I don't understand the precautions for the ternary operator in the PHP documentation

Note: Note that the ternary operator is a statement, so its evaluation is not a variable, but the result of the statement. This is important if you want to return a variable by reference. The statement return $var == 42 ? $a : $b; in a function that returns by reference will not work, and a future version of PHP will issue a warning about this.

return $var == 42 ? $a : $b; What does it mean if it doesn’t work? Is it impossible to return a value or something?

function test($var){
  return $var == 42 ? 1 : 2;
}

echo test(40);    //2

This way you can return it during testing...

Reply content:

Note: Note that the ternary operator is a statement, so its evaluation is not a variable, but the result of the statement. This is important if you want to return a variable by reference. The statement return $var == 42 ? $a : $b; in a function that returns by reference will not work, and a future version of PHP will issue a warning about this.

return $var == 42 ? $a : $b; What does it mean if it doesn’t work? Is it unable to return a value or something?

function test($var){
  return $var == 42 ? 1 : 2;
}

echo test(40);    //2

This way you can return it during testing...

Affects the scenario of "returning a variable by reference"

See the example, get2 cannot achieve the expected results

https://3v4l.org/2Q9ai

<?php

$data = new stdClass;
$data->a = 13;
$data->b = 42;

$var = &get1($data, true);
$var = 14;
var_dump($data);

$var2 = &get2($data, false);
$var2 = 43;
var_dump($data);

function &get1($data, $isA) {
    if($isA) {
        return $data->a;
    } else {
        return $data->b;
    }
}

function &get2($data, $isA) {
    return $isA ? $data->a : $data->b;
}