Representation of float in C

Question

I was trying to understand the floating point representation in C using this code (both float and int are 4 bytes on my machine):

int x = 3;
float y = *(float*) &x;
printf("%d %e \n", x, y);

We know that the binary representation of x will be the following

00000000000000000000000000000011

Therefore I would have expected y to be represented as follows

• Sign bit (first bit from left) = 0

• Exponent (bits 2-9 from left) = 0

• Mantissa (bits 10-32): 1 + 2^(-22)+2^(-23)

Leading to y = (-1)^0 * 2^(0-127) * (1+2^(-22) + 2^(-23)) = 5.87747E-39

My program however prints out

3 4.203895e-45

That is, y has the value 4.203895e-45 instead of 5.87747E-39 as I expected. Why does this happen. What am I doing wrong?

P.S. I have also printed the values directly from gdb so it is not a problem with the printf command.

Answer 1

IEEE floating point numbers with exponent fields of all 0 are 'denormalized'. This means that the implicit 1 in front of the mantissa no longer is active. This allows really small numbers to be represented. See This wikipedia article for more explanation. In your example the result would be 3 * 2^-149

Answer 2

-127 in the exponent is reserved for denormalised numbers. Your calculation is for normalized numbers while your float is a denormalised float.

Denormalised numbers are calculated using a similar method, but:

1. exponent is -126
2. implicit leading bit is no longer assumed

So this means the calculation is instead:

(-1)**0*2**(-126)*(2**(-22)+2**(-23)) = 4.2038953929744512e-45

The above is python, where ** means the same as ^

Answer 3

In details it is described http://en.wikipedia.org/wiki/IEEE_754-2008 This standard assumed that you shifting left mantissa until hiding first meaning bit (increasing exponent). In your case yo have expression 1+2^(-23) - then you get correct answer 4.9..E-32