Question

I’m having some trouble understanding why the following doesn’t result in a compiler error in 5.3.3 (errored-out correctly on my coworkers 5.2.5):

<?php
    echo "starting\n";

    switch(1) {
        case 2:
            echo "two\n";
            break;
        defalut:        // note the misspelling
            echo "deflaut\n";
    }

    echo "ending\n";

Instead of giving me a compiler error (or even a warning) it just gives this:

starting
ending

However, if I use it in an if-statement it gives me what I’d expect:

<?php
    if (1 == deflaut)
        echo "deflaut2\n";

gives:

PHP Notice:  Use of undefined constant deflaut - assumed 'deflaut' in ...

Why is this? Is there a setting somewhere I can disable to tell it to be strict about this sort of thing?

Answer 1

The problem is that your code isn’t doing what you think. A case block only ends when the next case block occurs, or when default: is found, or when the closing } is reached. This means that defalut is part of the case 2: block. So it is never even interpreted.

However, it doesn’t even fire a syntax error (not even if you do switch (2). This is because the goto operator was introduced in PHP 5.3. The syntax word: at the beginning of a PHP statement is now a target accessible via goto. So goto defalut; can be used to go to the label.

_{(Actually, it can’t, because of a restriction on targets inside switch blocks to avoid infinite loops, but this should illustrate the point…)}

You can make it force an error by doing case defalut, when the error that you expect is found.