Missing concatenation part in PHP

Question

Here is the code:

 <?php 

      for($i =0, $x = 100 ; $i<1; $i++){

      echo $x .  'y' . $i+1 . ' = '. $i*$x . ' <br>';
      }

    ?>

My Expected Output was: 100 y 1 = 0

But the actual result was: 101 = 0;

Where did 'y' go?

https://ideone.com/n9HYGp

there is no error in your code and your output will be : 100 y 1 = 0 — Syed mohamed aladeen, Oct 17 '15 at 07:54
It's just because of [operator precedence](http://php.net/manual/en/language.operators.precedence.php) of PHP that you've ignored!! — someOne, Oct 17 '15 at 08:08

Subin Thomas · Accepted Answer · 2015-10-18T10:27:17.607

5

. have more operator precedence than +.

echo $x .  'y' . $i+1 = 101

Because it will operate as

echo ($x .  "y" . $i)+1 ;

This is what happens.

$x3=  ($x .  "y" . $i); //100y0
$u = $x3+1 ;  //101

You are doing + operation on a string. So the first digits before any characters will be taken as integer value.

Eg:

10y0g8 = 10
t10  =0

By doing an Arithmetic Operation, interpreter will convert string to integer, and it will discard all other characters. so 100y+1 = 101 it won't be 101y

edited Oct 18 '15 at 10:27

answered Oct 17 '15 at 07:55

Subin Thomas

1,408
10
19

In your answer. $x3 = 100y0; then you add +1 in $x3 which becomes 101.. very confusing.. how 100y0 +1 = 101 ?? – Waqar Oct 17 '15 at 08:02
ok i can understand what you say. 100y0 +1 = 101... Fine ... why the answer is 101 and not 101y where did y go? – Waqar Oct 17 '15 at 08:14
ok beside this + operator, the compiler should have writter y after 101.. that what i say. – Waqar Oct 17 '15 at 08:17
1

is it what you said.. compiler will discard all other characters because it will convert string to integer.. .... CLEARED NOW THANKS MAN – Waqar Oct 17 '15 at 08:24
2

For the record, [it's not the compiler! it's the interpreter](http://stackoverflow.com/q/1514676/3709765) :) – someOne Oct 18 '15 at 03:45
thanks for the correction. I learned. :) One day i will be in a position to give better answers.. ONE DAY :) – Waqar Oct 18 '15 at 10:26

score 1 · Answer 2 · answered Oct 17 '15 at 08:03

1

as explained in the above (Subin Thomas) answer the addition of 1 with the varchar value like 100y0 will add 1 with 100(the first occurence of integer). the following code will work as you expected

<?php 

  for($i =0 ,$x = 100; $i<1; $i++){

  echo $x .  'y' . ($i+1) . ' = '. $i*$x . ' <br>';
  }

?>

answered Oct 17 '15 at 08:03

Syed mohamed aladeen

6,507
4
32
59

Man you are absolutely right. But again if you donot put brackets around ($i+1), then why it becomes 101 = 0 – Waqar Oct 17 '15 at 08:06
$x = 100. and y will concat to it as 100y when it added $i+1=1 then 100y+1. it will take first occurence of integer that is 100+1=101. – Syed mohamed aladeen Oct 17 '15 at 08:08
ok as u said it will take first occurrence of integer, ok fine, then where did y go? – Waqar Oct 17 '15 at 08:10
1

you are using + operator right. this will take any characters before it as integer y is not an integer so it removes "y". – Syed mohamed aladeen Oct 17 '15 at 08:12

score 0 · Answer 3 · answered Oct 17 '15 at 08:04

0

To test :

<?php 

  for($i =0, $x = 100 ; $i<1; $i++){

  echo $x .  'y' . ($i+1) . ' = '. $i*$x . ' <br>';
  }

?>

answered Oct 17 '15 at 08:04

scoolnico

3,055
14
24

1

Obviously because of higher precedence of ().. But what if we donot put brackets.. just give explanation to the actual answer.. – Waqar Oct 17 '15 at 08:08
Have a look to http://php.net/manual/en/language.operators.string.php post #56... – scoolnico Oct 17 '15 at 08:14

Missing concatenation part in PHP

3 Answers3