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Jordan Chong
8,173 PointsQuiz Question followup- Why Cakebatter instead of CakebatterCookie Dough
Given the following question:
What does the following block of code output? <?php
$flavors = array("Cake Batter","Cookie Dough");
foreach ($flavors as $a => $b) {
echo $b;
exit;
}
?>
My assumption would have been that both values were stored in the $b working variable and therefore the echo command would print both flavors. Am I missing something here as to why cookie dough isn't echo'd?
2 Answers
Robert Walker
17,146 PointsHi Jordan,
You are correct, both values are stored in $b, however because you have exit; inside your foreach loop after the echo statement, only the first iteration (First Value) is echo'd.
If for example you removed the exit; you would then echo out Cake BatterCookie Dough and for $a 01.
Hope this helps.
Jordan Chong
8,173 PointsYes! Didn't notice that, thanks!