3
$a = ((0.1 + 0.7) * 10) == (int)((0.1 + 0.7) * 10); 

PHP returns false.

Could anybody explain me, why that happens? First returns 8, second 7.

2

7 Answers 7

14

Quoting the big fat red warning in the PHP Manual on Floating Point Precision:

It is typical that simple decimal fractions like 0.1 or 0.7 cannot be converted into their internal binary counterparts without a small loss of precision. This can lead to confusing results: for example, floor((0.1+0.7)*10) will usually return 7 instead of the expected 8, since the internal representation will be something like 7.9.

This is due to the fact that it is impossible to express some fractions in decimal notation with a finite number of digits. For instance, 1/3 in decimal form becomes 0.3.

So never trust floating number results to the last digit, and never compare floating point numbers for equality. If higher precision is necessary, the arbitrary precision math functions and gmp functions are available.

Sign up to request clarification or add additional context in comments.

1 Comment

In particular, only fractional numbers that can be represented in the form p/q, where q is an integer power of 2 and p is any integer, can be expressed exactly, with a finite number of bits.
5

Floating point arithmetic is not precise. Instead of 8.0 exactly you can get 7.999... which gets truncated to 7 when cast to an integer.

echo number_format((0.1 + 0.7) * 10, 20); 

Result:

7.99999999999999911182 

1 Comment

It doesn't help that var_dump((0.1 + 0.7) * 10); rounds to float(8) for output.
2

http://en.wikipedia.org/wiki/Floating_point#Accuracy_problems

EDIT

$a = ((0.1 + 0.7) * 10) == 8; var_dump($a); echo '<br />'; define('PRECISION', 1.0e-08); $a = (abs(((0.1 + 0.7) * 10) - 8) < PRECISION); var_dump($a); 

Comments

1

Mark's response hits the nail on the head, but I think instead of casting you want the round PHP function:

http://php.net/manual/en/function.round.php

Comments

1

From The Flaoting-Point Guide (click for detailed explanations):

Because internally, computers use a format (binary floating-point) that cannot accurately represent a number like 0.1, 0.2 or 0.3 at all.

When the code is compiled or interpreted, your “0.1” is already rounded to the nearest number in that format, which results in a small rounding error even before the calculation happens.

Comments

0

Because operating on floating point numbers isn't 100% accurate, propably before casting value from expression is something like 7.9999...

Comments

0

You can do the comparison this way:

$a = round(((0.1 + 0.7) * 10), 1) == (int)round(((0.1 + 0.7) * 10), 1); 

Comments

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.