为什么 operator<< 不会隐式地将我的自定义 class 对象转换为字符串

Why operator<< does not implicitly converting my custome class object to string

我有一个 class 实现了一个 public string 转换成员函数。当与 operator<<(iostream &, xxx) 结合使用时，我期望我的 class 会自动（隐式）转换为 string 因此适合参数类型。

然而，事实并非如此。为什么，我不想写 operation<< 函数。

#include <string>
#include <iostream>

using namespace std;

struct A {
    operator string() { return "asd"; }
};

int main() {
    cout << A() << endl;  // error
    cout << string(A()) << endl; // ok
}

operator<< for std::string is a template, and implicit conversion won't be considered in template argument deduction，失败了。

Type deduction does not consider implicit conversions (other than type adjustments listed above): that's the job for overload resolution, which happens later.

正如您所展示的，您可以将 A 显式转换为 std::string，或者为 A.

编写 operator<<