Why is a `switch` considered a looping structure for the purposes of `continue`?

Question

I just got bit by assuming the following:

foreach ($arr as $key => $value) {
  switch($key) {
    // ... some other cases
    default:
      continue;
      // ^== assumption: move on to the next iteration of the foreach
      //     actual PHP: treat this continue just like a break
  }
  // ...
}

But in fact, according to the documentation for continue:

the switch statement is considered a looping structure for the purposes of continue.

Is there a reason for this choice on the part of PHP language designers? As far as I can tell, switch isn't a looping control structure, so why treat it like one in this case?

Answer 1

The best explanation I can think of is that PHP considers it a looping structure so it fits the model of something that uses continue and break. The switch documentation doesn't shed much further light on it other than

Note that unlike some other languages, the continue statement applies to switch and acts similar to break. If you have a switch inside a loop and wish to continue to the next iteration of the outer loop, use continue 2.

So perhaps it is because, like a loop, it stops the execution of the rest of the code in its structure.

However, when you use a number of levels, these two behave quite differently:

continue without a level

<?php
for($i=0;$i<5;$i++) {
    switch($i) {
        case 2:
            continue;
        default:
            echo $i, "\n";
    }
    echo "Finished with {$i}\n";
}
//0
//Finished with 0
//1
//Finished with 1
//Finished with 2
//3
//Finished with 3
//4
//Finished with 4

continue with a level

<?php
for($i=0;$i<5;$i++) {
    switch($i) {
        case 2:
            continue 2;
        default:
            echo $i, "\n";
    }
    echo "Finished with {$i}\n";
}
//0
//Finished with 0
//1
//Finished with 1
//3
//Finished with 3
//4
//Finished with 4

break without a level

<?php
for($i=0;$i<5;$i++) {
    switch($i) {
        case 2:
            break;
        default:
            echo $i, "\n";
    }
    echo "Finished with {$i}\n";
}
//0
//Finished with 0
//1
//Finished with 1
//Finished with 2
//3
//Finished with 3
//4
//Finished with 4

break with a level

<?php
for($i=0;$i<5;$i++) {
    switch($i) {
        case 2:
            break 2;
        default:
            echo $i, "\n";
    }
    echo "Finished with {$i}\n";
}
//0
//Finished with 0
//1
//Finished with 1

Answer 2

I think you won't find any real "reason" for this behavior.

The only real motivation behind this behavior was probably that implementing switch as if it were a looping structure allows PHP to reuse existing break and continue semantics of loops instead of reimplementing a special version for switch.

Or to phrase it more positively: It's for consistency.