bit shifting to pack float to 13 bit float in c?

Question

For example lets say 18.xxx is read in as the input of a function as a float value. It will be truncated down to 18.0. From then, encoded it to: 0 10011 0010000 which satisfies the 13 bit float desired, and will be returned as an int with decimal value 2448. Anyone know how this can be accomplished using shifts?

Answer 1

This might do what you want if your floating point number is represented in 32-bit IEEE 754 single-precision binary format with an unsigned exponent:

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

unsigned short float32to13(float f) {

    assert(sizeof(float) == sizeof(unsigned int));

    unsigned int g;
    memcpy(&g, &f, sizeof(float)); // allow us to examine a float bit by bit

    unsigned int sign32 = (g >> 0x1f) & 0x1; // one bit sign
    unsigned int exponent32 = ((g >> 0x17) & 0xff) - 0x7f; // unbias 8 bits of exponent
    unsigned int fraction32 = g & 0x7fffff; // 23 bits of significand 

    assert(((exponent32 + 0xf) & ~ 0x1f) == 0); // don't overflow smaller exponent

    unsigned short sign13 = sign32;
    unsigned short exponent13 = exponent32 + 0xf; // rebias exponent by smaller amount
    unsigned short fraction13 = fraction32 >> 0x10; // drop lower 16 bits of significand precision

    return sign13 << 0xc | exponent13 << 0x7 | fraction13; // assemble a float13
}

int main() {

    float f = 18.0;

    printf("%u\n", float32to13(f));

    return 0;
}

OUTPUT

> ./a.out
2448
>

I leave any endian issues and additional error checking to the end user. This example is provided only to demonstrate to the OP the types of shifts necessary to convert between floating point formats. Any resemblance to actual floating point formats is purely coincidental.