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

Question

为什么把std::basic_string作为函数模板的参数从const char*推导失败，直接构造却可以推导成功？

#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post ，但是答案没有解释背后的原因

Answer 1

因为在template argument deduction中没有考虑隐式转换（从const char*到std::basic_string<char>），这导致模板参数推导失败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");

或显式传递 std::basic_string。

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

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

std::basic_string as a parameter of a function template cannot be deduced from const char*

c++

templates

implicit-conversion

template-argument-deduction