Variable declaration with dollar sign in PHP - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this PHP code with variable variables?

Consider the following PHP code snippet. What will it output?

PHP

<?php
$a = 'hello';
$hello = 'world';
echo $$a;
?>

A $hello

B hello

C world

D Error: Undefined variable

💡 Hint

Remember that $$a means the variable whose name is the value of $a.

The variable $a contains the string 'hello'. So $$a means $hello, which contains 'world'. Thus, echo $$a outputs 'world'.

❓ Predict Output

intermediate

What will this PHP code output?

Analyze the following PHP code and select the output it produces.

PHP

<?php
$var = 'name';
$name = 'PHP';
echo "$var is $name";
?>

A PHP is name

B name is PHP

C $var is $name

D Error: Undefined variable

💡 Hint

Variables inside double quotes are replaced by their values.

Inside double quotes, $var and $name are replaced by their values 'name' and 'PHP' respectively. So the output is 'name is PHP'.

❓ Predict Output

advanced

What error does this PHP code raise?

Examine the code below. What error will it produce when run?

PHP

<?php
$1var = 'test';
echo $1var;
?>

A Parse error: syntax error, unexpected '1var'

B Notice: Undefined variable: 1var

C Fatal error: Cannot use numeric variable name

D test

💡 Hint

Variable names in PHP cannot start with a number.

PHP variable names must start with a letter or underscore. Starting with a number causes a syntax error.

❓ Predict Output

advanced

What is the output of this PHP code using variable variables and concatenation?

What will this PHP code print?

PHP

<?php
$foo = 'bar';
$bar = 'baz';
$baz = 'qux';
echo $$$foo;
?>

A qux

B Error: Undefined variable

C bar

D baz

💡 Hint

Evaluate from right to left: $$$foo means ${$foo} then ${${$foo}}.

First, $foo = 'bar', so $$foo = $bar = 'baz'. Then $$$foo = ${$bar} = $baz = 'qux'. So echo $$$foo outputs 'qux'.

🧠 Conceptual

expert

How many variables are declared and accessible after this PHP code runs?

Consider this PHP code snippet. How many distinct variables are declared and accessible after execution?

PHP

<?php
${'var'.'1'} = 10;
$var2 = 20;
$var3 = ${'var'.'2'} + 5;
?>

A 1

B 4

C 2

D 3

💡 Hint

Count all variables with names starting with $var and those created dynamically.

Variables declared: $var1 (via ${'var'.'1'}), $var2, and $var3. So total 3 variables accessible.

Practice

1. Which of the following is the correct way to declare a variable named age in PHP?

easy

A. $age = 25;

B. age = 25;

C. var age = 25;

D. int $age = 25;

Solution

1. Step 1: Recognize PHP variable syntax

   In PHP, all variables must start with a dollar sign ($) followed by the variable name.

2. Step 2: Check each option for correct syntax

   $age = 25; uses $age = 25; which is correct. Options B and C miss the dollar sign, and A uses a type declaration which is not standard in PHP variable declaration.

3. Final Answer:

   $age = 25; -> Option A

4. Quick Check:

   PHP variables start with $ = $age = 25; [OK]

Hint: Always start PHP variables with $ sign [OK]

Common Mistakes:

• Forgetting the $ sign before variable name
• Using other language syntax like var or int
• Trying to declare type explicitly in PHP

2. Which of the following is a syntax error in PHP variable declaration?

easy

A. $1number = 10;

B. $name = "John";

C. $total = 100;

D. $_score = 50;

Solution

1. Step 1: Understand valid variable names in PHP

   Variable names must start with a letter or underscore, not a number.

2. Step 2: Identify invalid variable name

   $1number = 10; uses $1number which starts with a number, causing a syntax error. Other options start with $ followed by letter or underscore, which is valid.

3. Final Answer:

   $1number = 10; -> Option A

4. Quick Check:

   Variable names cannot start with digits = $1number = 10; [OK]

Hint: Variable names cannot start with numbers [OK]

Common Mistakes:

• Starting variable names with digits
• Using spaces in variable names
• Using special characters other than underscore

3. What will be the output of the following PHP code?

$x = 5;
$y = $x;
$x = 10;
echo $y;

medium

A. Error

B. 10

C. 0

D. 5

Solution

1. Step 1: Understand variable assignment in PHP

   When $y = $x; runs, $y gets the current value of $x, which is 5.

2. Step 2: Check effect of changing $x after assignment

   Changing $x to 10 later does not affect $y because $y holds a copy of the value, not a reference.

3. Final Answer:

   5 -> Option D

4. Quick Check:

   Variable copy keeps old value = 5 [OK]

Hint: Variable assignment copies value, not reference [OK]

Common Mistakes:

• Assuming $y changes when $x changes
• Confusing references with copies
• Expecting runtime error for reassignment

4. Identify the error in this PHP code snippet:

$var1 = 10;
var2 = 20;
echo $var2;

medium

A. echo statement syntax error

B. Missing $ sign before var2

C. Missing semicolon after var1

D. Variable names cannot have numbers

Solution

1. Step 1: Check variable declarations

   $var1 is declared correctly with a $ sign. var2 is missing the $ sign, which is required in PHP.

2. Step 2: Analyze the echo statement

   echo uses $var2 which is undefined because var2 was not declared properly.

3. Final Answer:

   Missing $ sign before var2 -> Option B

4. Quick Check:

   All PHP variables need $ sign = Missing $ sign before var2 [OK]

Hint: Every PHP variable must start with $ [OK]

Common Mistakes:

• Forgetting $ sign before variable
• Assuming semicolon missing causes error here
• Thinking numbers are not allowed in variable names

5. You want to store a user's first name and age in PHP variables and then display them. Which code correctly declares and uses these variables?

1. $firstName = "Alice";
2. $age = 30;
3. echo "Name: $firstName, Age: $age";

hard

A. Declare variables without $ sign and use $ in echo

B. Use variables without $ sign in echo: echo "Name: firstName, Age: age";

C. Declare variables with $ and use them with $ in echo as shown

D. Use $ only in declaration, not in echo

Solution

1. Step 1: Declare variables with $ sign

   Lines 1 and 2 correctly declare variables $firstName and $age with $ sign.

2. Step 2: Use variables with $ sign inside double quotes

   In PHP, variables inside double quotes are parsed and replaced with their values, so echo "Name: $firstName, Age: $age"; works correctly.

3. Final Answer:

   Declare variables with $ and use them with $ in echo as shown -> Option C

4. Quick Check:

   Variables need $ in declaration and usage [OK]

Hint: Use $ sign in both declaration and usage [OK]

Common Mistakes:

• Omitting $ sign in echo
• Using single quotes which don't parse variables
• Declaring variables without $ sign