在c ++中将字符串乘以int

Question

我什么时候必须这样做

string s = ".";

如果我做

cout << s * 2;

它会是一样的吗？

cout << "..";

？

Answer 1

std :: string有一个表单的构造函数

std::string(size_type count, char c);

这将重复这个角色.例如

#include <iostream>

int main() {
   std::string stuff(2, '.');
   std::cout << stuff << std::endl;
   return 0;
}

将输出

..

Answer 2

我使用运算符重载来模拟 C++ 中的这种行为。

#include <iostream>
#include <string>
using namespace std;

/* Overloading * operator */
string operator * (string a, unsigned int b) {
    string output = "";
    while (b--) {
        output += a;
    }
    return output;
}


int main() {
    string str = "abc";
    cout << (str * 2);
    return 0;
}

输出：abcabc

Answer 3

不,std::string没有operator *.您可以将(char,string)添加到其他字符串.请看http://en.cppreference.com/w/cpp/string/basic_string

如果你想要这种行为(没有建议)你可以使用这样的东西

#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;
}

http://liveworkspace.org/code/52f7877b88cd0fba4622fab885907313

Answer 4

没有预定义的*运算符将字符串乘以a int,但您可以定义自己的:

#include <iostream>
#include <sstream>
#include <string>

using namespace std;

string operator*(const string& s, unsigned int n) {
    stringstream out;
    while (n--)
        out << s;
    return out.str();
}

string operator*(unsigned int n, const string& s) { return s * n; }

int main(int, char **) {
    string s = ".";
    cout << s * 3 << endl;
    cout << 3 * s << endl;
}