Question

我有一个变量char* etag="8a964ee2a5e88be344f36c22562a6486";。

如何将其转换为byte数组，其中byte实际为typedef unsigned char byte？

Answer 1

byte bytes[16];
int i=0;
for (;i<16;++i) {
  int value;
  sscanf(etag+2*i,"%02x",&value);
  bytes[i] = value;
}

Answer 2

如果我理解正确，你想要将16个字母的十六进制字符组组成字节，然后将这些字节存储在char数组中。试试这个：

char *str = "long hex string";

int i;
unsigned char bytes[16];
char tmp[3] = { 0 };
for(i = 0; i < 16; i++)
{
    memcpy(tmp, str + 2 * i, 2);
    bytes[i] = (unsigned char)strtol(tmp, NULL, 16);
}

Answer 3

您可能使用strtol

下面是示例程序：

/****************************************************************
 *
 * Purpose: To show examples of the strtol function.
 * Author:  M.J. Leslie
 * Date:    06-Nov-94
 *
 ****************************************************************/

#include <stdlib.h>

main()
{
  char num[10];

                /* Test a valid number      */
  strcpy(num,"13");

  printf("%s(Oct) is %i(Dec)\n", num, strtol(num, NULL,  8));
  printf("%s(Dec) is %i(Dec)\n", num, strtol(num, NULL, 10));
  printf("%s(hex) is %i(Dec)\n", num, strtol(num, NULL, 16));

  puts("----------------------------------");

                /* Test a slightly valid number
                 * Returns the same results as 
                 * above.           */
  strcpy(num, "13hzcd");

  printf("%s(Oct) is %i(Dec)\n", num, strtol(num, NULL,  8));
  printf("%s(Dec) is %i(Dec)\n", num, strtol(num, NULL, 10));
  printf("%s(hex) is %i(Dec)\n", num, strtol(num, NULL, 16));

  puts("----------------------------------");

                /* Test an invalid number
                 * Returns ZERO         */
  strcpy(num, "hzcd");

  printf("%s(Oct) is %i(Dec)\n", num, strtol(num, NULL,  8));
  printf("%s(Dec) is %i(Dec)\n", num, strtol(num, NULL, 10));
  printf("%s(hex) is %i(Dec)\n", num, strtol(num, NULL, 16));


  puts("----------------------------------");

                /* Test 0 base.
                 * This will look at the number 
                 * and decide the base for its self!
                 */
  strcpy(num, "13");
  printf("%s is %i(Dec)\n", num, strtol(num, NULL, 0));

  strcpy(num, "013");
  printf("%s is %i(Dec)\n", num, strtol(num, NULL, 0));

  strcpy(num, "0x13");
  printf("%s is %i(Dec)\n", num, strtol(num, NULL, 0));

}

/****************************************************************
 *
 * Results of the program:
 *
 *  13(Oct) is 11(Dec)
 *  13(Dec) is 13(Dec)
 *  13(hex) is 19(Dec)
 *  ----------------------------------
 *  13hzcd(Oct) is 11(Dec)
 *  13hzcd(Dec) is 13(Dec)
 *  13hzcd(hex) is 19(Dec)
 *  ----------------------------------
 *  hzcd(Oct) is 0(Dec)
 *  hzcd(Dec) is 0(Dec)
 *  hzcd(hex) is 0(Dec)
 *  ----------------------------------
 *  13 is 13(Dec)
 *  013 is 11(Dec)
 *  0x13 is 19(Dec)
 *
 ****************************************************************/

将十六进制数解析为数组

3 个答案: