奇怪的正则表达式问题

Question

I think I may have overlooked something here, but I'm not sure why this strange behaviour is occurring.

I know this is a nonsensical regex capture, but my question is why is this occurring.

<?php
// < 15 "a" characters returns "aaaaaaaaaaaaaa {" (no matches)
$code = 'aaaaaaaaaaaaaa {';
// 15 "a" characters returns NULL
$code = 'aaaaaaaaaaaaaaa {';

$code = preg_replace('#(a+)+\{#', 'b', $code);
var_dump($code);
?>

I realise that the regex has a \s missing. If I change the regex to #(a+)+\s\{# I get "b". But why don't I get "aaaaaaaaaaaaaaa {" with 15 "a" characters instead of a NULL? The manual states that NULL is returned when an error occurs, but nothing is being printed out by PHP (error checking is on).

EDIT:

The reason I am using (a+)+ is because I am trying to simplify the problem down to the basic level. I started off with this:

I want to capture a series of non-whitespace characters followed by spaces followed by an opening brace.

this is some text {

So I used the regex:

#([^\s\{\}]+\s*)+\{#msi

Which also uses the double +. Is there a better way, and why is this causing an issue in the first place?

EDIT 2: From the answers below, it appears that the double + causes a lot of backreferences and should be avoided. I have opted for this [^\s\{\}][^\{\}]*+\{.

Answer 1

<?php

ini_set('pcre.backtrack_limit',20000);
$code = 'aaaaaaaaaaaaaaa {';
$code = preg_replace('#(a+)+\{#', 'b', $code);
var_dump($code, preg_last_error());

ini_set('pcre.backtrack_limit',1000000);
$code = 'aaaaaaaaaaaaaaa {';
$code = preg_replace('#(a+)+\{#', 'b', $code);
var_dump($code, preg_last_error());

Result:

NULL
int(2)
string(17) "aaaaaaaaaaaaaaa {"
int(0)

But you really should use either (a+) or (a)+, not a combination of both.