计算16位校验和？ [英] Calculating a 16 bit checksum?

问题描述

使用c中的程序读取文件，然后我必须对程序进行8位和16位校验和。到目前为止，我只完成了8位校验和。

这是我的理解

我读取文件并将信息存储在一个字符数组中，最后它需要换行。因此，例如，要计算8位校验和，本质上就是这样

文件共有3个字母（3个字母和换行符）

因此数组可容纳4个字符aaa +（换行符）（97 + 97 + 97 + 10）

据我所知，我在数组，然后做％256，那是我的校验和。

97 * 3 = // 3从我的理解中从ascii表中拉出一个（小a）

291 + 10 = 301 // +换行符

301％256 = cc以十六进制表示// //

但是我对于如何计算16位校验和感到困惑，因为如果单个字符数组不能一次添加2个字符，那么我将无法添加？

任何帮助将不胜感激

解决方案

要计算16位校验和，请以2为增量处理数组，然后将一个字节放入低字节。

  uint8_t array [MAX] ]; //将数据复制到
 size_t长度； //这是数据
 uint16_t校验和= 0的长度； 
 size_t even_length =长度-长度％2; //对于（int i = 0; i< even_length; i + = 2），向下舍入为2 
的倍数{
 uint16_t val = array [i] + 256 * array [i + 1] ; 
校验和+ = val; 
} 
 if（i< length）{//最后一个字节，如果它是奇数长度
校验和+ = array [i]; 
} 
  

不需要模数，因为无符号整数会自动实现模块化算法。 / p>

Working with a program in c that reads in a file and then I have to do both 8 bit and 16 bit checksum for a program.I only have 8 bit checksum done so far.

This is what I understand

I read the file and store the information in an array of characters and at end it takes the newline feed. so for example to calculate 8 bit check sum this is what happens essentially

File has 3 letters total ( 3 a's and a newline feed)

so array holds 4 chars aaa+(newline) (97+97+97+10)

To my understanding I add all the bytes in the array then do % 256 and that is my checksum.

97 * 3 = //3 a's (small a ) pulled from ascii table from what I understand

291 + 10 = 301 // + newline

301 % 256 = cc in hex //

however I am getting confused on how to calculate the 16 bit checksum because I can't add 2 characters at a time if its a single character array?

any help would be greatly appreciated

解决方案

To calculate a 16-bit checksum, you process the array in increments of 2, and put one byte into the low-order byte of the value that you're adding, and the other byte into the high-order byte.

uint8_t array[MAX]; // The data gets copied into here
size_t length; // This is the length of the data
uint16_t checksum = 0;
size_t even_length = length - length%2; // Round down to multiple of 2
for (int i = 0; i < even_length; i += 2) {
    uint16_t val = array[i] + 256 * array[i+1];
    checksum += val;
}
if (i < length) { // Last byte if it's odd length
    checksum += array[i];
}

There's no need to use modulus, since unsigned integers implement modular arithmetic automatically.

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