C ++整数类型是给定类型宽度的两倍 [英] C++ integer type twice the width of a given type

问题描述

在此示例中， coord_squared_t 是整数类型的别名，其大小至少是整数类型 coord_t 的两倍:

In this example, coord_squared_t is the alias for an integer type with at least twice the size of the integer type coord_t:

typedef int_least32_t coord_t;

coord_squared_t CalculateSquaredHypothenuse(coord_t x, coord_t y){
    coord_squared_t _x=x;
    coord_squared_t _y=y;
    return _x*_x+_y*_y;
}

用什么来表达 coord_squared_t ?在标准库中有什么可以让我做类似 double_width< coord_t> :: type 的事情来获取正确的宽度，而不是显式地选择类型吗?

What could be used to express coord_squared_t in terms of coord_t? Is there anything in the standard library that allows me to do something like double_width<coord_t>::type to get the correct width, instead of explicitly choosing the type?

C ++ 11或C ++ 14很好.

C++11 or C++14 are fine.

推荐答案

您可以使用

You could use boost::int_t:

using coord_squared_t = boost::int_t<sizeof(coord_t)*CHAR_BIT*2>::least;

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