转换为ASCII字符串从用C十六进制字符串 [英] Convert to ASCII String from hex String in C

问题描述

我想一个十六进制字符串转换为其对应的ASCII。我给出一个字符串的六角的价值作为一个字符串，即不是ABCD我得到41424344。我只需要提取41这将是我的hexvalue和再code为ABCD。下面是我。

  INT主（INT ARGC，CHAR *的argv []）{    char *之海峡=ABCD;
    unsigned int类型VAL = 0;
    INT I = 0;
    INT MAX = 4;
    对于（i = 0; I＆LT; MAX;我++）{
        VAL =（STR [1]  - 安培; 0xFF的）;
        //的printf（DEC VAL =％d个\\ N，VAL）;
        //的printf（六角VAL =％02X \\ n，VAL）;
    }
    VAL = 0;
    字符* hexstr =41424344;
    字符* SUBSTR =（字符*）malloc的（3）;
    字符* PTR = hexstr;    对于（i = 0; I＆LT; 8;我++）{
        函数strncpy（SUBSTR，PTR，2）;
        的printf（SUBSTR =％S \\ n，SUBSTR）;
        INT S =的atoi（SUBSTR）;
        的printf（S =％d个\\ N，S）;
        PTR = PTR + 2;
        I = I + 2;
    }
    返回0;}
 

的事情是从这里开始，我不得不做出这个S值是一个十六进制值，而不是一个int。如何才能做到这一点？

更新：

下面是我你的答案后：

 的#include＆LT;＆stdio.h中GT;
＃包括LT＆;＆stdlib.h中GT;
＃包括LT＆;＆string.h中GT;
INT主（INT ARGC，CHAR *的argv []）{    char *之海峡=ABCD;
    无符号整型VAL;
    VAL = 0;
    INT I = 0;
    INT MAX = 4;
    对于（i = 0; I＆LT; MAX;我++）{
        VAL =（STR [1]  - 安培; 0xFF的）;
        //的printf（DEC VAL =％d个\\ N，VAL）;
        //的printf（六角VAL =％02X \\ n，VAL）;
    }
    VAL = 0;
    字符* hexstr =41424344;
    字符* SUBSTR =（字符*）malloc的（3）;
    字符* PTR = hexstr;
    字符* retstr =（字符*）malloc的（5）;
    字符* retptr = retstr;    对于（I = 0; I＆下; 8; I + 1）{
        函数strncpy（SUBSTR，PTR，2）;
        的printf（SUBSTR =％S \\ n，SUBSTR）;
        INT S =与strtol（SUBSTR，NULL，16）;
        的printf（S =％d个\\ N，S）;
        PTR = PTR + 2;
        I = I + 2;
        sprintf的（retptr，％C，S）;
        retptr = retptr +1;
    }
    的printf（retstr =％S \\ n，retstr）;
    返回0;}
 

解决方案

从

更改的变线

int类型=的atoi（SUBSTR）;

到

int类型=与strtol（SUBSTR，NULL，16）;

I am trying to convert a hex string to its equivalent ASCII. I am given a "hex" value of a string as a string, i.e instead of "ABCD" I get "41424344". I just need to extract 41 which will be my hexvalue and recode as "ABCD". Here is what I have.

   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;

}

The thing is from here on, I have to make this "s" value to be a hex value and not an int. How can this be done?

UPDATE:

Here is what I have after your answers:

#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;

}

解决方案

Change your "s" variable line from

int s = atoi(substr);

to

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

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