Why would the first $i be echoed out?

Am I reading this wrongly? Isn’t the $i being updated before the end of the statement in the loop?

1st step: $i = 0; 2nd step: check wether $i is still smaller than 10; if it is go to step 3 3rd step: update the value of the variable $i by incrementing it with 1.

The result should then be: 123456789 instead of 0123456789, no?

Or is the initial setting of the variable $i a statement for the loop to know where to start?

“where to start the loop;”

Answer 51922466eb29bf64740027e0

1 vote

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If you set $i =0 at the begining of the for-loop, the echo will print out the 0. Maybe I can explain it to you by a while-loop:

$i =0;
while(i<10){
    echo $i;
    $i++; 
}

This is how a for-loop works, first it checks if the statement is true, if so it continues the code and after this, it increments the loop-variable. I hope you got it and sorry for my grammar ;>

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