std :: endl和可变参数模板 [英] std::endl and variadic template

查看:131
本文介绍了std :: endl和可变参数模板的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

代码:

#include <iostream>

void out()
{
}

template<typename T, typename... Args>
void out(T value, Args... args)
{
    std::cout << value;
    out(args...);
}

int main()
{
    out("12345", "  ", 5, "\n"); // OK
    out(std::endl);              // compilation error
    return 0;
}

生成错误:

g++ -O0 -g3 -Wall -c -fmessage-length=0 -std=c++11 -pthread -MMD -MP -MF"main.d" -MT"main.d" -o "main.o" "../main.cpp"
../main.cpp: In function ‘int main()’:
../main.cpp:17:15: error: no matching function for call to ‘out(<unresolved overloaded function type>)’
../main.cpp:17:15: note: candidates are:
../main.cpp:3:6: note: void out()
../main.cpp:3:6: note:   candidate expects 0 arguments, 1 provided
../main.cpp:8:6: note: template<class T, class ... Args> void out(T, Args ...)
../main.cpp:8:6: note:   template argument deduction/substitution failed:
../main.cpp:17:15: note:   couldn't deduce template parameter ‘T’

> std :: endl 。我如何解决这个问题(除了使用\\\
)?

So, everything is OK except std::endl. How can I fix this (except of using "\n")?

推荐答案

std :: endl 是一个重载函数,(在许多STL实现中,一个模板),编译器没有关于选择什么的信息。

std::endl is an overloaded function, (in many STL implementations, a template) and the compiler has no info about what to choose from.

转换为 static_cast< std :: ostream&(*)(std :: ostream&)>(std :: endl)

这篇关于std :: endl和可变参数模板的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆