For loop in C++ - Step-by-Step Execution

Concept Flow - For loop

Initialize i=0

↓

Check: i < 5?

No→EXIT

Yes↓

Execute body

↓

Update: i = i+1

↩Back to Check

The for loop starts by setting i to 0, then checks if i is less than 5. If yes, it runs the loop body, then increases i by 1 and repeats. If no, it stops.

Execution Sample

C++

for (int i = 0; i < 5; i++) {
    std::cout << i << std::endl;
}

Prints numbers from 0 to 4, one per line.

Execution Table

Iteration	i	Condition (i < 5)	Action	Output
1	0	true	Print 0, i++	0
2	1	true	Print 1, i++	1
3	2	true	Print 2, i++	2
4	3	true	Print 3, i++	3
5	4	true	Print 4, i++	4
6	5	false	Exit loop

💡 i reaches 5, condition 5 < 5 is false, loop stops

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	Final
i	0	1	2	3	4	5	5

Key Moments - 3 Insights

Why does the loop stop when i is 5 even though we print numbers starting from 0?

The loop condition checks if i < 5. When i becomes 5, this condition is false, so the loop stops before printing 5. See execution_table row 6.

When exactly does i increase in the loop?

i increases after the loop body runs, before the next condition check. This is shown in execution_table where 'i++' happens after printing.

What happens if we change i++ to i += 2?

i would increase by 2 each time, so the loop runs fewer times. The variable_tracker would show i values jumping by 2 instead of 1.

Visual Quiz - 3 Questions

Look at the execution table, what is the value of i during the 3rd iteration?

A 1

B 3

C 2

D 0

💡 Hint

Check the 'i' column in the execution_table row labeled '3'

At which iteration does the loop condition become false?

A 5

B 6

C 4

D 1

💡 Hint

Look at the 'Condition (i < 5)' column in the execution_table where it is false

If we change i++ to i += 2, how would the variable_tracker change after the first iteration?

A i would be 2 instead of 1

B i would be 0 instead of 1

C i would be 3 instead of 1

D i would not change

💡 Hint

Refer to variable_tracker and think how i increments by 2 instead of 1

Concept Snapshot

for loop syntax: for (init; condition; update) { body }
Runs body while condition is true.
Init runs once at start.
Update runs after each body execution.
Stops when condition is false.

Full Transcript

A for loop in C++ starts by initializing a variable, here i=0. Then it checks if i is less than 5. If yes, it runs the loop body which prints i. After that, i increases by 1. This repeats until i reaches 5, when the condition becomes false and the loop stops. The variable i changes from 0 up to 5, but 5 is not printed because the loop stops before printing it.

Practice

1.

What does a for loop in C++ do?

easy

A. Defines a new function

B. Runs code only once

C. Stops the program immediately

D. Repeats a block of code a fixed number of times

Solution

1. Step 1: Understand the purpose of a for loop

   A for loop is designed to repeat code multiple times based on a condition.

2. Step 2: Identify the correct description

   Among the options, only repeating code a fixed number of times matches the for loop's behavior.

3. Final Answer:

   Repeats a block of code a fixed number of times -> Option D

4. Quick Check:

   For loop = repeat fixed times [OK]

Hint: For loops repeat code a set number of times [OK]

Common Mistakes:

• Thinking for loops run only once
• Confusing for loops with if statements
• Believing for loops stop the program

2.

Which of the following is the correct syntax for a for loop in C++?

for (___; i < 5; i++) {
    // code
}

easy

A. int i = 0

B. i = 0

C. int i

D. int i == 0

Solution

1. Step 1: Check initialization syntax in for loop

   The initialization part must declare and initialize the loop variable, e.g., int i = 0.

2. Step 2: Compare options for correctness

   Only int i = 0 correctly declares and initializes the variable.

3. Final Answer:

   int i = 0 -> Option A

4. Quick Check:

   Initialization needs declaration and assignment [OK]

Hint: Initialization needs declaration and assignment [OK]

Common Mistakes:

• Using 'int i' without initialization
• Using 'int i == 0' which is a comparison
• Omitting the type declaration

3.

What is the output of this code?

for (int i = 1; i <= 3; i++) {
    std::cout << i * 2 << " ";
}

medium

A. 1 2 3

B. 2 4 6

C. 0 2 4

D. 2 3 4

Solution

1. Step 1: Identify loop iterations

   The loop runs with i = 1, 2, 3 because condition is i <= 3.

2. Step 2: Calculate output for each iteration

   Outputs are i*2: 2, 4, 6 printed with spaces.

3. Final Answer:

   2 4 6 -> Option B

4. Quick Check:

   Multiply i by 2 each loop [OK]

Hint: Multiply loop variable by 2 each iteration [OK]

Common Mistakes:

• Confusing loop range (off by one)
• Printing i instead of i*2
• Missing spaces in output

4.

Find the error in this for loop:

for (int i = 0; i > 5; i++) {
    std::cout << i << " ";
}

medium

A. Variable i is not declared

B. Missing semicolon after i++

C. The condition 'i > 5' is incorrect, loop never runs

D. The loop will run infinitely

Solution

1. Step 1: Analyze the loop condition

   The condition is i > 5 but i starts at 0, so it is false immediately.

2. Step 2: Understand loop execution

   Since condition is false at start, loop body never executes.

3. Final Answer:

   The condition 'i > 5' is incorrect, loop never runs -> Option C

4. Quick Check:

   Condition false at start means no loop runs [OK]

Hint: Check if loop condition is true initially [OK]

Common Mistakes:

• Assuming loop runs despite false condition
• Thinking missing semicolon causes error here
• Confusing variable declaration errors

5.

How can you use a for loop to print all even numbers from 2 to 10 inclusive?

Choose the correct code snippet:

hard

A. for (int i = 2; i <= 10; i += 2) { std::cout << i << ' '; }

B. for (int i = 1; i < 10; i++) { if (i % 2 == 0) std::cout << i << ' '; }

C. for (int i = 2; i < 10; i++) { std::cout << i * 2 << ' '; }

D. for (int i = 0; i <= 10; i += 3) { std::cout << i << ' '; }

Solution

1. Step 1: Understand the goal

   We want to print even numbers 2,4,6,8,10 inclusive.

2. Step 2: Check each option

   for (int i = 2; i <= 10; i += 2) { std::cout << i << ' '; } increments by 2 starting at 2, printing 2,4,6,8,10 correctly.
   for (int i = 1; i < 10; i++) { if (i % 2 == 0) std::cout << i << ' '; } prints 2,4,6,8 but misses 10.
   for (int i = 2; i < 10; i++) { std::cout << i * 2 << ' '; } prints 4,6,8,10,12,14,16,18 (wrong range).
   for (int i = 0; i <= 10; i += 3) { std::cout << i << ' '; } increments by 3, not even numbers.

3. Final Answer:

   for (int i = 2; i <= 10; i += 2) { std::cout << i << ' '; } -> Option A

4. Quick Check:

   Increment by 2 from 2 to 10 [OK]

Hint: Increment loop variable by 2 starting at 2 [OK]

Common Mistakes:

• Using wrong increment step
• Forgetting to include 10
• Using condition inside loop unnecessarily