it in ordinary C++:

#include <iostream>
#include <utility>

void log(){}

template<typename First, typename ...Rest>
void log(First && first, Rest && ...rest)
{
    std::cout << std::forward<First>(first);
    log(std::forward<Rest>(rest)...);
}

int main()
{
    log("Hello", "brave","new","world!\n");
    log("1", 2,std::string("3"),4.0,'\n');
}

输出：

Hellobravenewworld!
1234

您应该研究可变参数模板
和参数包
而不是可变参数宏，这在现代C ++中很少有用。

You should research variadic templates and parameter packs rather than variadic macros, which are rarely useful in modern C++.

（gcc 4.9.2 -std = c ++ 11）

(gcc 4.9.2 -std=c++11)

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