从 char 数组到 std::string 的类型推导 [英] type deduction from char array to std::string

问题描述

我正在尝试使用可变参数模板编写 sum 函数.在代码中，我会写一些类似

I'm trying to write sum function using variadic template. In the code I would write something like

sum(1, 2., 3)

并且它将返回最通用的总和类型(在这种情况下 - double).问题出在字符上.当我称之为

and it will return most general type of the sum (in this case - double). The problem is with characters. When I call it like

sum("Hello", " ", "World!") 

模板参数被推导为例如const char [7]，因此它不会编译.我找到了将最后一个参数指定为 std::string("World!") 的方法，但它并不漂亮，有没有办法实现对 std::string<的自动类型推导/code> 或正确重载 sum?

template parameters are deduces as const char [7] for example, thus it won't compile. I found the way to specify last argument as std::string("World!"), but it's not pretty, is there any way to achieve automatic type deduction to std::string or correctly overload sum?

我到目前为止的代码:

template<typename T1, typename T2>
auto sum(const T1& t1, const T2& t2) {
    return t1 + t2;
}

template<typename T1, typename... T2>
auto sum(const T1& t1, const T2&... t2) {
    return t1 + sum(t2...);
}

int main(int argc, char** argv) {
    auto s1 = sum(1, 2, 3);
    std::cout << typeid(s1).name() << " " << s1 << std::endl;

    auto s2 = sum(1, 2.0, 3);
    std::cout << typeid(s2).name() << " " << s2 << std::endl;

    auto s3 = sum("Hello", " ", std::string("World!"));
    std::cout << typeid(s3).name() << " " << s3 << std::endl;

    /* Won't compile! */
    /*
    auto s4 = sum("Hello", " ", "World!");
    std::cout << typeid(s4).name() << " " << s4 << std::endl;
    */

    return 0;
}

输出:

i 6
d 6
Ss Hello World!

推荐答案

我只想写一个简单的重载标识函数，专门处理 const char*.

I would just write a simple overloaded identity function which handles const char* specially.

template<typename T>
decltype(auto) fix(T&& val)
{
    return std::forward<T>(val);
} 

auto fix(char const* str)
{
    return std::string(str);
}

template<typename T1, typename T2>
auto sum(const T1& t1, const T2& t2) {
    return fix(t1) + t2;
}

template<typename T1, typename... T2>
auto sum(const T1& t1, const T2&... t2) {
    return t1 + sum(t2...);
}

这篇关于从 char 数组到 std::string 的类型推导的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！