一次将Byte数组转换为Integer 2字节 [英] Convert Byte array to Integer 2 bytes at a time

问题描述

我有一个字节数组，看起来像

01 00 02 00 73 45 69 A5

所以我必须读取前2个字节并将其转换为整数以获取其值.
接下来的2个字节相同，然后接下来的4个字节相同.

在这里，前2个字节(01 00)的值为1，后2个字节(02 00)的值为2.

I have a byte array which looks something like

01 00 02 00 73 45 69 A5

So i have to read first 2 bytes and convert it into integer to get its value.
Same for next 2 bytes and then next 4 bytes.

Here the value for first 2 bytes (01 00) is 1, next 2 bytes (02 00) is 2.

So could some one help me on this.

推荐答案

该框架有一个类，用于将字节数组转换为原始类型，反之亦然.

System.BitConverter [

The framework has a class to convert byte array to primitive types and vice versa.

The System.BitConverter[^] class is worth a passing glance as it''s very useful and the ToInt16 method is the solution to your problem.

Alan.

我假定无符号数字.

byte [] b

int i = b [x] + b [x + 1] * 256

围绕它抛出一个带有x的2增量的循环.

欢呼

I assume unsigned numbers.

byte[] b

int i = b[x] + b[x + 1] * 256

Throw a loop with an increment of 2 of x around it.

Cheers

无论如何，将两个字节转换为整数非常容易.将第一个字节(作为整数)乘以256，然后加上第二个字​​节.

Anyway, converting two bytes to an integer is very easy. Take the first byte (as an integer), multiply it by 256 and add the second byte.

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