获取整型模板参数的有符号/无符号变量，无需显式traits [英] Get the signed/unsigned variant of an integer template parameter without explicit traits

问题描述

我想要定义一个模板类，其模板参数将始终是一个整数类型。类将包含两个成员，一个类型 T ，另一个作为 T 类型的无符号变量 - 即if T == int ，然后 T_Unsigned == unsigned int 。我的第一本能是这样做：

I am looking to define a template class whose template parameter will always be an integer type. The class will contain two members, one of type T, and the other as the unsigned variant of type T -- i.e. if T == int, then T_Unsigned == unsigned int. My first instinct was to do this:

template <typename T> class Range {
    typedef unsigned T T_Unsigned; // does not compile
public:
    Range(T min, T_Unsigned range);
private:
    T m_min;
    T_Unsigned m_range;
};

但它不工作。然后我考虑使用部分模板专门化，像这样：

But it doesn't work. I then thought about using partial template specialization, like so:

template <typename T> struct UnsignedType {}; // deliberately empty
template <> struct UnsignedType<int> {
    typedef unsigned int Type;
};

template <typename T> class Range {
    typedef UnsignedType<T>::Type T_Unsigned;
    /* ... */
};

只要你部分专门化 UnsignedType 每个整数类型。这是一个额外的复制粘贴工作（斜线明智的使用宏），但可服务。

This works, so long as you partially specialize UnsignedType for every integer type. It's a little bit of additional copy-paste work (slash judicious use of macros), but serviceable.

但是，我现在好奇 - 是否有另一种方法来确定整数类型的有符号类型，和/或使用类型的无符号变量，而不必手动定义每个类型的Traits类？

However, I'm now curious - is there another way of determining the signed-ness of an integer type, and/or using the unsigned variant of a type, without having to manually define a Traits class per-type? Or is this the only way to do it?

推荐答案

答案在< type_traits>

要确定类型的签名类型，请使用std :: is_signed和std :: is_unsigned

For determining the signed-ness of a type use std::is_signed and std::is_unsigned

对于添加/删除签名，有std :: make_signed和std :: make_unsigned

For adding/removing signed-ness, there is std::make_signed and std::make_unsigned

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