cout的行为<< uint8和uint16的十六进制 [英] Behavior of cout << hex with uint8 and uint16

问题描述

我注意到cout << hex给了我奇怪的结果，而我找不到任何能回答原因的地方.我正在做的只是将一些值同时分配给uint8_t和uint16_t，然后尝试将它们写入stdout.当我运行此命令时:

I'm noticing that cout << hex is giving me strange results, and I cannot find anywhere that answers why. What I am doing is simply assigning some values to both a uint8_t and uint16_t and then attempting to write them to stdout. When I run this:

uint8_t a = 0xab;
uint16_t b = 0x24de;
cout << hex << a << endl;
cout << hex << b << endl;

我得到结果:

$./a.out

24de
$

，未显示uint8_t的值.是什么原因造成的?我不认为不会为一种类型提供cout实现，而不是为另一种类型实现.

with no value displayed for the uint8_t. What could be causing this? I didn't think there wouldn't be a cout implementation for one type for not the other.

推荐答案

std::uint8_t是unsigned char的别名:

typedef unsigned char uint8_t;

因此选择了使用char&的插入程序的重载，并编写了0xab的ASCII表示形式，由于0xab在扩展ASCII范围内，因此在技术上可能会因操作系统而异.

So the overload of the inserter that takes a char& is chosen, and the ASCII representation of 0xab is written, which could technically vary by your operating system, as 0xab is in the range of Extended ASCII.

您必须将其转换为整数:

You have to cast it to an integer:

std::cout << std::hex << static_cast<int>(a) << std::endl;

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