[Solved] Identify the error in the following PHP code:$var = 123;settype($var, integer); — Ans: Missing quotes around 'integer' type name. | PHP

PHP - Type Handling

Identify the error in the following PHP code:

$var = 123;
settype($var, integer);

AMissing quotes around 'integer' type name.

BVariable $var is not initialized.

Csettype cannot convert to integer type.

DSyntax error due to missing semicolon.

Step-by-Step Solution

Solution:

Step 1: Check settype syntax
The type name must be a string literal, so it requires quotes around 'integer'.
Step 2: Identify error cause
Without quotes, PHP treats integer as a constant, causing an error.
Final Answer:
Missing quotes around 'integer' type name. -> Option A
Quick Check:
Type names in settype must be quoted strings [OK]

Quick Trick: Always quote the type name in settype [OK]

Common Mistakes:

Master "Type Handling" in PHP

9 interactive learning modes - each teaches the same concept differently

More PHP Quizzes

Identify the error in the following PHP code: