how to convert array of integers into single float variable?

Question

I have uint8_t size 4 integers stored in a array: uint8_t sdo_rx_data_buffer[OD_entry_len];. The length of this array is 4 bytes (OD_entry_len = 4), so is equal to float size (32 bits). I need to convert this array into float variable. For exampe, I have these values: int array[4] = {0xd0, 0x69, 0x83, 0x3f}; And i should get float i_motor = 1.02666664; Any ideas how to convert it?

Answer 1

Assuming that you know that the binary representation gives a valid floating point number on your system and you got the endianess right, then you can use a union:

#include <stdint.h>
#include <stdio.h>

typedef union
{
  uint8_t raw [sizeof(float)];
  float   f;
} float_converter;

int main (void)
{
  float_converter fc = { .raw = {0xd0, 0x69, 0x83, 0x3f} };
  printf("%f\n", fc.f);
}

Output on x86_64 Linux:

1.026667

Answer 2

One incorrect way which may seem to work would be to reinterpret the array using a pointer.

uint8_t array[4] =  {0xd0, 0x69, 0x83, 0x3f};
float *p = (float*)array;

printf( "%f\n", *p );

However, this code has undefined behavior, because it violates the strict aliasing rule. It may also have alignment issues.

On the compilers gcc and clang, you can use __attribute__((__may_alias__)) on the pointer p, so that there is no strict aliasing violation:

uint8_t array[4] =  {0xd0, 0x69, 0x83, 0x3f};
float __attribute__((__may_alias__)) *p = (float*)array;

printf( "%f\n", *p );

However, there still may be alignment issues.

A different way, which fully complies with the ISO C standard (and therefore should work on all compilers), would be to use memcpy instead:

uint8_t array[4] =  {0xd0, 0x69, 0x83, 0x3f};
float f;

memcpy( &f, array, sizeof f );

printf( "%f\n", f );

Most compilers will optimize away the memcpy when compiler optimizations are active.

Beware of endianness issues, though. The posted code will only work on little-endian platforms.