Why is 1 not being outputted? (Simple php problem)

Question

I have a question which I don't know the answer of. I've been thinking about it for a while.

The following code:

$i = 1;
while($i < 10) 
    if(($i++) % 2 == 0)
    echo $i;

It correctly outputs 3579, but why isnt 1 also included in the output?

I'm a beginner with PHP and am looking forward for someone to help me.

Thank you very much! :D

It correctly outputs 3579.. How can this be? It throws a Fatal Error because of what @RiggsFolly mentioned — B001ᛦ
– B001ᛦ, Commented Mar 11, 2019 at 12:12
@B001ᛦ I am assuminmg thats a accident in the writing of the question :) — RiggsFolly
– RiggsFolly, Commented Mar 11, 2019 at 12:14
why isnt 1 also included in the output? because you are incrementing it i++ so it becomes 2. use $i=0; at the start — Ash-b
– Ash-b, Commented Mar 11, 2019 at 12:16

Pupil · Accepted Answer · 2019-03-11 12:21:58Z

3

Two modifications:

$i = 0; // Make it 0 from 1
while($i < 10)
if(($i++) % 2 == 0)
echo "<br/>".$i; // Make $i instead of $1

Output:


1
3
5
7
9

Program hand run:

1) Set $i to 0.

2) If it is greater than 10, go ahead.

3) Increment it by 1

4) So, for $i => 0->1, 1->2

4) if new $i is even, print it. (So for first loop iteration, you are have $i -> 1 instead of 0 because of ++$i

edited Mar 11, 2019 at 12:21

answered Mar 11, 2019 at 12:13

Pupil

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3 Comments

Frankeyx Over a year ago

Hmm. Yes, now it does correctly output it, thank you! But, I still do not understand why. I'm not very good at PHP... trying to learn :). Could you explain in text, why it starts at 3 and not 1?

RiggsFolly Over a year ago

Because $i++ gets run before the % modulo therefore as you started %i at 1 it never saw a 1 the first time $i is tested it is already 2

Frankeyx Over a year ago

Yep, got it. Thanks! :)

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