php regex remove inline comment only - php

Question

I have simple code look like this
function session(){
return 1; // this default value for session
}
I need regex or code to remove the comment // this is default value for session, And only remove this type of comment, which starts by a space or two or more, then //, then a newline after it.
All other types of comment and cases are ignored.

Answer 1

UPDATED (1)
And only remove this type of comment, which starts by a space or two or more, then //, then a newline after it
Try this one:
regex101 1
PHP Fiddle 1 -hit "run" or F9 to see the result
/\s+\/\/[^\n]+/m
\s+ starts by a space or two or more
\/\/ the escaped //
[^\n]+ anything except a new line
UPDATE: to make sure -kinda-this only applied to code lines, we can make use of the lookbehind (2) regex to check if there is a semicolon ; before the space[s] and the comment slashes //, so the regex will be this:
regex101 2
PHP Fiddle 2
/(?<=;)\s+\/\/[^\n]+/m
where (?<=;) is the lookbehind which basically tells the engine to look behind and check if there's a ; before it then match.
-----------------------------------------------------------------------
(1) The preg_replace works globally, no need for the g flag
(2) The lookbehind is not supported in javascript

Answer 2

A purely regex solution would look something like this:
$result = preg_replace('#^(.*?)\s+//.*$#m', '\1', $source);
but that would still be wrong because you could get trapped by something like this:
$str = "This is a string // that has a comment inside";
A more robust solution would be to completely rewrite the php code using token_get_all() to actually parse the PHP code into tokens that you can then selectively remove when you re-emit the code:
foreach(token_get_all($source) as $token)
{
if(is_array($token))
{
if($token[0] != T_COMMENT || substr($token[1] != '//', 0, 3))
echo $token[1];
}
else
echo $token;
}