可变参数模板的正确语法

Question

我正在尝试将一些 ascii 字符转换为另一个字符。但是，由于对可变参数模板语法缺乏经验，我无法做到。

template<typename T>
void despecialize(const T& original, T& totransform, const int& charnum)
{
    if (static_cast<int>(totransform) == charnum) {
        totransform = original;
    }
}

template<typename T>
void despecialize(const T& original, T& totransform, const int& charnum, const int& charother...)
{
    despecialize(original, totransform, charnum);
    despecialize(original, totransform, charother...); //C3546 there are no parameter packs available to expand
}


std::string removePortugueseSpecialLetter(std::string& data)
{
    std::string transformed;
    for (auto& c : data)
    {
        despecialize('a', c, 176, 131, 132, 133, 134);
        transformed += c;
    }
    return transformed;
}

正确的语法应该是什么？

Answer 1

为了拥有可变模板，您需要一个可变模板参数。看起来像

template <typename... VariadicTypeParater>
// or
template <some_type... VariadicNonTypeParater>

将其应用到您的函数中会给您

template<typename T, typename... Chars>
void despecialize(const T& original,
                  T& totransform,
                  const int& charnum,
                  const Chars&... charother)
{
    despecialize(original, totransform, charnum);
    despecialize(original, totransform, charother...);
}

虽然这将允许为 Chars 传递任何类型，而不仅仅是 int。要限制您需要使用 enable_if_t 非类型模板参数添加 SFINAE，例如

template <typename T,
          typename... Chars,
          std::enable_if_t<(std::is_same_v<Chars, int> && ...), bool> = true>
void despecialize(const T& original,
                  T& totransform,
                  const int& charnum,
                  const Chars& charother...)
{
    despecialize(original, totransform, charnum);
    despecialize(original, totransform, charother...);
}

可变参数模板的正确语法

Correct syntax for variadic template

c++

variadic-templates