我正在开发一个硬件通信应用程序,我从外部硬件发送或需要数据。我已经完成了所需的数据部分。
我刚刚发现我可以使用一些帮助来计算校验和。
一个包被创建为 NSMutableData,然后在发送之前将其转换为字节数组。一个包看起来像这样:
0x1E 0x2D 0x2F 数据校验和
我在想我可以将十六进制转换为二进制来一一计算。但我不知道这是否是个好主意。如果这是唯一的方法,或者有一些我不知道的内置函数,请告诉我。任何建议将不胜感激。
顺便说一句,我刚刚从其他人的帖子中找到了 C# 的代码,我会尝试让它在我的应用程序中工作。如果可以的话,我会和你分享。仍然有任何建议将不胜感激。
package org.example.checksum;
public class InternetChecksum {
/**
* Calculate the Internet Checksum of a buffer (RFC 1071 - http://www.faqs.org/rfcs/rfc1071.html)
* Algorithm is
* 1) apply a 16-bit 1's complement sum over all octets (adjacent 8-bit pairs [A,B], final odd length is [A,0])
* 2) apply 1's complement to this final sum
*
* Notes:
* 1's complement is bitwise NOT of positive value.
* Ensure that any carry bits are added back to avoid off-by-one errors
*
*
* @param buf The message
* @return The checksum
*/
public long calculateChecksum(byte[] buf) {
int length = buf.length;
int i = 0;
long sum = 0;
long data;
// Handle all pairs
while (length > 1) {
// Corrected to include @Andy's edits and various comments on Stack Overflow
data = (((buf[i] << 8) & 0xFF00) | ((buf[i + 1]) & 0xFF));
sum += data;
// 1's complement carry bit correction in 16-bits (detecting sign extension)
if ((sum & 0xFFFF0000) > 0) {
sum = sum & 0xFFFF;
sum += 1;
}
i += 2;
length -= 2;
}
// Handle remaining byte in odd length buffers
if (length > 0) {
// Corrected to include @Andy's edits and various comments on Stack Overflow
sum += (buf[i] << 8 & 0xFF00);
// 1's complement carry bit correction in 16-bits (detecting sign extension)
if ((sum & 0xFFFF0000) > 0) {
sum = sum & 0xFFFF;
sum += 1;
}
}
// Final 1's complement value correction to 16-bits
sum = ~sum;
sum = sum & 0xFFFF;
return sum;
}
}