Question

I am testing IEEE 754 floating format with VS2008 using the example below:

int main(int argc, char *argv[])
{
    float i = 0.15625;
}

I put &i to the VS2008 watch and I see the address is 0x0012FF60 and I can see address’s content is 00 00 20 3e from Memory debug window, see below:

0x0012FF60 00 00 20 3e cc cc cc cc

BTW I have the basic knowledge of IEEE754 floating format and I know IEEE 754 floating format consist of three fields: sign bit, exponent, and fraction. The fraction is the significand without its most significant bit.

But how did I calcuate exactly from little endian 00 00 20 3e to 0.15625 ?

Many thanks

Answer 1

You are printing out something broken. We only need 32 bits, which are:

00 00 20 3E

Your variable in binary:

00000000 00000000 00100000 00111110

Logical value accounting for little endian:

00111110 00100000 00000000 00000000

According to IEEE:

0 01111100 01000000000000000000000
S E - 127  M - 1

So now it’s clear:

• the sign is +1 (S = 0)
• the exponent is 124 – 127 = -3
• the mantissaa is 1.01b, which is 5/4

So the value is 5/4 / 8 = 5/32 = 0.15625.