可以将 operator* 重载为多个 int 和 char* 吗?

Possible to overload operator* to multiple an int and a char*?

我想获得功能,以便我可以这样做:

std::cout << "here's a message" << 5*"\n";

我尝试了以下方法:

std::string operator* (int lhs, const char* rhs) {
  std::string r = "";
  for(int i = 0; i < lhs; i++) {
    r += rhs;
  }
  return r;
}

我收到了这条错误信息:

error: ‘std::string operator*(int, const char*)’ must have an argument of class or enumerated type

根据此 SO post What does 'must have an argument of class or enumerated type' actually mean 中的答案,我似乎无法完成这段时间。真的是这样吗?如果没有,我该如何解决这个问题或安排解决方法?

我知道我能做的是将 rhs 作为 std::string,但是整个练习的重点都已成定局,因为 5*std::string("\n") 非常笨重。

来自[over.oper]:

An operator function shall either be a non-static member function or be a non-member function that has at least one parameter whose type is a class, a reference to a class, an enumeration, or a reference to an enumeration.

所以你不能重载参数都是内置的运算符。此外,为了找到 operator*(int, std::string),它必须在 namespace std 中,并且向该命名空间添加定义是不正确的。

相反,您可以简单地提供一个小包装器:

struct Mult { int value; };

并为其提供重载:

std::string operator*(const Mult&, const char* );
std::string operator*(const char*, const Mult& );

来自 C++ 常见问题 here,

C++ language requires that your operator overloads take at least one operand of a “class type” or enumeration type. The C++ language will not let you define an operator all of whose operands / parameters are of primitive types.

您既不能也不必重载该操作;

string ctor (2) 为您代劳

#include <iostream>
#include <string>

int main() {
    std::cout << "here's a message:\n" 
              << std::string(5, '\n') 
              << "EOF" << std::endl;
} 

输出:

here's a message:






EOF

(live at Coliru)

您应该能够使用用户定义的文字来实现它。例如:

#include <iostream>
#include <string>


std::string operator"" _s(const char* s) { return std::string(s); }

std::string operator"" _s(const char* s, std::size_t len) { return std::string(s, len); }

std::string operator* (unsigned int k, std::string s) {
    std::string t;
    for (unsigned int i = 0; i < k; ++i) 
        t += s;

    return t;
}

std::string operator* (std::string s, unsigned int k) { return k * s; }

int main() {
    std::cout << "Jump!"_s * 5 << "\n";
}