CODEWITHSUNDEEP
c#
Richard
Richard

Reputation: 167

mod operator behavior

When I run this code:

double d = 0.0;
for (int i = 0; i < 90; i++)
{
    d += .01;
    d %= 1;
    Console.WriteLine(d);
}

I would expect the output to be

0.01
0.02
0.03
...
0.9

It acts like that until it gets to what should be 0.81. This is the output I see:

...
0.8
0.810000000000001
0.820000000000001
...
0.900000000000001

So what's going on here?

Upvotes: 0

Views: 183

Answers (3)

burning_LEGION
burning_LEGION

Reputation: 13450

Use decimal:

decimal d = 0.0m;
for (int i = 0; i < 90; i++)
{
    d += .01m;
    d %= 1;
    Console.WriteLine(d);
}

Upvotes: 0

phoog
phoog

Reputation: 43046

0.01 is a decimal fraction: 1/100. Binary floating-point numbers can only represent fractions with a finite number of digits if the denominator is a power of two. One hundred, of course, is not a power of two. (Decimal fractions can be finite if the denominator's only prime factors are 2 and 5, which are of course the prime factors of 10.)

Because the binary representation of 0.01 is infinite, the finite number that your program adds repeatedly is not exactly equal to 0.01; after a time, the inaccuraccy accumulates to the point that you see it in the formatted output.

See http://en.wikipedia.org/wiki/Double_precision_floating-point_format for more information.

The best way to do this particular task would be to change the repeated addition into integer addition. For example, you could change d += 0.01; into d = (double)(i + 1) / 100;.

Upvotes: 5

Shortys Oberto Dutari
Shortys Oberto Dutari

Reputation: 556

try this

double d = 0.0;
for (int i = 0; i < 90; i++)
    {
     d += .01;
     d %= 1;
     Console.WriteLine(Math.Round(d, 2));
    }

I think you do not get the expected result due to rounding

Upvotes: 1

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