铸造成字节在C# [英] Casting to byte in C#
问题描述
可能重复:结果
的会发生什么,当你从短期转换为C#?
块引用>
$字节b
$ b有人能解释铸造值一个字节时发生了什么,如果它的最小/最大字节的范围之外?这似乎是取整数值,并与255,我试图理解为什么这并不抛出异常的原因模了。
INT I = 5000;
字节B =(字节)I;
Console.WriteLine(二); //输出136
解决方案5000被表示为4个字节( INT)
(十六进制)
| 00 | 00 | 13 | 88 |
现在,当你将它转换为字节,它只是需要在最后1个字节
的理由:在IL级别的conv.u1 运营商将被用来在发生转换的int字节溢出,这将截断高位。 (见 conv.u1 文档中的备注部分)。 的
| 88 |
这是136十进制表示
Possible Duplicate:
What happens when you cast from short to byte in C#?Can someone explain what's happening when casting a value to a byte, if it's outside the range of min/max byte? It seems to be taking the integer value and modulo it with 255. I'm trying to understand the reason for why this doesn't throw an exception.
int i = 5000; byte b = (byte)i; Console.WriteLine(b); // outputs 136
解决方案5000 is represented as 4 bytes (int) (hexadecimal)
|00|00|13|88|
Now, when you convert it to byte, it just takes the last 1-byte.
Reason: At the IL level, conv.u1 operator will be used which will truncate the high order bits if overflow occurs converting int to byte. (See remarks section in the conv.u1 documentation).
|88|
which is 136 in decimal representation
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