Ç - 字符到六

Question

我有STR：

char *str = "lala"; 

現在，我想在STR轉換任何字符爲十六進制，例如：

str = convert(str); 
print str: 0x6C 0x61 0x6C 0x61 
      ^l^a^l^a 

我可怎麼辦呢？

Answer 1

char* x = str - 1; 
while(*++x) printf("%02x ", (int) *x); // Print one character in hex 
printf("\n") // Finish with a carriage return 

Answer 2

char *convert(char const *str) { 
    int len = strlen(str); 
    char *retVal = (char *)malloc(5 * len); 
    char *pos = retVal; 
    int i; 
    for(i = 0; i < len; ++i, pos += 5) sprintf(pos, i? " 0x%x" : "0x%x", str[i]); 
    retVal[5 * len - 1] = '\0'; 
    return retVal; 
} 

可能會錯過一些東西，沒有使用下八年。

Answer 3

只問printf的做到這一點：

void convert(char* str, size_t length) { 
    size_t i; 
    for(i = 0; i < length; i++) 
     printf("0x%02x ", str[i]); 
} 

Answer 4

您可以使用此實現實現它....

#define MAX 100 

char *convert(char *str) 
{ 
    char *hexStr = (char *)malloc(strnlen(str, MAX) * 5); 
    if (hexStr == NULL) 
     return NULL; 
    int i,j; 

    for (i=0, j=0; str[i]; j+=5, i++) 
     sprintf(hexStr + j, "0x%02x ", str[i]); 

    hexStr[--j] = '\0'; 

    return hexStr; 
}