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Full Discussion: Floating point error in C
Top Forums Programming Floating point error in C Post 302100690 by Perderabo on Thursday 21st of December 2006 01:48:14 PM
Old 12-21-2006
Floating point numbers have a complex representation using base 2 rather than base 10. Scientific notation is close enough to understand what is happening. With scientific notation, we want a number between 1 and 10 multiplied by a power of 10: 8080.9940 = 8.0809940 * 10^3 and we can check that out with bc:
Code:
$ bc
bc 1.03 (Nov 2, 1994)
Copyright (C) 1991, 1992, 1993, 1994 Free Software Foundation, Inc.
This is free software with ABSOLUTELY NO WARRANTY.
For details type `warranty'.
8.0809940 * 10^3
8080.9940000
^D$

But now we need to do that in binary... We want a number between 1 and 10 in binary (which means 1 and 2 in decimal) that can be multiplied by a power of 10 in binary to yield the decimal number 8080.9940. Fortunately, you picked an easy number! It's is obvious that dividing it by 4096 will yield a number between 1 and 2. So let's use bc to do this...
Code:
$ bc -l
bc 1.03 (Nov 2, 1994)
Copyright (C) 1991, 1992, 1993, 1994 Free Software Foundation, Inc.
This is free software with ABSOLUTELY NO WARRANTY.
For details type `warranty'.
obase=2
scale=40
8080.9940/4096
1.111110010000111111100111011011001000101101000011100101011000000100\
0001100010010011011101001011110001101010011111101111100111011011001
l(4096)/l(2)
1100.000000000000000000000000000000000000000000000000000000000000000\
00000000000000000000000000000000000000000000000000000000000000000000\
00
obase=10
ibase=2
1.1111100100001111111001110110110010001011010000 * 10^1100
8080.9939999999478459358215332031250000000000000000
1.11111001000011111110011101101100100 * 10^1100
8080.99399995803833007812500000000000000

Now repeat the last operation, but replace 1.11111001000011111110011101101100100 with some other close-by binary number. No matter what you try, you cannot get exactly 8080.9940 Floating point numbers are bit more complex than this, but the exact same issue arises with them. There is no way to exactly represent 8080.9940 as a floating point number.
 

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