Floating point does not print accurately

Question

Possible Duplicate:
Floating point issue in C

#include<stdio.h>
main()
{
        int a,b;
        float f;
        scanf("%2d%3d%4f",&a,&b,&f);
        printf("%d %d %f",a,b,f);
}

When i run this program and input 2 4 56.8 ,it gives output 2 4 56.799999.....but I would expect 2 4 56.8....why is it so???

@FlorianSowade, yes but float arguments get promoted to double for functions without prototypes and variadic arguments. — AProgrammer, Jun 12 '12 at 11:07

score 6 · Accepted Answer · answered Jun 12 '12 at 06:32

That is correct. Floating point numbers are approximations. Just as 0.33333333 is an approximation to 1/3, 56.7999999 is an approximation for 56.8. There is no exact floating point representation for 0.1.

See some of what has been written:

score 0 · Answer 2 · answered Jun 12 '12 at 06:33

0

56.8 is not exactly-representable in the floating-point format used by your C compiler. See What Every Computer Scientist Should Know About Floating Point Arithmetic (1991) by David Goldberg.

answered Jun 12 '12 at 06:33

mlp

809
7
21

score 0 · Answer 3 · edited Jun 12 '12 at 13:14

0

It might have to do with floating point inaccuracy.

edited Jun 12 '12 at 13:14

Peter DeWeese

18,141
8
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answered Jun 12 '12 at 06:33

cody

61
1
4

score 0 · Answer 4 · answered Jun 12 '12 at 07:08

All floating point numbers cannot be stored as its exact value, as the continuous space of the floating point number domain is mapped into a discrete levels. For example the floating point number can be stored in the IEEE754 single precision format. It stores the floating number in three parts in 32 bits of memory. Assume that [a, b] are the two closest floating point numbers that can be represented exactly with IEEE754 format, but we will have infinite number of floating point numbers in between a and b, in that case these numbers will be approximated with either a or b. This induces the imprecision you are experiencing.

The number is stored in normalized format, although some very small number can be stored in unnormalized format. Just have a look at the documents.

For example in your case 56.8 IEEE754 single point precision representation will be as follows:

sign: 0 (1 bit)
biased exponent: 10000100 (8 bits)
mantissa: 11000110011001100110011 (23 bits)

Now, if you convert back this number into decimal then you get

sign is 0 ie the value is positive
10000100 = 132 bias amount 127 therefore exponent = 132 - 127 = 5
mantissa with hidden bit 1.11000110011001100110011
after adjusting decimal point with exponent (move 5 places right): 111000.110011001100110011
- The integer part : 111000 = 2^5 + 2^4 + 2^3 = 56
- The fraction part : .110011001100110011 = 2^-1 + 2^-2 + 2^-5 + 2^-6 + 2^-9 + 2^-10 + 2^-13 + 2^-14 + 2^-17 + 2^-18 = 0.799999237

Therefore when you retrieve the value from the IEEE754 single precision format it becomes: 56.799999237 where you have stored 56.8

You can play with IEEE754 representation here: http://www.h-schmidt.net/FloatConverter/IEEE754.html

Floating point does not print accurately

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