Convert float to uint8 buffer

Question

A have a float variable, and I need it's value to be stored in a uint8 buffer as characters. To be more exact, given something like this

float f = 123.45
uint8_t buffer[11];
memset(buffer, 0x30, sizeof(buffer));   // I set it at 0x30 because it is character '0' 

Given this example, my buffer needs to get the values:

0x30 0x30 0x30 0x30 0x30 0x31 0x32 0x33 0x2E 0x34 0x35   // 0x2E is the character '.'

I unfortunately need it this way to integrate it with a existing functionality, so there is no way around the size 11 buffer.

Any suggestion on how to go about this would be appreciated.

Answer 1

An easy way is:

1. Allocate buffer with one more element (buffer_buffer here)
2. Convert the float to string via snprintf()
3. Copy the conversion result to buffer

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

int main(void) {
    float f = 123.45;
    uint8_t buffer[11];
    char buffer_buffer[12];
    snprintf(buffer_buffer, sizeof(buffer_buffer), "%011.2f", f);
    memcpy(buffer, buffer_buffer, sizeof(buffer));
    for (int i = 0; i < 11; i++) printf(" 0x%02X", buffer[i]);
    putchar('\n');
    return 0;
}

Directly using snprintf() to buffer won't work well because there are no room for terminating null-character there.