如何將十六進制字符緩衝區轉換爲ASCII字符串[C]

Question

我需要將十六進制字符串轉換爲Ascii字符串。不僅可以用printf顯示，還可以將其保存在內存中。 我使用此代碼來顯示它，但我想保存它。我該怎麼辦？：

#include <stdio.h> 
#include <string.h> 

int hex_to_int(char c){ 
     int first = c/16 - 3; 
     int second = c % 16; 
     int result = first*10 + second; 
     if(result > 9) result--; 
     return result; 
} 

int hex_to_ascii(char c, char d){ 
     int high = hex_to_int(c) * 16; 
     int low = hex_to_int(d); 
     return high+low; 
} 

int main(){ 
     const char st[12] = "48656C6C6F3B"; 
     int length = strlen(st); 
     int i; 
     char buf = 0; 
     for(i = 0; i < length; i++){ 
       if(i % 2 != 0){ 
         printf("%c", hex_to_ascii(buf, st[i])); 
       }else{ 
         buf = st[i]; 
       } 
     } 
} 

我的第一個解決方案是使用sprintf(buf,"%c",hex_to_ascii(buf,st[i]) INT for循環，但這種方法行不通，因爲sprintf需要的char的指針。

Answer 1

好吧，讓我們有一個鏡頭：

#include <ctype.h> 
#include <stdio.h> 

static unsigned char hexdigit2int(unsigned char xd) 
{ 
    if (xd <= '9') 
    return xd - '0'; 
    xd = tolower(xd); 
    if (xd == 'a') 
    return 10; 
    if (xd == 'b') 
    return 11; 
    if (xd == 'c') 
    return 12; 
    if (xd == 'd') 
    return 13; 
    if (xd == 'e') 
    return 14; 
    if (xd == 'f') 
    return 15; 
    return 0; 
} 

int main(void) 
{ 
    const char st[] = "48656C6C6F3B", *src = st; 
    char text[sizeof st + 1], *dst = text; 

    while (*src != '\0') 
    { 
    const unsigned char high = hexdigit2int(*src++); 
    const unsigned char low = hexdigit2int(*src++); 
    *dst++ = (high << 4) | low; 
    } 
    *dst = '\0'; 
    printf("Converted '%s', got '%s'\n", st, text); 
    return 0; 
} 

的非常詳細的前瞻性digit2int()功能是試圖不ASCII依賴，這總是好的。請注意，main()中的循環假設輸入正確，它僅檢查每兩個字符的終止並且不處理非十六進制數據。

哦，這個打印：

Converted '48656C6C6F3B', got 'Hello;' 

Answer 2

你可能知道每個character is represented by an int - > ...'a'= 97，'b'= 98 ...因此，一旦你將一個十六進制轉換爲int，就不需要創建另一個函數，將它轉換爲ascii - 只要將整數存儲到一個字符中就可以做到這一點（除非我錯過理解練習的東西）。

作爲用於存儲回存儲器：有很多種選擇： 1.除去constconst char st[12] = "48656C6C6F3B";和 之前，從函數的返回值分配給期望的細胞或使用的sscanf

看一看here

Answer 3

你做得很好，但有一些問題。有以下比較：

const char* st = "48656C6C6F3B"; // I don't want to calculate storage size 
char r[12] = { 0 }; // Here is I want to store the result 
char* h = r; // pointer to write position in the result 
int length = strlen(st); 
for (int i = 0; i < length; i+=2) { 
    assert((h - r) < sizeof(r)); // check that there is not problem with out of range 
    *h++ = hex_to_ascii(st[i], st[i + 1]); 
} 
printf("%s", r); // now the r contains "ansi string" 

Answer 4

「拯救」 的結果，簡單的增加一個緩衝區，dest，存儲結果。

代碼中包含的其他建議。

// add void to signature 
int main(void) { 
    const char st[12] = "48656C6C6F3B"; 
    int length = strlen(st); 
    int i; 
    char buf = 0; 

    // add destination buffer 
    char dest[10]; 

    // Add test 
    // for (i = 0; i < length; i++) { 
    for (i = 0; i < length && (i/2 + 1) < sizeof(dest); i++) { 
    if (i % 2 != 0) { 

     // printf("%c", hex_to_ascii(buf, st[i])); 
     dest[i/2] = hex_to_ascii(buf, st[i]); 

    } else { 
     buf = st[i]; 
    } 
    } 

    // Add termination 
    dest[i/2] = '\0'; 
    // Do someting with dest 
    puts(dest); 
    return 0; 
} 

---

可替換地，一些代碼，處理各種可能的問題：下/上殼體十六進制數字，無效字符，奇數，小緩衝器，嵌入空字符。

#include <stdlib.h> 
#include <string.h> 
// There are _many_ ways to do this step. 
unsigned char2digit(int ch) { 
    static const char Hex[] = "ABCDEFabcdef"; 
    char *p = memchr(Hex, ch, 32); 
    if (p) { 
    return (unsigned) (p - Hex) % 16; 
    } 
    return (unsigned) -1; // error value 
} 

// Return NULL with ill-formed string 
char *HexStringToString(char *dest, size_t size, const char *src) { 
    char *p = dest; 
    if (size <= 0) { 
    return NULL; 
    } 
    size--; 
    while (*src) { 
    if (size == 0) return NULL; 
    size--; 

    unsigned msb = char2digit(*src++); 
    if (msb > 15) return NULL; 
    unsigned lsb = char2digit(*src++); 
    if (lsb > 15) return NULL; 
    char ch = (char) (msb * 16 + lsb); 

    // Optionally test for embedded null character 
    if (ch == 0) return NULL; 

    *p++ = ch; 
    } 
    *p = '\0'; 
    return dest; 
} 

void testHex(const char *s) { 
    char buf[10]; 
    char *dest = HexStringToString(buf, sizeof buf, s); 
    printf("%-24s --> '%s'\n", s, dest ? dest : "NULL"); 
} 

#include <stdio.h> 
int main(void) { 
    testHex("48656C6C6F3B");  // upper case 
    testHex("48656c6c6f3b");  // lower case 
    testHex(""); 
    testHex("48656C6C6F3B48656C"); 
    // fails 
    testHex("48656C6C6F3B48656C6C"); // Too long 
    testHex("48656C6C6F3B0");  // Odd character count 
    testHex("48656C6C6F3Bxx");  // Non-hex character 
    testHex("48006C6C6F3B");   // null character 
    return 0; 
}   

輸出

48656C6C6F3B    --> 'Hello;' 
48656c6c6f3b    --> 'Hello;' 
         --> '' 
48656C6C6F3B48656C  --> 'Hello;Hel' 
48656C6C6F3B48656C6C  --> 'NULL' 
48656C6C6F3B0   --> 'NULL' 
48656C6C6F3Bxx   --> 'NULL' 
48006C6C6F3B    --> 'NULL'