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I am using Arduino. I have the following code which converts binary data to its ASCII equivalent. It uses the String object.

static uint16_t index = 0;
static char buffer[1600]; //contains binary data 0x11, 0x22, 0x1, 0xa ...

String msg;
index = strlen(buffer);
for (i=0; i < index; i++)
{
    //Produce a zero in front for single digits. Examples, 0x5 transforms into 05, 0xa transforms into 0a
    if (buffer[i] <= 0x0F)
    {
        msg += "0";
    }

    msg += String(buffer[i], HEX); //msg contains the ASCII equivalent of buffer    
}

How can the code be modified such that the String object is not used but the same objective is accomplished?

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2
  • 2
    static uint16_t index = 0; and for (i=0; i < index; i++) are you sure that's correct? Commented Aug 16, 2016 at 11:03
  • THanks for catching the error. index should be length of the buffer. Commented Aug 16, 2016 at 11:06

3 Answers 3

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Simply convert each digits.

static char digits[] = "0123456789abcdef"; // characters used to represent digits

static uint16_t index = 0;
static char buffer[1600]; //contains binary data 0x11, 0x22, 0x1, 0xa ...

static char msg[3201]; // 2 digits for each bytes and 1 terminating null-character

for (i=0; i < index; i++)
{
    //Produce a zero in front for single digits. Examples, 0x5 transforms into 05, 0xa transforms into 0a
    msg[i * 2] = digits[((unsigned char)buffer[i] >> 4) & 0xf];

    msg[i * 2 + 1] = digits[(unsigned char)buffer[i] & 0xf]; //msg contains the ASCII equivalent of buffer
}

msg[index * 2] = '\0'; // terminate the string

Arduino may not be capable to store 4KB data (SRAM on ATmega328P is only 2KB), so reduce the buffer size if they are too much.

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Comments

1

Pure C solultion:

#include <stdio.h>
#include <stdlib.h>

int main()
{
const unsigned char bytes[] = {0x04, 0x55, 0x56, 0xce , 0xdf };
int i;
int sz = sizeof(bytes);
char *result = (char*)malloc(sz*4+1);
char *current = result;

for (i = 0; i < sz; i++)
{
    sprintf(current,"%02x",bytes[i]);
    current += 2;
}
printf("Result : %s\n",result);
free(result);
}

result:

045556cedf

You can also change "%02x" format by "%02X" to get uppercase hex-digits.

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1

You can use something like this:

char * append_hex(char *out, uint8_t value)
{
  static const char digits[] = "0123456789abcdef";
  *out++ = digits[value >> 4];
  *out++ = digits[value & 0xf];
  return out;
}

Then just call that in a loop, passing the returned value from it on each successive call. You can add separators between the calls, if desired.

Remember to 0-terminate the string when done.

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