在c ++中用int乘以一个字符串

问题描述

我必须这样做，当我

string s = ".";

如果我这样做

cout << s * 2;

是否与

cout << "..";

？

推荐答案

否， std :: string 没有运算符* 。您可以将（char，string）添加到其他字符串。查看此

No, std::string has no operator *. You can add (char, string) to other string. Look at this http://en.cppreference.com/w/cpp/string/basic_string

如果你想要这个行为（没有建议这个），你可以使用这样

And if you want this behaviour (no advice this) you can use something like this

#include <iostream>
#include <string>

template<typename Char, typename Traits, typename Allocator>
std::basic_string<Char, Traits, Allocator> operator *
(const std::basic_string<Char, Traits, Allocator> s, size_t n)
{
   std::basic_string<Char, Traits, Allocator> tmp = s;
   for (size_t i = 0; i < n; ++i)
   {
      tmp += s;
   }
   return tmp;
}

template<typename Char, typename Traits, typename Allocator>
std::basic_string<Char, Traits, Allocator> operator *
(size_t n, const std::basic_string<Char, Traits, Allocator>& s)
{
   return s * n;
}

int main()
{
   std::string s = "a";
   std::cout << s * 5 << std::endl;
   std::cout << 5 * s << std::endl;
   std::wstring ws = L"a";
   std::wcout << ws * 5 << std::endl;
   std::wcout << 5 * ws << std::endl;
}

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