因此,我重载了ostream <<运算符,以便它可以使用大多数STL容器并进行打印。但是,它不适用于传递字符串(错误是“ cout << s;”行的“歧义过载”)。如何使它在字符串上好像没有重载一样起作用?
#include <bits/stdc++.h>
using namespace std;
template<typename T>
ostream& _containerprint(ostream &out, T const &val) {
return (out << val << " ");
}
template<typename T1, typename T2>
ostream& _containerprint(ostream &out, pair<T1, T2> const &val) {
return (out << "{" << val.first << " " << val.second << "} ");
}
template<template<typename, typename...> class TT, typename... Args>
ostream& operator<<(ostream &out, TT<Args...> const &cont) {
for(auto&& elem : cont) {
_containerprint(out, elem);
}
return out;
}
int main() {
string s = "help me";
cout << s;
}
编辑:请不要害怕,#include <bits/stdc++.h>
是因为它是用于编程比赛的设置;真的没关系!
答案 0 :(得分:4)
std::string
是模板类(std::basic_string<char>
的别名)
因此std::cout << s
与您的过载匹配,而std::basic_string<T>
与您的过载匹配。
没有一个过载比另一个过载更专业。
可能的解决方法是为std::string
添加额外的重载:
std::ostream& operator << (std::ostream& out, const std::string& s)
{
return operator << <char>(out, s);
// Select template <class CharT, class Traits, class Allocator>
// std::basic_ostream<CharT, Traits>&
// operator<<(std::basic_ostream<CharT, Traits>& os,
// const std::basic_string<CharT, Traits, Allocator>& str);
}
答案 1 :(得分:0)
使您的ostream
参数更通用,从而使您的重载比std::string
的标准参数更通用。
template<typename CharT, typename Traits, typename T>
ostream& _containerprint(std::basic_ostream<CharT, Traits> &out, T const &val) {
return (out << val << " ");
}
template<typename CharT, typename Traits, typename T1, typename T2>
ostream& _containerprint(std::basic_ostream<CharT, Traits> &out, pair<T1, T2> const &val) {
return (out << "{" << val.first << " " << val.second << "} ");
}
template<typename CharT, typename Traits, template<typename, typename...> class TT, typename... Args>
ostream& operator<<(std::basic_ostream<CharT, Traits> &out, TT<Args...> const &cont) {
for(auto&& elem : cont) {
_containerprint(out, elem);
}
return out;
}