Re: Bug #1452 Updated: $i = $i++ fails

Not really. In fact, in C, it's undefined. In PHP, it's always nada. A C compiler can understand it as: result = rvalue inc(rvalue) lvalue = result or lvalue = rvalue inc(rvalue) Compilers generally do it the first way and not the second way, even though they can choose either. It's undefined which part of an assignment expression is evaluated first. Zeev At 01:36 26/05/99 , Rasmus Lerdorf wrote:

In fact, if that loop hangs for you, than it succeeds doing what it's supposed to do. $i=$i++ is equivalent to an empty statement. Consult the documentation regarding what post-increment is to understand why. Uh? $i = $i++; This should be equivalent to doing: $i=$i; $i++; at least from my understanding of what post-increment is supposed to do. Try this C program: int main() {

    int i=0;
    while(i < 50) {
        printf("%d\n",i);
        i = i++;
    }

} It is the equivalent to his example and it correctly counts from 0 to 49, so I do believe the submitter of this bug is correct. -Rasmus

-- Zeev Suraski <zeev@zend.com> http://www.zend.com/ For a PGP public key, finger bourbon@netvision.net.il -- PHP Development Mailing List http://www.php.net/ To unsubscribe send an empty message to php-dev-unsubscribe@lists.php.net For help: php-dev-help@lists.php.net