How to manually (bitwise) perform (float)x?

Question

Now, here is the function header of the function I\'m supposed to implement:

/*
 * float_from_int - Return bit-level equivalent of expression (float) x
 *            


        

Answer 1

The problem is that the lowest int is -2147483648, but the highest is 2147483647, so there is no absolute value of -2147483648. While you could work around it, I would just make a special case for that one bit pattern (like you do for 0):

if (x == 0)
    return 0;
if (x == -2147483648)
    return 0xcf000000;

The other problem is that you copied an algorithm that only works for numbers from 0 to 32767. Further down in the article they explain how to expand it to all ints, but it uses operations that you're likely not allowed to use.

I would recommend writing it from scratch based on the algorithm mentioned in your edit. Here's a version in C# that rounds towards 0:

uint float_from_int(int x)
{
    if (x == 0)
        return 0; // 0 is a special case because it has no 1 bits

    // Save the sign bit of the input and take the absolute value of the input.
    uint signBit = 0;
    uint absX = (uint)x;
    if (x < 0)
    {
        signBit = 0x80000000u;
        absX = (uint)-x;
    }

    // Shift the input left until the high order bit is set to form the mantissa.
    // Form the floating exponent by subtracting the number of shifts from 158.
    uint exponent = 158;
    while ((absX & 0x80000000) == 0)
    {
        exponent--;
        absX <<= 1;
    }

    // compute mantissa
    uint mantissa = absX >> 8;

    // Assemble the float from the sign, mantissa, and exponent.
    return signBit | (exponent << 23) | (mantissa & 0x7fffff);
}