if I try to print
echo ((0.1 + 0.7) * 10);
output (http://codepad.org/m3mNNO77)
8
but if I try to print
echo (int)((0.1 + 0.7) * 10);
output (http://codepad.org/8OTCnlVG)
7
why two different results ?
If you perform the echo without the (int) cast, it resulits in:
echo ((0.1 + 0.7) * 10);
8
So what happens is that the floating point actually represents it as 7.99999....
When you perform an (int) cast, the default behavior is to take the integral part, so that's 7. Although it is extremely close to 8, it is not the behavior of a cast to round off.
The errors are due to the fact that floating point numbers are represented with a binary radix. Representing 0.8 correctly is thus impossible. Floating points round off the answers, and hope it doesn't bother that much. When you represent the floating point like 0.8, the result is 0.80000...
Proof of concept using the PHP interactive shell (php -a):
$ php -a
Interactive mode enabled
php > echo number_format(((0.1 + 0.7) * 10),20)."\n";
7.99999999999999911182
php > echo number_format(0.8,20);
0.80000000000000004441
As you may be aware, floating point math is not exact due to restrictions in their representation.
(0.1 + 0.7) * 10 works out to something like 7.99999999999....
When echo'd out, the number is converted to a string using rounding, which uses the php.ini precision setting (15 by default, I believe) to limit the number of decimal numbers. It happens that this rounding gives exactly 8.
However, (int) casts the float to an integer instead, and this is done with truncation. It doesn't matter how close to the next integer up you are, it will always round down. This is why you get 7 here.
Related
This question already has answers here:
Why does `intval(19.9 * 100)` equal `1989`?
(5 answers)
Closed 4 years ago.
i am doing some calculation where i noticed intval weird behavior
which is making a whole lot of mess because i just want int part of whole variable but intval is stabbing me in the back in comparisons. Any ideas why its behaving like that and what i can use for my requirement?
intval( 9.62 * 100 ) //gives 961
(int)( 9.62 * 100 ) // gives 962.0
my tried methods to achieve my goals were:
floor(9.62 * 100) // but its giving 962.0 no acceptable
This should work for you.
intval(strval( 9.62 * 100));
It seems to be a problem with the precision of floating points, and converting the number to string and then to int seems to fix the problem.
Here is a link to the documentation, http://php.net/manual/en/language.types.float.php
As it states,
Additionally, rational numbers that are exactly representable as floating point numbers in base 10, like 0.1 or 0.7, do not have an exact representation as floating point numbers in base 2, which is used internally, no matter the size of the mantissa. Hence, they 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 7.9999999999999991118....
echo intval((0.1 + 0.7) * 10); //returns 7
echo intval(strval(0.1 + 0.7) * 10); //returns 8
Since (int) always rounds the number down, a small error in the representation makes the cast round it one number down that you would otherwise expect.
I tried sprintf('%.40F', 9.62 * 100.0);, and get this: "961.9999999999998863131622783839702606201172".
Thats why intval( 9.62 * 100 ) gives you 961.
You can try to use bcmul()
(int) bcmul(9.62, 100);
Please try this
ceil(9.62*100);
This question already has answers here:
Is floating point math broken?
(31 answers)
Closed 6 years ago.
I found somewhere in book
echo (int) ((0.1 + 0.7) * 10);
output is : 7
echo ((0.1 + 0.7) * 10);
output : 8
why both out are different ? I think answer should be 8
When you write
echo ((0.1 + 0.7) * 10);
the result of this simple arithmetic
expression is stored internally as 7.999999 instead of 8.
Now when the value is converted to int,
echo (int) ((0.1 + 0.7) * 10); // 7.999999 when typecasted to int becomes 7
PHP simply truncates away the fractional part, resulting in a rather
significant error (12.5%, to be exact).
It's because float point at this scenario does not fit to memory and is truncated when converting to integer.
Read about that in PHP manual about float
Warning
Floating point precision
Floating point numbers have limited precision. Although it depends on the system, PHP typically uses the IEEE 754 double precision format, which will give a maximum relative error due to rounding in the order of 1.11e-16. Non elementary arithmetic operations may give larger errors, and, of course, error propagation must be considered when several operations are compounded.
Additionally, rational numbers that are exactly representable as floating point numbers in base 10, like 0.1 or 0.7, do not have an exact representation as floating point numbers in base 2, which is used internally, no matter the size of the mantissa. Hence, they 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 7.9999999999999991118....
So never trust floating number results to the last digit, and do not compare floating point numbers directly for equality. If higher precision is necessary, the arbitrary precision math functions and gmp functions are available.
Please check http://php.net/manual/en/language.types.integer.php and find this
Warning : Never cast an unknown fraction to integer, as this can sometimes lead to unexpected results.
<?php
echo (int) ( (0.1+0.7) * 10 ); // echoes 7!
?>
See also the warning about float precision.
CLARIFYING: This isn't asking why I'm getting rounding errors. I understand this is a mistake or an oversight. The question asks why it prints as whole in the first var_dump, but casting acts as if it were 57916.9repeating and truncates said .9repeating.
The following occurs:
You take a string (or float -- does not matter) that contains the value 579.17 and multiply it 100. It var_dumps the expected 57917. Not 57916.99999999999999999999999 or similar. var_dump should not be rounding anything as a debugging function in my opinion. It may have to truncate, but rounding is unexpected in a debugging function.
However, if one then casts that to an integer, you get an unexpected 57916 from var_dump.
I'm aware of issues with floating point numbers, but the act of casting a floating point number that prints as exactly 57917 in PHP apparently effectively subtracts 1. This is a very small number.
This only appears to happen for some numbers, such as 579.17. It does not occur for others I've tested. All we're doing is multiplying a number by 100 to send to an API that expects cents. The API library understandably casts to integer since the API doesn't accept fractional cents.
Test case:
php -r '$n = ("579.17" * 100); var_dump($n, (int)$n);'
Output:
float(57917)
int(57916)
Environment:
x86-32,
x86-64 both.
var_dump uses precision from php.ini to display float value. You could raise it to see what happens.
php -r 'ini_set("precision", 20); $n = ("579.17" * 100); var_dump($n, (int)$n);'
// double(57916.999999999992724)
// int(57916)
Also. There is no matter x86 or x64. PHP uses 64 bits for floats.
http://php.net/manual/en/language.types.float.php
Use round() instead of int(). The actual value of 579.17 * 100 is something like 57916.99999. var_dump() shows this as 57917, but when you use int() it truncates the fraction. Using round() will go to the nearest integer, rather than always truncating down.
I believe this is because hardware cannot truly and accurately express floating point numbers. So what appears as 579.17 is actually more like 579.16999999. So when you multiply it and cast it as an int it truncates the decimal leaving you with 57916.
How come the result for
intval("19.90"*100)
is
1989
and not 1990 as one would expect (PHP 5.2.14)?
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.
You can also use bc* function for working with float :
$var = bcmul("19.90", "100");
echo intval($var);
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
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
I believe the php doc at http://de2.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.
I recently start to learn PHP.
<?php
echo (int) ( (0.1+0.7) * 10 ); // prints '7', why not '8' ?
?>
Please convince me of this type-conversion process.
From PHP.net
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.
You are running into floating point inaccuracy.
0.1 + 0.7 is not exactly 0.8, but slightly less. The cast to int simply truncates the value and yields 7 as the result.
To get the correct result use rounding:
<?php
echo (int) round( (0.1+0.7) * 10 );
?>
From http://php.net/manual/en/language.types.float.php
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.
As others have said, it's due to a loss of precision, as revealed by:
<?php
printf("%30.27f\n", (0.1 + 0.7) );
printf("%d\n", (int) (0.1 + 0.7) );
printf("%30.27f\n", ( (0.1+0.7) * 10 ) );
printf("%d\n", (int) ((0.1 + 0.7) * 10) );
?>
which outputs:
0.799999999999999933386618522
0
7.999999999999999111821580300
7
You're probably seeing a floating point rounding error, and when you int() it, it goes from being 7...9 to 7.
I am not sure, but this is probably because the int cast is truncating.
.1 + .7 might result in .799999999999999999999999 and
multiplying this by 10 = 7.9999999999999999999
(int) truncates this to 7.