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Home/ Questions/Q 1086109
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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T22:46:21+00:00 2026-05-16T22:46:21+00:00

$a = ’35’; $b = ‘-34.99’; echo ($a + $b); Results in 0.009999999999998 What

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$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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  1. Editorial Team
    Editorial Team
    2026-05-16T22:46:21+00:00Added an answer on May 16, 2026 at 10:46 pm

    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.

    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:

    1. Use round($result, 2) on the result to round it to 2 decimal places.

    2. 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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