How to make a variadic macro for std::cout?

问题

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.

Conceptual Example:

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

would output:

exampleoutputfillertext

回答1:

You do not need preprocessor macros to do this. You can write it in ordinary C++. In C++11/14:

#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');
}

Live demo

In C++17:

template<typename ...Args>
void log(Args && ...args)
{
    (std::cout << ... << args);
}

is all it takes. Live demo

Output:

Hellobravenewworld!
1234

Research variadic templates, parameter packs and fold expressions rather than variadic macros, which are rarely useful in modern C++.

回答2:

When using variadic macros you need __VA_ARGS__ to expand all the arguments. The problem however is that those arguments are comma-separated. Presumabely it just separates arguments to a function call, but since macros works with just text you can actually use __VA_ARGS__ in other contexts as well, where a comma-separated list makes sense.

The trick you can employ is to define your own comma operator for std::ostream (the type of std::cout). For example:

#include<iostream>
#define LOG(...) std::cout , __VA_ARGS__ , std::endl

template <typename T>
std::ostream& operator,(std::ostream& out, const T& t) {
  out << t;
  return out;
}

//overloaded version to handle all those special std::endl and others...
std::ostream& operator,(std::ostream& out, std::ostream&(*f)(std::ostream&)) {
  out << f;
  return out;
}

int main() {
  LOG("example","output","filler","text");
  return 0;
}

Now, the LOG invocation will expand to:

std::cout , "example" , "output" , "filler" , "text" , std::endl;

and the commas will invoke the overloaded comma operators.

If you don't like overloaded operator, polluting all std::ostream-s, you can encapsulate std::cout with your own special logger class.

回答3:

Not sure there is any way of defining a variadic macro in C++ (at least, not a portable way). Why don't you use a variadic template approach? Something like

#include <iostream>

void LOG() {}

template<typename Head, typename... Args>
void LOG(const Head& head, const Args&... args )
{
    std::cout << head << " ";
    LOG(args...);
}

int main()
{
    LOG("This", "is" , "the", 3, "rd test");
}

回答4:

You can try this one

#include <iostream>

#define LOG() std::cout

int main()
{
    LOG() << "Hello! " << "how " << "are " << "you " << std::endl;
    return 0;
}

Output:

Hello!  how are  you

来源：https://stackoverflow.com/questions/29326460/how-to-make-a-variadic-macro-for-stdcout

标签

c++

c++11

c-preprocessor