PHP not rounding floats properly?

Question

I am trying to round the double 0.0045. When I call...

echo round($theFloat, 3);

I get 0.004, instead of 0.005 which is the expected response.

Here is the code:

$increase = 1.1; $previousPrice = round(0.11 / $increase, 2); $nextPrice = round(0.11 * $increase, 2); $afterCut = round(0.11 * 0.95, 6); $willSend = $afterCut - $previousPrice; echo round($willSend, 3);

Can you echo $willSend to verify what is getting passed to round()? — dethtron5000
– dethtron5000, Commented Jan 29, 2013 at 4:36
I know it looks fine, but I can't figure out what's wrong with it and why it rounds it to 0.004! — apscience
– apscience, Commented Jan 29, 2013 at 4:38

Ignacio Vazquez-Abrams · Accepted Answer · 2013-01-29 04:46:16Z

Although what you have looks like 0.0045, the value is actually approximately 0.00449999999999999966, since IEEE 754 floating point values cannot specify exactly that number. And that value, when rounded to 3 places, is 0.004.

Sujit Singh · Accepted Answer · 2013-01-29 04:57:04Z

1

Try like this

$increase = 1.1; $previousPrice = 0.11 / $increase; $nextPrice = 0.11 * $increase; $afterCut = round(0.11 * 0.95, 6); $willSend = $afterCut - $previousPrice; echo round($willSend, 3);

edited Jan 29, 2013 at 4:57

Sujit Singh

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answered Jan 29, 2013 at 4:42

GautamD31

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2 Comments

GautamD31 Over a year ago

But i dont knw what is the value of $r[1] which you have used

GautamD31 Over a year ago

Already gladoscc says $r[1] is 0.11 so there is no need to put static in place of $r[1]....so sujit singh no need to edit my ans right..??

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