我如何加快十六进制字符到二进制字符的转换

Question

I have already a function that convert hex char(input) to binary char(output). it works perfect, for small amount of data(input length).But when the input is too big, it stuck/not working. May be strcat take too much time. Is there some alternate solution, So i can convert big hex input characters into equivalent binary. My function is:

void fun_hex_ch_2bin(int len_hex_str,uint8_t *hex,uint8_t *bin){
  /* Extract first digit and find binary of each hex digit */
  int i=0,j=0;

   char array_hex[16]={'0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f'};
   uint8_t *new_hex=malloc(len_hex_str*2);
   char hex_char1,hex_char2;
   j=0;
   for(i=0;i<len_hex_str;i++)
   {
      hex_char1=array_hex[hex[i]&0x0f];
      hex_char2=array_hex[(hex[i]>>4)&0x0f];
      //printf("%c  %c\n",hex_char1,hex_char2);
      new_hex[j]=hex_char2;
      new_hex[j+1]=hex_char1;
      j=j+2;
   }

    for(i=0; i<len_hex_str*2; i++)
    {
        switch(new_hex[i])
        {
            case '0':
                strcat(bin, "0000");
                break;
            case '1':
                strcat(bin, "0001");
                break;
            case '2':
                strcat(bin, "0010");
                break;
            case '3':
                strcat(bin, "0011");
                break;
            case '4':
                strcat(bin, "0100");
                break;
            case '5':
                strcat(bin, "0101");
                break;
            case '6':
                strcat(bin, "0110");
                break;
            case '7':
                strcat(bin, "0111");
                break;
            case '8':
                strcat(bin, "1000");
                break;
            case '9':
                strcat(bin, "1001");
                break;
            case 'a':
            case 'A':
                strcat(bin, "1010");
                break;
            case 'b':
            case 'B':
                strcat(bin, "1011");
                break;
            case 'c':
            case 'C':
                strcat(bin, "1100");
                break;
            case 'd':
            case 'D':
                strcat(bin, "1101");
                break;
            case 'e':
            case 'E':
                strcat(bin, "1110");
                break;
            case 'f':
            case 'F':
                strcat(bin, "1111");
                break;
            default:
                printf("Invalid hexadecimal input.");
        }
    }

}

Answer 1

Just use sprintf() instead of strcat()

char *bin; // points to a long enough buffer
int binlen = 0;
binlen += sprintf(bin + binlen, "something"); // strcat(bin, "something");
binlen += sprintf(bin + binlen, "otherthing"); // strcat(bin, "otherthing");
binlen += sprintf(bin + binlen, "foobar"); // strcat(bin, "foobar");
//...
// you can even do
binlen += sprintf(bin + binlen, "%.2f", 2.71828); // strcat(bin, "2.72");

Answer 2

16 is a power of 2 so converting it to binary is pretty simple. Each hex digit corresponds to exactly 4 binary digits - and you can rely on this fact. As a first step you need to convert the input characters from ASCII to numeric values. That's easily done in one pass over the input hex string and subtracting 48 if the character is between '0' and '9' or subtracting 88 if the character is between 'a' and 'f' (take a look at the ASCII table if need an explanation on why). After that the conversion is straightforward - go over the hex array and for each hex value take a look at the last bit and proceed to the next bit exactly 4 times, move to the next hex value and repeat the procedure. Something like this:

int i = 0, j = 0;
while(i < len_hex_str)
{
   bin[j++]=hex[i] & 1;
   hex[i] >>= 1;
   if(j % 4 == 0) i++;
}

And since you seem to be in need to have it ASCII representation, just pass over the output string and add 48 to each digit.

Answer 3

How can i speed up hexadecimal characters conversion to binary characters (?)
May be strcat take too much time.

Yes. Each call to strcat() takes longer and longer time as code does not take advantage of data already converted.

strcat() take n time to traverse the first characters.

1st strcat call, n = 0    
2st strcat call, n = 8    
3rd strcat call, n = 16    
4th strcat call, n = 24
ith strcat call, n = 8*(i-1)

See how the sum (0+8+16+24+...) goes up by order of i*i as i increases?

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Note that the first call to strcat(bin, ...) is suspect as bin[0] is not certainly a null character - something required when concatenating to a string .

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Is there some alternate solution (?)

I recommend a re-write. Directly read from hex as binary and skip the in-between conversion to hexadecimal.

void fun_hex_ch_2bin(int len_hex_str, uint8_t *hex, uint8_t *bin) {
  while (len_hex_str > 0) {
    len_hex_str--;
    // Start with the MSBit
    for (uint8_t mask = 0x80; mask; mask >>=1) {
      *bin++ = mask & *hex ? '1' : '0';
    }
    hex++;
  }
  // Append a null character as `bin` is to point to a _string_.
  *bin = '\0';
}

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I'd expect bin , as a string to be char* and not unit8_t * .