c++ - std::basic_string 作为函数模板的参数不能从 const char* 推导出

Question

为什么放std::basic_string作为函数模板的参数将无法从 const char* 推导出来，但是直接构造就可以推导出成功？

#include <string>
#include <iostream>
template<class Char/*, class Traits, class Allocator*/>   
                      //^doesn't matter whether the second and third template parameter is specified
void printString(std::basic_string<Char/*, Traits, Allocator*/> s)
{
    std::cout << s;
}
int main()
{
    printString("hello");    //nope
    std::basic_string s{ "hello" };//works
}

我找到了一个相关的帖子 here ，但答案并没有解释背后的原因

Answer 1

因为在 template argument deduction 中不考虑隐式转换(从 const char* 到 std::basic_string<char> ) , 无法推导出模板参数 Char .

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

您可以明确指定模板参数，

printString<char>("hello");

或通过 std::basic_string明确地。

printString(std::basic_string("hello"));