CODEWITHSUNDEEP
phpfloating-point
Andris
Andris

Reputation: 27875

Strange number conversion error in PHP

How come the result for

intval("19.90"*100)

is

1989

and not 1990 as one would expect (PHP 5.2.14)?

Upvotes: 3

Views: 1670

Answers (5)

andig
andig

Reputation: 13868

I believe the php doc at https://www.php.net/manual/en/function.intval.php is omitting the fact that intval will not deliver "the integer value" but the integer (that is non-fractional) part of the number. It does not round.

Upvotes: 1

grunk
grunk

Reputation: 14938

You can also use bc* function for working with float : 

$var = bcmul("19.90", "100");
echo intval($var);

Upvotes: 3

Artefacto
Artefacto

Reputation: 97805

That's because 19.90 is not exactly representable in base 2 and the closest approximation is slightly lower than 19.90.

Namely, this closest approximation is exactly 2^-48 × 0x13E66666666666. You can see its exact value in decimal form here, if you're interested.

This rounding error is propagated when you multiply by 100. intval will force a cast of the float to an integer, and such casts always rounds towards 0, which is why you see 1989. Use round instead.

Upvotes: 13

Jordan
Jordan

Reputation: 5058

Why are you using intval on a floating point number? I agree with you that the output is a little off but it has to do with the relative inprecision of floating point numbers.

Why not just use floatval("19.90"*100) which outputs 1990

Upvotes: 1

ayush
ayush

Reputation: 14568

intval converts doubles to integers by truncating the fractional component of the number. When dealing with some values, this can give odd results. Consider the following:

print intval ((0.1 + 0.7) * 10);

This will most likely print out 7, instead of the expected value of 8.

For more information, see the section on floating point numbers in the PHP manual

Upvotes: 2

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