Rounding in PHP

Question

$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.

Answer 1

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.

Answer 2

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

Answer 3

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);

Answer 4

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

Answer 5

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.

Answer 6

因为在浮点数上运算不是100％准确，所以在从表达式转换值之前可能是类似于7.9999 ......

Answer 7

你可以这样做比较：

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