php string number concatenation messed up

Question

I got some php code here:

<?php
echo 'hello ' . 1 + 2 . '34';
?> 

which outputs 234,

but when I add a number 11 before "hello":

<?php
echo '11hello ' . 1 + 2 . '34';
?> 

It outputs 1334 rather than 245(which I expected it to), why is that?

Answer 1

you have to use () in mathematical operation

echo 'hello ' . (1 + 2) . '34'; // output hello334
echo '11hello ' . (1 + 2) . '34'; // output 11hello334

Answer 2

That's strange...

But

<?php
echo '11hello ' . (1 + 2) . '34';
?>

OR

<?php
echo '11hello ', 1 + 2, '34';
?>

fixing issue.

---

UPDv1:

Finally managed to get proper answer:

'hello' = 0 (contains no leading digits, so PHP assumes it is zero).

So 'hello' . 1 + 2 simplifies to 'hello1' + 2 is 2, because no leading digits in 'hello1' is zero too.

---

'11hello ' = 11 (contains leading digits, so PHP assumes it is eleven).

So '11hello ' . 1 + 2 simplifies to '11hello 1' + 2 as 11 + 2 is 13.

---

UPDv2:

http://www.php.net/manual/en/language.types.string.php

The value is given by the initial portion of the string. If the string starts with valid numeric data, this will be the value used. Otherwise, the value will be 0 (zero). Valid numeric data is an optional sign, followed by one or more digits (optionally containing a decimal point), followed by an optional exponent. The exponent is an 'e' or 'E' followed by one or more digits.

Answer 3

The dot operator has the same precedence as + and -, which can yield unexpected results.

That technically answers your question... if you want numbers to be treated as numbers during concatination just wrap them in parenthesis.

<?php
echo '11hello ' . (1 + 2) . '34';
?>

Answer 4

If you hate putting operators in between assign them to vaiable

$var = 1 + 2;

echo 'hello ' . $var . '34';

Answer 5

You should check PHP type conversion table to get better idea what's happen behind the scenes: http://php.net/manual/en/types.comparisons.php