将十六进制数组 (char*) 转换为整数

Question

我需要在 C 中将字节转换为 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 header，后跟一个 ID 和一个值（每个两个 little-endian 字节）。如何获取 ID 和值？

尝试失败

我尝试制作指向整数的指针：

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

我也试过用

改变字节顺序

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

我也尝试了 sscanf 和 strol 解决方案，但我得到了奇怪的结果。

Answer 1

让我提供一个您正在寻找的示例。

 #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("\n");
   printf("Header: 0x%02x 0x%02x 0x%02x 0x%02x.\n", 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).\n", id, id, val, val);

   return 0;
 }

在我的电脑上，下面是输出：

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

将十六进制数组 (char*) 转换为整数

Convert hexa array (char*) into integer

c

int

endianness

binary-data

尝试失败