将十六进制数组（char *）转换为整数

Posted 2021-04-15

我需要将字节转换为C中的int（或uint8 / 16/32）。这是我的代码的解释：

char* out;
//init
//out=read_udp (from equipment)
for(int i=0; i<n; ++i)
  printf("out[%d]=%x ", i, out[i]);

输出：

out [0] = 0x00 out [1] = 0xAA out [2] = 0x44 out [3] = 0x12 out [4] = 0x2B out [5] = 0x00 out [6] = 0x7E out [7] = 0x3B

前四个字节是标题，后跟一个ID和一个值（每个字节有两个小端字节）。我怎样才能获得ID和价值？

Failed attempts

我试着指向整数：

char* p_ID;
char* p_val;
//malloc
p_ID=&out[4];
p_val=&out[6];
printf("%d %d", p_ID, p_val);

我也试过改变字节顺序

p_ID[0]=out[1]; p_ID[1]=out[2];

我也尝试过sscanf和strol解决方案，但我得到了奇怪的结果。

答案

让我举一个您正在寻找的例子。

 #include <stdio.h>
 #include <string.h>
 #include <stdint.h>

 #define SWAP2BYTES(val) (((val>>8)&0x00FF)|((val<<8)&0xFF00))
 bool isLittleEndian() { // just to check host endian 
   uint32_t val = 1;
   uint8_t *c = (uint8_t*)&val;
   return (1 == (uint32_t)*c);
 }
 // if the same endian, then no need to swap, otherwise swap
 #define CHECKSWAP(val) (isLittleEndian()?val:SWAP2BYTES(val))
 int main(int argc, char **argv) {
   uint8_t out[8] = {0x00,0xAA,0x44,0x12,0x2B,0x00,0x7E,0x3B};
   uint16_t id, val;

   // to prevent from alignment issue, memcpy used insted of assignment
   memcpy((uint8_t*)&id, &out[4], sizeof(id));
   memcpy((uint8_t*)&val, &out[6], sizeof(val));

   id = CHECKSWAP(id);
   val = CHECKSWAP(val);

   printf("outdata = ");
   for (int i = 0; i < sizeof(out); i++)
     printf("0x%02x ", out[i]);
   printf("
");
   printf("Header: 0x%02x 0x%02x 0x%02x 0x%02x.
", out[0], out[1], out[2], out[3]);
   // just to make sure whether or not intended values are retrieved
   printf("ID: %d(0x%04x), Value: %d(0x%04x).
", id, id, val, val);

   return 0;
 }

在我的电脑上，以下是输出：

outdata = 0x00 0xaa 0x44 0x12 0x2b 0x00 0x7e 0x3b
Header: 0x00 0xaa 0x44 0x12.
ID: 43(0x002b), Value: 15230(0x3b7e).