If statement in Java - Step-by-Step Execution

Concept Flow - If statement

Start

↓

Evaluate condition

↓

Execute 'if' block

↓

Continue

↓

Skip 'if' block

↓

Continue

The program checks a condition. If true, it runs the code inside the if block. If false, it skips it and continues.

Execution Sample

Java

int x = 10;
if (x > 5) {
    System.out.println("x is greater than 5");
}
System.out.println("Done");

Checks if x is greater than 5 and prints a message if true, then prints 'Done'.

Execution Table

Step	Condition (x > 5)	Result	Branch Taken	Output
1	10 > 5	true	Execute if block	x is greater than 5
2	N/A	N/A	Continue after if	Done
3	N/A	N/A	Program ends	No more output

💡 Program ends after printing both messages.

Variable Tracker

Variable	Start	After Step 1	After Step 2	Final
x	10	10	10	10

Key Moments - 2 Insights

Why does the code inside the if block run only when the condition is true?

Because the if statement checks the condition (x > 5). If it is true (see execution_table step 1), it runs the code inside. If false, it skips it (not shown here).

What happens if the condition is false?

The program skips the if block and continues with the next statements after it, as shown in the 'Branch Taken' column for false (not in this example but explained in concept_flow).

Visual Quiz - 3 Questions

Look at the execution table, what is the output at step 1?

A x is greater than 5

B Done

C No output

D Error

💡 Hint

Check the 'Output' column in execution_table row for step 1.

At which step does the program print "Done"?

A Step 1

B Step 2

C Step 3

D Never

💡 Hint

Look at the 'Output' column for step 2 in execution_table.

If x was 3 instead of 10, what would happen at step 1?

A Condition true, print "x is greater than 5"

B Error because x is less than 5

C Condition false, skip if block

D Program stops

💡 Hint

Refer to concept_flow and key_moments about condition false behavior.

Concept Snapshot

If statement syntax:
if (condition) {
    // code runs if condition is true
}

Checks condition once.
Runs block only if true.
Skips block if false.

Full Transcript

This example shows how an if statement works in Java. The program sets x to 10. It checks if x is greater than 5. Since 10 is greater than 5, it prints "x is greater than 5". Then it prints "Done". If the condition was false, it would skip the first print and only print "Done". The flow diagram shows the decision path. The execution table tracks each step, condition result, branch taken, and output. The variable tracker shows x stays 10 throughout. Key moments clarify why the if block runs only when true and what happens if false. The quiz tests understanding of outputs and condition effects. The snapshot summarizes the if statement basics.

Practice

1. What does an if statement do in Java?

easy

A. It stores data in a variable.

B. It repeats code multiple times.

C. It creates a new class.

D. It runs code only if a condition is true.

Solution

1. Step 1: Understand the purpose of an if statement

   An if statement checks a condition and decides whether to run some code based on that condition.

2. Step 2: Match the description to the options

   It runs code only if a condition is true. correctly says it runs code only if the condition is true, which matches the if statement's behavior.

3. Final Answer:

   It runs code only if a condition is true. -> Option D

4. Quick Check:

   If statement = runs code if true [OK]

Hint: If checks true condition to run code [OK]

Common Mistakes:

• Confusing if with loops
• Thinking if stores data
• Mixing if with class creation

2. Which of these is the correct syntax for an if statement in Java?

easy

A. if (condition) System.out.println("Yes");

B. if condition { System.out.println("Yes"); }

C. if condition: System.out.println("Yes");

D. if (condition) then { System.out.println("Yes"); }

Solution

1. Step 1: Recall Java if statement syntax

   Java requires parentheses around the condition and braces are optional if only one statement follows.

2. Step 2: Check each option

   if (condition) System.out.println("Yes"); uses parentheses and a single statement without braces, which is valid syntax.

3. Final Answer:

   if (condition) System.out.println("Yes"); -> Option A

4. Quick Check:

   Parentheses needed around condition [OK]

Hint: Use parentheses around condition in if [OK]

Common Mistakes:

• Omitting parentheses around condition
• Using colons instead of braces
• Adding 'then' keyword

3. What will be the output of this code?

int x = 10;
if (x > 5) {
  System.out.println("Big number");
} else {
  System.out.println("Small number");
}

medium

A. Small number

B. Big number

C. No output

D. Syntax error

Solution

1. Step 1: Evaluate the condition x > 5

   Since x is 10, the condition 10 > 5 is true.

2. Step 2: Determine which block runs

   The if block runs and prints "Big number" because the condition is true.

3. Final Answer:

   Big number -> Option B

4. Quick Check:

   10 > 5 is true, prints "Big number" [OK]

Hint: Check if condition true to pick output [OK]

Common Mistakes:

• Choosing else output when condition is true
• Thinking no output if condition true
• Confusing syntax with output

4. Find the error in this code:

int num = 7;
if num > 5 {
  System.out.println("Greater");
}

medium

A. Braces not allowed in if

B. Missing semicolon after if

C. Missing parentheses around condition

D. Wrong variable name

Solution

1. Step 1: Check if statement syntax

   Java requires parentheses around the condition in an if statement.

2. Step 2: Identify the missing parentheses

   The code has if num > 5 without parentheses, which causes a syntax error.

3. Final Answer:

   Missing parentheses around condition -> Option C

4. Quick Check:

   If condition must be in parentheses [OK]

Hint: Always put condition in parentheses [OK]

Common Mistakes:

• Forgetting parentheses around condition
• Adding semicolon after if line
• Misnaming variables

5. You want to print "Even" if a number is even and "Odd" if it is odd. Which code correctly uses an if statement to do this?

int n = 4;
// Your code here

hard

A. if (n % 2 == 0) {
System.out.println("Even");
} else {
System.out.println("Odd");
}

B. if (n / 2 == 0) {
System.out.println("Even");
} else {
System.out.println("Odd");
}

C. if (n % 2 = 0) {
System.out.println("Even");
} else {
System.out.println("Odd");
}

D. if (n % 2 != 0) {
System.out.println("Even");
} else {
System.out.println("Odd");
}

Solution

1. Step 1: Understand how to check even numbers

   A number is even if dividing by 2 leaves remainder 0, so use n % 2 == 0.

2. Step 2: Check each option's condition

   if (n % 2 == 0) { System.out.println("Even"); } else { System.out.println("Odd"); } correctly uses n % 2 == 0 to check evenness and prints "Even" or "Odd" accordingly.

3. Final Answer:

   if (n % 2 == 0) { System.out.println("Even"); } else { System.out.println("Odd"); } -> Option A

4. Quick Check:

   Use modulus == 0 for even check [OK]

Hint: Use n % 2 == 0 to check even numbers [OK]

Common Mistakes:

• Using assignment = instead of comparison ==
• Checking division instead of remainder
• Reversing even and odd conditions