通过显式构造函数初始化数组 [英] Initializing array through explicit constructor

问题描述

我正在编写一个类，该类具有带const char*参数的显式构造函数.出于这个问题的意图和目的，它看起来像这样:

I'm writing a class that has an explicit constructor taking a const char* argument. For the intents and purposes of this question it looks like this:

struct Symbol
{
    Symbol()=default;
    explicit Symbol(const char*);
};

现在，出于文档目的，我想编写一个示例来初始化数组(array/vector/list-我不在乎确切的类型)，并且我需要使示例尽可能简洁明了.理想情况下，它看起来应该像这样:

Now I want to write an example for documentation purposes that initializes an array (array/vector/list - I don't care about the exact type) and I need the example to be as clear and concise as possible. Ideally it would look like this:

Symbol symbols[] = { "a", "b", "c"};

由于显式关键字而无法编译，因此我不准备使构造函数隐式化.

That does not compile because of the explicit keyword and I am not prepared to make the constructor implicit.

如何使这项工作有效，重点在于使示例代码尽可能具有表现力?

How can I make this work, with the focus of making the example code as expressive as possible?

在Caleth的一点帮助下，我寻求了Bolov的解决方案:

I went for Bolov's solution with a little help from Caleth:

struct Symbol
{
    Symbol();
    explicit Symbol(const char*);

    template <class... Args> 
    static std::array<Symbol, sizeof...(Args)> Array(Args... args)
    {
        return {Symbol{args}...}; 
    } 
};

int main()
{
    auto symbols = Symbol::Array("a", "b", "c");
}

推荐答案

好，您的构造函数是显式的，因此您需要这样使用它:

Well, your constructor is explicit so you need to use it as such:

Symbol symbols[] = {Symbol{"a"}, Symbol{"b"}, Symbol{"c"}};

gcc和clang都是copy/move构造函数，并且由于C ++ 17是必需的行为，因此没有性能开销.

Both gcc and clang both the copy/move constructor and since C++17 that is the required behavior so there is no performance overhead.

如果您真的想保留构造函数explicit并能够创建一个数组而无需为每个元素明确声明，则可以创建一个辅助函数:

If you really want to keep the constructor explicit and be able to create an array without explicitly stating it for every element then you can create a helper function:

template <class... Args,
          class Enable = std::enable_if_t<(... && std::is_same_v<Args, const char*>)>>
auto make_symbols(Args... args) -> std::array<Symbol, sizeof...(Args)>
{
    return {Symbol{args}...};
}

并像这样使用它:

auto symbols = make_symbols("a", "b", "c");

再次取消移动/副本.

make_symbols函数使用C ++ 17功能来检查参数类型.如果您需要以前的标准版本(包括C ++ 11)的约束，请参见以下答案限制可变参数模板参数.或者，根据您的需要，也可以选择取消支票.

The make_symbols function uses C++17 features for checking the arguments types. If you need the constraint for a previous standard version (including C++11) see this answer Restrict variadic template arguments. Or, depending on your needs, removing the check can also be a choice.

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