CODEWITHSUNDEEP
php
user9715097
user9715097

Reputation:

how it is executing for $a++ + $a++ + $a vs $a + $a++ + $a++

I've seen a question on SO like this and confused :( , as I know $a++ will increment value that will available when I'll print $a, itll print 2 . but if I print $a++ it is 1 not 2

<?php
 $a=1;
 echo $a + $a++ + $a++; // it returns 5
?>

but

<?php
 $a=1;
 echo $a++ + $a++ + $a ;// it returns 6
?>

I want to know why the later in prints 6 instead if 5? not sure about associativity or precedency for now. can anyone explain?

Upvotes: 1

Views: 82

Answers (3)

user10463097
user10463097

Reputation:

1-read code left to right 2-note: what is $a++ , it's a "post-increment" operator, meaning the initial value is returned (return to sum machine) before incrementing. 

echo $a++ + $a++ + $a;

same as: echo ($a++ + $a++) + $a; (p1) (p2) (p3) 1 + 2 + 3 = 6

first a = 1

in (p1) a is 1:

1 enter in sum machine then a increase

current value of a is 2

in (p2) a is 2 :

2 enter in sum machine then a increase

current value of a is 3

in (p3) a is 3 :

3 enter in sum machine then no change

current value of a is 3

output of sum machine is 6

Upvotes: 1

msg
msg

Reputation: 8171

The explanation might be a bit confusing: $a++ construct returns the variable current value and then increments.

So what it's happening in the first case is that you are echoing: 1 + 1 + 2 + 1 = 5. The last 1 is the result of the pending ++ operation.

In the second case you have 1 + 2 + 3 = 6. 2 is the value after the increment that in that iteration becomes 3.

There is no precedence here since are all additions.

The counter example is the preincrement:

$a = 1;
echo ++$a + $a;

It should print 4. The first operation increments and then returns so 2 and the second is still 2.

How the operator behaves is explained here

Upvotes: 2

B. Fleming
B. Fleming

Reputation: 7220

When ++ occurs after the variable, it's a "post-increment" operator, meaning the initial value is returned before incrementing. When it occurs before the variable, it's a "pre-increment" operator, which increments before returning the newly-incremented value.

Additionally, when evaluating left + right++, you need to evaluate right++ first because post-increment takes precedence (you need to return the value before you can evaluate the addition).

Thus $a + $a++ + $a++ will evaluate as ($a + (1++)) + $a++ -> (2) + 1 + ($a++) -> 2 + 1 + (2++) -> 5, and $a++ + $a++ + $a will evaluate as (1++) + $a++ + $a -> 1 + (2++) + $a -> 1 + 2 + (3) -> 6.

Upvotes: 1

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