从 C 中的十六进制字符串转换为 ASCII 字符串

Question

我正在尝试将十六进制字符串转换为其等效的 ASCII。我得到一个字符串的“十六进制”值作为字符串，即我得到“41424344”而不是“ABCD”。我只需要提取 41 这将是我的十六进制值并重新编码为“ABCD”。这是我的。

   int main(int argc, char *argv[]){

    char *str = "ABCD";
    unsigned int val = 0;
    int i = 0;
    int MAX = 4;
    for (i = 0; i<MAX; i++){
        val = (str[i] & 0xFF);
        //printf("dec val= %d\n", val);
        //printf("hex val= %02x\n", val);
    }
    val = 0;
    char *hexstr = "41424344";
    char *substr = (char*)malloc(3);
    char *ptr = hexstr;

    for (i = 0; i<8; i++){
        strncpy(substr, ptr, 2);
        printf("substr = %s\n", substr);
        int s = atoi(substr);
        printf("s= %d\n", s);
        ptr= ptr+2;
        i = i+2;
    }
    return 0;

}

从现在开始，我必须让这个“s”值成为一个十六进制值，而不是一个整数。如何做到这一点？

更新:

这是您回答后我得到的:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char *argv[]){

    char *str = "ABCD";
    unsigned int val;
    val = 0;
    int i = 0;
    int MAX = 4;
    for (i = 0; i<MAX; i++){
        val = (str[i] & 0xFF);
        //printf("dec val= %d\n", val);
        //printf("hex val= %02x\n", val);
    }
    val = 0;
    char *hexstr = "41424344";
    char *substr = (char*)malloc(3);
    char *ptr = hexstr;
    char *retstr = (char *)malloc(5);
    char *retptr = retstr;

    for (i = 0; i<8; i+1){
        strncpy(substr, ptr, 2);
        printf("substr = %s\n", substr);
        int s = strtol(substr, NULL, 16);
        printf("s= %d\n", s);
        ptr= ptr+2;
        i = i+2;
        sprintf(retptr, "%c", s);
        retptr = retptr +1;
    }
    printf("retstr= %s\n", retstr);
    return 0;

}

Answer 1

从以下位置更改“s”变量行

int s = atoi(substr);

到

int s = strtol(substr, NULL, 16);