note 101465 deleted from control-structures.switch by danbrown

Note Submitter: cory at watchdogserver dot com ---- Somewhere above, this example below was shown and said to be an error within PHP where more than one default would trigger. <?php switch(9) { case 1: echo 1 . "\n"; case 2: echo 2 . "\n"; break; case 3: echo 3 . "\n"; case 4: echo 4 . "\n"; case 5: echo 5 . "\n"; default: echo 6 . "\n"; default: echo 7 . "\n"; default: echo 8 . "\n"; } // Outputs: // 8 // // Did you expect 6 and 7 to print too? switch(3) { case 1: echo 1 . "\n"; case 2: echo 2 . "\n"; break; case 3: echo 3 . "\n"; case 4: echo 4 . "\n"; case 5: echo 5 . "\n"; default: echo 6 . "\n"; default: echo 7 . "\n"; default: echo 8 . "\n"; } // Outputs: // 3 // 4 // 5 // 6 // 7 // 8 ?> CORRECTION: The reason it outputs 8 on the first example is because the last default listed is triggered. Yes, this part is correct. However the reason the second example outputs 3 4 5 6 7 8 is not because of an error within PHP rather an error in your code. It is triggering all of the cases simply because you did not place a break after each case. Call it a "run-on-case" or lack of break/stop point definition. Second example re-written: (Correctly) <?php switch(3) { case 1: echo 1 . "\n"; break; case 2: echo 2 . "\n"; break; case 3: echo 3 . "\n"; break; case 4: echo 4 . "\n"; break; case 5: echo 5 . "\n"; break; default: echo 6 . "\n"; break; default: echo 7 . "\n"; break; default: echo 8 . "\n"; break; } ?> // outputs // 3 I think the point you were trying to make is with more than one default listed only the last default is triggered. This is correct. The rest is simply a mistake in coding. Now, whether should PHP allow/permit run-on-cases to occur could be considered an error within php, yes, but not the error as you have described. Tested with PHP 5.3.4