Can somebody explain me this code?

Question

<?php

Function fun(){
$x=0;
Echo $x;
$x++;
}
Fun();
Fun();
Fun();

?>

It will output 000 but again i have no ideea why. Thanks

Answer 1

It doesn't matter that $x++ is placed after the echo (I mean: it doesn't mean that it's not achived) but after it increments $x then in the next loop the line $x=0; will set $x to zero again (btw, It wouldn't if $x was a static variable but it's not). joshua johns and razvan you can read Eduardo's answer again he is right and for me that answer is clear, but I decided to write my own after seeing that it looks like you are paying to much attention that $x++ is after the line with echo (of course you won't get incremented variable in the output if $x++ will be written after the echo (echo $x;) it's obvious... but the tricky part here (in this question) that you are re_setting $x to zero each time the loop starts and so if you will write $x++ before the line with "echo" you will get incremented value (variable) 1 of course... but it doesn't mean that you'll get 123 as an output of those 3 function calls, because you'll...

Answer 2

well you call the function fun 3 times and always your function set the x value to 0.
Basically you are print 0 the first time, then x increase but the function ends, and the same after.

Answer 3

...because you'll get 111 as an output of 3 function calls. Can I explain something else here? (Or was it not needed at all as you do understand it that way?) ☺

Answer 4

Andrew Harchenko (Tomsk) thank you. i understood now. no further questions :) thanks again.

Answer 5

right..i understand now..so $x++ will not count because this code line is writen after the echo..and then the function ends without any echo..thank you

Answer 6

The code outputs 000 because there is no condition set so as to increment to be achieved.i.e the increment comes after echo statement which makes no sense