PHP dynamic typing behavior - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of adding string and integer

What is the output of this PHP code?

$a = "5 apples";
$b = 3;
echo $a + $b;

PHP

$a = "5 apples";
$b = 3;
echo $a + $b;

A 8

B 53

C 5 apples3

D Error

💡 Hint

PHP converts strings to numbers when using arithmetic operators.

PHP converts the string "5 apples" to the number 5 when used with +. So 5 + 3 = 8.

❓ Predict Output

intermediate

Boolean comparison with string

What is the output of this PHP code?

$x = "0";
if ($x == false) { echo "Yes"; } else { echo "No"; }

PHP

$x = "0";
if ($x == false) { echo "Yes"; } else { echo "No"; }

A Yes

B Error

C 0

D No

💡 Hint

In PHP, the string "0" is considered false in loose comparisons.

When comparing with ==, "0" is treated as false, so the condition is true and "Yes" is printed.

🔧 Debug

advanced

Why does this string concatenation fail?

Consider this PHP code:

$a = 5;
$b = " apples";
echo $a - $b;

What error or output will this produce?

PHP

$a = 5;
$b = " apples";
echo $a - $b;

A Fatal error

B 5

C 0

D Notice: A non-numeric value encountered

💡 Hint

PHP tries to convert strings to numbers in arithmetic but " apples" has no leading digits.

PHP issues a notice about non-numeric value because " apples" cannot be converted to a number, so it treats it as 0 and outputs 5 - 0 = 5 but with a notice.

❓ Predict Output

advanced

Result of loose equality with different types

What is the output of this PHP code?

$a = 0;
$b = "any string";
var_dump($a == $b);

PHP

$a = 0;
$b = "any string";
var_dump($a == $b);

A bool(false)

B bool(null)

C bool(true)

D Error

💡 Hint

When comparing number and string, PHP converts string to number. Non-numeric strings become 0.

The string "any string" converts to 0, so 0 == 0 is true, var_dump prints bool(true).

🧠 Conceptual

expert

Understanding PHP type juggling in array keys

Given this PHP code:

$arr = [];
$arr["1"] = "one";
$arr[1] = "ONE";
echo count($arr);

What is the output and why?

PHP

$arr = [];
$arr["1"] = "one";
$arr[1] = "ONE";
echo count($arr);

A 2 because string and integer keys are different

B 1 because PHP treats string "1" and integer 1 as the same key

C 0 because keys overwrite each other and array is empty

D Error due to key type mismatch

💡 Hint

PHP converts string keys that are numeric to integers in arrays.

PHP treats "1" and 1 as the same key in arrays, so the second assignment overwrites the first. The array has one element.