Oh, and how do you imagine 1/3 being stored in your model? There's no pair of integers m and e where m * 10^e gives 1/3. Your model doesn't work for periodic numbers.

You could approximate it. You could use m = 3333333333 and e = -10, but it's not quite exact.

And guess what. That's exactly the problem faced here.

The number is stored as two integers. But it's not a power of 10; it's a power of 2.

So, in this case, you need a pair of integers m and e where m * 2^e = 895/100. There is no such pair of integers because 895/100 is periodic in binary.

So the computer used m = 0x11E66666666666 and e = -49, but it's not quite exact. 

$ perl -Mv5.14 -e'say 8.95 == 0x11E66666666666 * 2**(-49) ? 1 : 0'
1

$ perl -Mv5.14 -e'say sprintf "%.751g", 0x11E66666666666 * 2**(-49)'
8.949999999999999289457264239899814128875732421875

$ perl -Mv5.14 -e'say sprintf "%.751g", 8.95'
8.949999999999999289457264239899814128875732421875

[download]

If you want to be picky, an IEEE double is actually stored as follows to get a couple of extra bits:

( -1 )^s * ( 1 + ( m * 2^( -52 ) ) * 2^( e - 1023 ) )

In reply to Re^3: Behaviour of int() unexpected by ikegami
in thread Behaviour of int() unexpected by ceade1000

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