Why does adding a small float to a large float just drop the small one?

Question

Say I have:

float a = 3            // (gdb) p/f a   = 3
float b = 299792458    // (gdb) p/f b   = 299792448

then

float sum          


        

Answer 1

Floating-point number have limited precision. If you're using a float, you're only using 32 bits. However some of those bits are reserved for defining the exponent, so that you really only have 23 bits to use. The number you give is too large for those 23 bits, so the last few digits are ignored.

To make this a little more intuitive, suppose all of the bits except 2 were reserved for the exponent. Then we can represent 0, 1, 2, and 3 without trouble, but then we have to increment the exponent. Now we need to represent 4 through 16 with only 2 bits. So the numbers that can be represented will be somewhat spread out: 4 and 5 won't both be there. So, 4+1 = 4.