通常,人们会在数据流的末尾添加CRC。 CRC校验将包括CRC本身,如果CRC正确则返回0。
我需要添加CRC来验证我的嵌入代码。需要对其进行检查,但内存空间中的首字是中断向量。是否可以在中间放置一个键值,以便CRC校验为整个代码返回0? (或者这是无法解决的?)
答案 0 :(得分:2)
这绝对有可能。您可以向后运行CRC,这将是快速和简单的。下面是示例代码。
实际上,您可以在流中的任何位置给我分散位的位置,如果您给我足够的位数,我可以告诉您如何设置它们以在结尾处获得零CRC或任何其他CRC这个问题的价值。我的spoof code解决了线性方程式,得出了答案。
但是我想知道你为什么要这样做。为什么不知道CRC的存储位置并计算除此之外的所有内容的CRC,然后根据存储的CRC检查结果?
// Example of the generation of a "middle" CRC, which is inserted somewhere in
// the middle of a sequence, where the CRC is generated such that the CRC of
// the complete sequence will be zero. This particular CRC has no pre or post
// processing.
//
// Placed into the public domain by Mark Adler, 11 May 2016.
#include <stddef.h> // for size_t
#include <stdint.h> // for uint32_t and uint64_t
#define POLY 0xedb88320 // CRC polynomial
// Byte-wise CRC tables for forward and reverse calculations.
uint32_t crc_forward_table[256];
uint32_t crc_reverse_table[256];
// Fill in CRC tables using bit-wise calculations.
void crc32_make_tables(void) {
for (uint32_t n = 0; n < 256; n++) {
uint32_t crc = n;
for (int k = 0; k < 8; k++)
crc = crc & 1 ? (crc >> 1) ^ POLY : crc >> 1;
crc_forward_table[n] = crc;
crc_reverse_table[crc >> 24] = (crc << 8) ^ n;
}
}
// Return the forward CRC of buf[0..len-1], starting with crc at the front.
uint32_t crc32(uint32_t crc, unsigned char *buf, size_t len) {
for (size_t n = 0; n < len; n++)
crc = (crc >> 8) ^ crc_forward_table[(crc ^ buf[n]) & 0xff];
return crc;
}
// Return the reverse CRC of buf[0..len-1], starting with crc at the end.
uint32_t crc32_reverse(uint32_t crc, unsigned char *buf, size_t len) {
while (len)
crc = (crc << 8) ^ crc_reverse_table[crc >> 24] ^ buf[--len];
return crc;
}
// Put a 32-bit value into a byte buffer in little-endian order.
void put4(uint32_t word, unsigned char *pos) {
pos[0] = word;
pos[1] = word >> 8;
pos[2] = word >> 16;
pos[3] = word >> 24;
}
#include <stdlib.h> // for random() and srandomdev()
// Fill dat[0..len-1] with uniformly random byte values. All of the bits from
// each random() call are used, except for possibly a few leftover at the end.
void ranfill(unsigned char *dat, size_t len) {
uint64_t ran = 1;
while (len) {
if (ran < 0x100)
ran = (ran << 31) + random();
*dat++ = ran;
ran >>= 8;
len--;
}
}
#include <stdio.h> // for printf()
#define LEN 1024 // length of the message without the CRC
// Demonstrate the generation of a middle-CRC, using the forward and reverse
// CRC computations. Verify that the CRC of the resulting sequence is zero.
int main(void) {
crc32_make_tables();
srandomdev();
unsigned char dat[LEN+4];
ranfill(dat, LEN/2);
put4(0, dat + LEN/2); // put zeros where the CRC will go
ranfill(dat + LEN/2 + 4, (LEN+1)/2);
put4(crc32(0, dat, LEN/2) ^ crc32_reverse(0, dat + LEN/2, (LEN+1)/2 + 4),
dat + LEN/2); // replace the zeros with the CRC
printf("%08x\n", crc32(0, dat, LEN+4));
return 0;
}