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.");
}
}
}
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");