dcmoody - 1 year ago 101

PHP Question

`$a = '35';`

$b = '-34.99';

echo ($a + $b);

Results in 0.009999999999998

What is up with that? I wondered why my program kept reporting odd results.

Why doesn't PHP return the expected 0.01?

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Answer Source

Because floating point arithmetic != real number arithmetic. An illustration of the difference due to imprecision is, for some floats `a`

and `b`

, `(a+b)-b != a`

. This applies to any language using floats. See an example in Python.

Since floating point are binary numbers with finite precision, there's a finite amount of representable numbers, which leads accuracy problems and surprises like this. Here's another interesting read: What Every Computer Scientist Should Know About Floating-Point Arithmetic.

Back to your problem, basically there is no way to accurately represent 34.99 or 0.01 in binary (just like in decimal, 1/3 = 0.3333...), so approximations are used instead. To get around the problem, you can:

Use

`round($result, 2)`

on the result to round it to 2 decimal places.Use integers. If that's currency, say US dollars, then store $35.00 as 3500 and $34.99 as 3499, then divide the result by 100.

It's a pity that PHP doesn't have a decimal datatype like other languages do.

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