我不明白如何在 C 中将十六进制字符串转换为无符号字符？

Question

我有输入

unsigned char hex[64] =  9f86d081884c7d659a2feaa0c55ad015a3bf4f1b2b0b822cd15d6c15b0f00a0"

我想要这样的输出：

unsigned char tmpInHash[32] = { 0x9f, 0x86, 0xd0, 0x81, 0x88, 0x4c, 0x7d, 0x65, 0x9a, 0x2f, 0xea, 0xa0, 0xc5, 0x5a, 0xd0, 0x15, 0xa3, 0xbf, 0x4f, 0x1b, 0x2b, 0x0b, 0x82, 0x2c, 0xd1, 0x5d, 0x6c, 0x15, 0xb0, 0xf0, 0x0a, 0x08 }

我到处寻找答案，但我找到的答案不合适。

编辑

当我写的时候，我想要这样：

for(i=0; i< strlen(tmpInHash); i++ {
    printf("%c ", tmpInHash);
}

我明白这个：

0x9f 0x86 0xd0 0x81 0x88 0x4c 0x7d 0x65 0x9a ...

可能吗？

Answer 1

转换它的一种方法是以下方法：

• 循环访问字符串的每个项目，其中每 2 个字符表示一个数字。
• 分解每个这样的字符对并将它们存储在一个临时字符串中：char tmp {str[i], str[i+1], ''}; .
• 调用此字符串上的strtol(tmp, NULL, 16)以获取整数。

Answer 2

没有"高级"函数

#include <stdio.h>
#include <string.h>
int main(void)
{
    unsigned char hex[] = "9f86d081884c7d659a2feaa0c55ad015a3bf4f1b2b0b822cd15d6c15b0f00a08";
    size_t stringLength = (sizeof(hex)/sizeof(hex[0]))-1;
    unsigned char tmpInHash[stringLength/2];
    int j=0;
    // reset the char to 0. This grants that next or operation works on reset buffer
    tmpInHash[0] = 0;
    // loop that parse the whole string
    for (size_t i = 0; i < stringLength; i++)
    {
        // check if the char is a to f
        if ((hex[i] >= 'a') && (hex[i] <= 'f'))
        {
            tmpInHash[j] |= hex[i] -'a' + 10;
        }
        // che if is a digit
        else if ((hex[i] >= '0') && (hex[i] <= '9'))
        {
            tmpInHash[j] |= hex[i] -'0';
        }
        // character not allowed
        else
        {
            fprintf(stderr, "Character not allowed: %c position: %zun", hex[i], i);
            return 1;
        }
        // even index chars are hig 4 bits of unsigned char
        if ((i%2) == 0)
        {
            tmpInHash[j]<<=4;
        }
        else
        {
            // nex unsigned char
            j++;
            // reset the char to 0. This grants that next or operation works on reset buffer
            if (j < stringLength/2)
               tmpInHash[j] = 0;
        }
    }
    for (size_t i = 0; i < stringLength/2; i++)
    {
        printf("0x%02X ", tmpInHash[i]);
    }
    printf("n");

    return 0;
}

输出

0x9F 0x86 0xD0 0x81 0x88 0x4C 0x7D 0x65 0x9A 0x2F 0xEA 0xA0 0xC5 0x5A 0xD0 0x15 0xA3 0xBF 0x4F 0x1B 0x2B 0x0B 0x82 0x2C 0xD1 0x5D 0x6C 0x15 0xB0 0xF0 0x0A 0x08

编辑另一个例子可以是

#include <stdio.h>
#include <string.h>
unsigned char hex[] = "9f86d081884g7d659a2feaa0c55ad015a3bf4f1b2b0b822cd15d6c15b0f00a08";
#define HEX_LEN (sizeof(hex)-1)
unsigned char tmpInHash[HEX_LEN/2]={0};
int main(void)
{
    size_t i = 0;
    size_t j = 0;
    // parse all charcaters of hex
    while(hex[i] != '')
    {
        // check if the char is a to f
        if ((hex[i] >= 'a') && (hex[i] <= 'f'))
        {
            tmpInHash[j] |= hex[i] -'a' + 10;
        }
        // che if is a digit
        else if ((hex[i] >= '0') && (hex[i] <= '9'))
        {
            tmpInHash[j] |= hex[i] -'0';
        }
        // character not allowed
        else
        {
            fprintf(stderr, "Character not allowed: %c position: %zun", hex[i], i);
            return 1;
        }
        // even index chars are hig 4 bits of unsigned char
        if ((i%2) == 0)
        {
            tmpInHash[j]<<=4;
        }
        else
        {
            // nex unsigned char
            j++;
        }
        // next hex char
        i++;
    }
    // print loop
    for (i = 0; i < (HEX_LEN/2); i++)
    {
        printf("0x%02X ", tmpInHash[i]);
    }
    printf("n");
    return 0;
}

Answer 3

unsigned char hex[64] =  "9f86d081884c7d659a2feaa0c55ad015a3bf4f1b2b0b822cd15d6c15b0f00a0";
 unsigned char *ptrHex =hex;
    unsigned char tmpInHash[32];
    int i ,tmp ;
    for(i=0 ; i < 32 ; i++)
    {
         if((*ptrHex)<=9 && (*ptrHex) >=0)
                  tmp=(*ptrHex)-'0';
         else tmp=(*ptrHex)-'a'+10;
         tmpInHash[i]=(tmp<<4);
         ptrHex++;
       if((*ptrHex)<=9 && (*ptrHex) >=0)
                  tmp=(*ptrHex)-'0';
         else tmp=(*ptrHex)-'a'+10;
         tmpInHash[i]|=tmp ;
         ptrHex++;
    }