Increment Operator in PHP and C Language

Question

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Could anyone explain these undefined behaviors (i = i++ + ++i , i = i++, etc…)

I have encountered with a strange problem regards increment operator.

I get different output of same expression in PHP and C.

In C language

main()
{
    int i = 5;
    printf("%d", i++*i++); // output 25;
}

In PHP

$i = 5;
echo $i++*$i++; // output 30

Can anyone explain this strange behavior? Thanks.

In C it is Undefined Behavior, So technically you can get any output.Good Read: [Undefined Behavior and Sequence Points](http://stackoverflow.com/questions/4176328/undefined-behavior-and-sequence-points) — Alok Save, Sep 11 '12 at 12:44
@Ashwini - Why does it matter? What would you use code like this for? — Bo Persson, Sep 11 '12 at 12:50
In PHP, this is also undefined. See [Example 1](http://php.net/manual/en/language.operators.precedence.php). — netcoder, Sep 11 '12 at 13:06

score 3 · Accepted Answer · answered Sep 11 '12 at 12:45

3

In C the result is undefined because either of the two operands could be evaluated first, thus reading it a second time is erroneous.

And, well, in PHP I wouldn't be surprised if the result was 42 pending some changes to php.ini.

answered Sep 11 '12 at 12:45

Luchian Grigore

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My religion would want me to upvote because of the mention of 42 but I'd rather not on a closed duplicate ^^" – Eregrith Sep 11 '12 at 13:26

score 1 · Answer 2 · answered Sep 11 '12 at 12:45

1

The behaviour of ++ is undefined when used in this style, as you don't know exactly when the ++ operation will occur and when the values will be "returned" from x++.

answered Sep 11 '12 at 12:45

slugonamission

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score 0 · Answer 3 · answered Sep 11 '12 at 12:44

0

It a matter of precedence. Have a look at http://php.net/manual/en/language.operators.precedence.php

answered Sep 11 '12 at 12:44

JvdBerg

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Keith Miller · Answer 4 · 2012-09-11T13:00:37.987

This is undefined behavior since i++ or ++i or --i or i-- do not increment/decrement in any particular order when passed as function parameter.

Not only that but if I'm not mistaken I believe that printf("%d", i++*i++); is just outputting 5*5 and then increments i twice.

remember ++i increments before operation, and i++ increments after operation. Consider this code:

int i, x = 5;

int i = x++;   // i is now equal to 5 and x is equal to 6 because the increment happened after the = operation.
x = 5;         //set x back to 5
i = ++x;       //i is now equal to 6 and x is equal to 6 because the increment happened before the = operation.

This is the case for C I however cannot vouch for PHP.

Increment Operator in PHP and C Language

4 Answers4