Why is $a + ++$a == 2?

Question

If I try this:

$a = 0;     echo $a + ++$a, PHP_EOL; echo $a; 

I get this output:

2 1 

Demo: http://codepad.org/ncVuJtJu

Why is that?

I expect to get this as an output:

1 1 

My understanding:

$a = 0;                    // a === 0     echo $a + ++$a, PHP_EOL;   // (0) + (0+1) === 1 echo $a;                   // a === 1 

But why isn't that the output?

Answer 1

All the answers explaining why you get 2 and not 1 are actually wrong. According to the PHP documentation, mixing + and ++ in this manner is undefined behavior, so you could get either 1 or 2. Switching to a different version of PHP may change the result you get, and it would be just as valid.

See example 1, which says:

// mixing ++ and + produces undefined behavior $a = 1; echo ++$a + $a++; // may print 4 or 5 

Notes:

1. Operator precedence does not determine the order of evaluation. Operator precedence only determines that the expression $l + ++$l is parsed as $l + (++$l), but doesn't determine if the left or right operand of the + operator is evaluated first. If the left operand is evaluated first, the result would be 0+1, and if the right operand is evaluated first, the result would be 1+1.

2. Operator associativity also does not determine order of evaluation. That the + operator has left associativity only determines that $a+$b+$c is evaluated as ($a+$b)+$c. It does not determine in what order a single operator's operands are evaluated.

Also relevant: On this bug report regarding another expression with undefined results, a PHP developer says: "We make no guarantee about the order of evaluation [...], just as C doesn't. Can you point to any place on the documentation where it's stated that the first operand is evaluated first?"

Answer 2

A preincrement operator "++" takes place before the rest of the expression it's in evaluates. So it is actually:

echo $l + ++$l; // (1) + (0+1) === 2