[英]Behavior of cout << hex with uint8 and uint16
I'm noticing that cout << hex
is giving me strange results, and I cannot find anywhere that answers why. 我注意到
cout << hex
给了我奇怪的结果,我无法找到解决原因的地方。 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. 我正在做的只是为
uint8_t
和uint16_t
分配一些值,然后尝试将它们写入stdout。 When I run this: 当我运行这个:
uint8_t a = 0xab;
uint16_t b = 0x24de;
cout << hex << a << endl;
cout << hex << b << endl;
That I get the result: 我得到了结果:
$./a.out
24de
$
with no value displayed for the uint8_t. 没有显示uint8_t的值。 What could be causing this?
可能是什么导致了这个? I didn't think there wouldn't be a cout implementation for one type for not the other.
我认为不会有一种类型的cout实现而不是另一种类型。
std::uint8_t
is an alias for unsigned char
: std::uint8_t
是unsigned char
的别名:
typedef unsigned char uint8_t;
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. 因此选择了
char&
的插入器的重载,并且写入了0xab
的ASCII表示,这在技术上可能因操作系统而异,因为0xab
在扩展ASCII的范围内。
You have to cast it to an integer: 你必须将它强制转换为整数:
std::cout << std::hex << static_cast<int>(a) << std::endl;
The other answers are correct about the reason. 其他答案是正确的。 The simplest fix is:
最简单的解决方法是:
cout << hex << +a << endl;
Demonstration: http://ideone.com/ZHHAHX 演示: http : //ideone.com/ZHHAHX
It works because operands undergo integer promotion. 它的工作原理是操作数经过整数提升。
uint8_t
is an alias for unsigned char
. uint8_t
是unsigned char
的别名。 You're essentially printing a character with the value 0xab
, which might be an invalid character depending on your encoding. 您实际上是打印一个值为
0xab
的字符,根据您的编码,该字符可能是无效字符。
This would be solvable by casting it to another integer type, converting its value to a string in advance or writing some sort of a wrapper class that implements std::ostream& operator<<(std::ostream&, const ClassName&)
. 这可以通过将其转换为另一个整数类型,将其值转换为字符串或编写某种实现
std::ostream& operator<<(std::ostream&, const ClassName&)
的包装类来解决。
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