二进制数据累加求和,取低八位

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This topic created in 1816 days ago, the information mentioned may be changed or developed.

以下是 16 进制表示,我把每一个 16 进制转换成 10 进制,然后进行累加,得到 2551

echo 2551&0xff;

转换成 16 进制是f7

5a 54 08 1f 01 01 61 00 50 53 0a 09 4e 01 64 00 00 00 00 00 00 00 00 00 00 00 00 02 00 00 00 00 00 00 00 00 50 53 0a 09 4f 03 64 00 00 00 00 00 00 00 50 53 0a 09 46 04 64 00 00 00 00 00 00 00 50 53 0a 0a 09 05 64 00 00 00 00 00 00 00 50 53 0a 0a 08 06 64 00 00 00 00 00 00 00 50 53 0a 09 34 07 64 00 00 00 00 00 00 00 50 53 0a 0a 0a 08 64 00 00 00 00 00 00 00

但是硬件那边传过来的求和校验是0x47，是我哪里算错了吗

累加

二进制

x47

xff

4 replies • 2021-05-11 11:58:05 +08:00

aa65535

May 10, 2021

明明加起来是 2631，不知道你为什么能得到 2551

SingeeKing

PRO

May 10, 2021

>>> sum([int(x, 16) for x in '''5a 54 08 1f 01 01 61 00 50 53 0a 09 4e 01 64 00 00 00 00 00 00 00 00 00 00 00 00 02 00 00 00 00 00 00 00 00 50 53 0a 09 4f 03 64 00 00 00 00 00 00 00 50 53 0a 09 46 04 64 00 00 00 00 00 00 00 50 53 0a 0a 09 05 64 00 00 00 00 00 00 00 50 53 0a 0a 08 06 64 00 00 00 00 00 00 00 50 53 0a 09 34 07 64 00 00 00 00 00 00 00 50 53 0a 0a 0a 08 64 00 00 00 00 00 00 00'''.strip().split()])
2631

takitooru

May 11, 2021

uint8_t test_list[]={0x5a,0x54,0x08,0x1f,0x01,0x01,0x61,0x00,0x50,0x53,0x0a,0x09,0x4e,0x01,0x64,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x02,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x50,0x53,0x0a,0x09,0x4f,0x03,0x64,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x50,0x53,0x0a,0x09,0x46,0x04,0x64,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x50,0x53,0x0a,0x0a,0x09,0x05,0x64,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x50,0x53,0x0a,0x0a,0x08,0x06,0x64,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x50,0x53,0x0a,0x09,0x34,0x07,0x64,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x50,0x53,0x0a,0x0a,0x0a,0x08,0x64,0x00,0x00,0x00,0x00,0x00,0x00,0x00};
int main(int argc,char *argv[])
{
uint8_t rst = 0;
for(int i=0;i<sizeof(test_list);i++)
{
rst += test_list[i];
}
printf("%d 0x%02x",rst,rst);
}
gcc .\test6_sig.c -m32
.\a.exe
-> 71 0x47

cs5117155

May 11, 2021

谢谢大家，原来我自已程序过滤了一个字节，导致结果错误了，2631 是对的