PHP equation output confusion

Question

$x = 5; echo "$x"; echo "<br>"; echo $x+++$x++; echo "<br>"; echo $x;

Wouldn't the output of the code above be "5,12,5"? PHP outputs "5,11,7"?

Why? I am confuse.

But the original value of $x is 5. So isn't it always going to be 5 when $x is called upon? — Iamwhoiam, Oct 19 '15 at 18:53

MohitC · Answer 1 · 2015-10-19T19:08:29.883

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The first reference to $x is when its value is still 5 (i.e., before it is incremented) and the second reference to $x is then when its value is 6 (i.e., before it is again incremented), so the operation is 5 + 6 which yields 11. After this operation, the value of $x is 7 since it has been incremented twice.

So it's basically

Supply 5 as one of the operands to an addition operation.
Increment x after supplying it to addition operation (post increment), which makes it 6.
Supply this previous incremented x as second operand to the addition operation, which is 6. So it makes it 5+6, yields 11
Lastly increment x after addition operation making it 7

edited Oct 19 '15 at 19:08

answered Oct 19 '15 at 18:57

MohitC

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Sorry, wrote in hurry. Corrected – MohitC Oct 19 '15 at 18:58
Actually it would be `$x(5)++(6)+$x(6+5 = 11)++ (7)` to make sense, considering operation precedence. – al'ein Oct 19 '15 at 18:59
You can prove it trying backwards: `echo $x(5)++(6) + ++(7)(5+7=12)$x(7);` – al'ein Oct 19 '15 at 19:02
Read now @Alan machado – MohitC Oct 19 '15 at 19:14

score 1 · Answer 2 · edited May 23 '17 at 10:28

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See notes in comments

$x = 5; 
echo $x . PHP_EOL;
// now x=5

echo ++$x + $x . PHP_EOL;
// left side of the operation makes x=6 then the right side adds x to it, meaning 6+6

echo ++$x . PHP_EOL;
// now x=7

Also check this What's the difference between ++$i and $i++ in PHP?
to get some idea about pre vs. post incrementation.

edited May 23 '17 at 10:28

Community

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answered Oct 19 '15 at 19:01

Alex Andrei

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It's not ++$x+$x. His is $x+++$x++ which prints 11 – MohitC Oct 19 '15 at 19:04
you are right, in OP's form the result is 11. As I wrote it, gives 12. It's just an example on how to use incrementation. – Alex Andrei Oct 19 '15 at 19:08
He's explaining the logic of precedence. – al'ein Oct 19 '15 at 19:08
But it still doesn't solve op's doubt about $x+++$x++ – MohitC Oct 19 '15 at 19:13

score 0 · Answer 3 · answered Oct 19 '15 at 19:01

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$x ++ means the current value of x is 5, and immediately after that, x will be 6.

for the second x, x = 6, then immediately after that, x = 7.

answered Oct 19 '15 at 19:01

Shiji.J

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PHP equation output confusion

3 Answers3