Understanding PHP & (ampersand, bitwise and) operator
I often use ($var & 1)
in my code, which returns true if $var
is an odd number and false if it's an even number.
But what does "&" actually do?
& is binary and
. If you have a binary value, and you and
with another binary value, then the result will be the bitwise and
of the two. An example:
01101010
& 01011001
= 01001000
The rightmost bit is either a 1 (and in that case the number is an odd number) or it is a 0, in which case the number is even. If you &
a number with 1, you only look at the least significant bit, and the if checks if the number is a 1 or a 0. As others have mentioned, look at the bitwise operators for info on how they work.
Two operations which are fundamental to binary systems are OR and AND.
OR means 'if either A is on or B is on'. A real world example would be two switches in parallel. If either is allowing current through, then current passes through.
AND means 'if both A and B is on'. The real world example is two switches in series. Current will only pass through if both are allowing current through.
In a computer, these aren't physical switches but semiconductors, and their functionality are called logic gates. They do the same sorts of things as the switches - react to current or no current.
When applied to integers, every bit in one number is combined with every bit in the other number. So to understand the bitwise operators OR and AND, you need to convert the numbers to binary, then do the OR or AND operation on every pair of matching bits.
That is why:
00011011 (odd number)
AND
00000001 (& 1)
==
00000001 (results in 1)
Whereas
00011010 (even number)
AND
00000001 (& 1)
==
00000000 (results in 0)
The (& 1) operation therefore compares the right-most bit to 1 using AND logic. All the other bits are effectively ignored because anything AND nothing is nothing. An even number in binary is also an even number in decimal notation (10 is a multiple of 2).
Other fundamental operations to binary systems include NOT and XOR. NOT means 'if A is off' and is the only form of logic gate that takes only one signal or 'parameter' instead of two. XOR means 'if either A or B is on, but not both'. And then there are NAND, NOR, and NXOR, which are basically just NOT combined with AND, OR, and XOR, ie NAND means 'if A and B are not both on'.
In programming, the operator
& means AND,
| means OR,
~ means NOT, and
^ means XOR.
The others can be made up by combining these, for example:
~ (a & b) is equivalent to a NAND operation
PHP specific note
Bitwise operators do not work on floating-point values, and in PHP float values will be implicitly converted to integers first. Numbers outside the range that can be expressed as integers will be truncated to zero - that is, all numbers over PHP_INT_MAX will look "even" in the expression ($num & 1)
). If you want to support numbers outside of PHP_INT_MIN/PHP_INT_MAX, you'll need fmod($num, 2)
. If, however, you're on 64-bit PHP your integers will have greater precision than floats anyway.
This is also interesting to know about bitwise and PHP:
/**
* Regular
*/
echo (true && true); // 1
echo (true && false); // nothing
echo (true || false); // 1
echo (false || false); // nothing
echo (true xor false); // 1
echo (false xor false); // nothing
/**
* Bitwise
*/
echo (true & true); // 1
echo (true & false); // 0
echo (true | false); // 1
echo (false | false); // 0
echo (true ^ false); // 1
echo (false ^ false); // 0
I know your question is about understanding the bitwise operator and the accepted answer explains it well. But for the example you give, I cannot help but recommending you use the modulo operator instead:
($var % 2) /* instead of */ ($var & 1)
Because it makes the intent clear that you are checking that the number is odd (not divisible by two), and it is more generic, so you can use ($var % 3) the same way and deduce how it works for any N.
In addition to the other answers, it's worth noting that
if(func1() && func2())
Will only call func2()
if func1()
returns true ("lazy evaluation"), whereas
if(func1() & func2())
Will call both functions regardless, but the truth tables for both will be the same (assuming they return booleans).
thomasrutter points out (in the comments below) that you probably shouldn't do the latter in practice. (A & B)
won't necessarily have the same truthiness as (A && B)
, particularly when A
and B
are integers. e.g., if A=1 and B=2 (both truthy), A & B will be falsey, whereas A && B is truthy. Also, another developer may think this is a typo and 'correct' it to two ampersands.