Loop execution flow in C++ - Step-by-Step Execution

Concept Flow - Loop execution flow

Initialize loop variable

↓

Check loop condition

No→Exit loop

Yes↓

Execute loop body

↓

Update loop variable

↩Back to Check loop condition

This flow shows how a loop starts by setting a variable, checks a condition, runs the loop body if true, updates the variable, and repeats until the condition is false.

Execution Sample

C++

for (int i = 0; i < 3; i++) {
    std::cout << i << std::endl;
}

This code prints numbers 0, 1, and 2 by looping while i is less than 3.

Execution Table

Iteration	i	Condition (i < 3)	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	false	Exit loop

💡 i reaches 3, condition 3 < 3 is false, loop ends

Variable Tracker

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

Key Moments - 3 Insights

Why does the loop stop when i equals 3?

Because the condition i < 3 becomes false at iteration 4 (see execution_table row 4), so the loop exits.

When is the variable i updated during the loop?

i is updated after the loop body executes in each iteration (see execution_table Action column), before the next condition check.

What happens if the initial condition is false?

The loop body does not run at all and the loop exits immediately (not shown here but implied by the flow).

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of i at iteration 2?

A 0

B 2

C 1

D 3

💡 Hint

Check the 'i' column in execution_table row for iteration 2.

At which iteration does the loop condition become false?

A Iteration 4

B Iteration 3

C Iteration 1

D Iteration 2

💡 Hint

Look at the Condition column in execution_table where it changes to false.

If we change the loop to i < 5, how many times will the loop run?

A 3 times

B 5 times

C 4 times

D 6 times

💡 Hint

The loop runs while i is less than the limit; check variable_tracker for i values.

Concept Snapshot

for (init; condition; update) {
  // loop body
}

1. Initialize variable
2. Check condition
3. Run body if true
4. Update variable
5. Repeat until condition false

Full Transcript

This example shows a for loop in C++ starting with i=0. Each step checks if i is less than 3. If yes, it prints i, then increases i by 1. This repeats until i reaches 3, when the condition fails and the loop stops. The variable i changes from 0 to 3 over the iterations. Key points are when the condition is checked and when i updates. The loop runs exactly three times, printing 0, 1, and 2.

Practice

1. What is the main purpose of a loop in C++ programming?

easy

A. To repeat a block of code multiple times

B. To store multiple values in a variable

C. To create a new data type

D. To stop the program execution immediately

Solution

1. Step 1: Understand the role of loops

   Loops are used to repeat a set of instructions multiple times without writing the code again and again.

2. Step 2: Identify the correct purpose

   Among the options, only repeating code matches the loop's purpose.

3. Final Answer:

   To repeat a block of code multiple times -> Option A

4. Quick Check:

   Loop purpose = repeat code [OK]

Hint: Loops repeat code blocks to save effort [OK]

Common Mistakes:

• Confusing loops with variables
• Thinking loops create new data types
• Believing loops stop program immediately

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

easy

A. for (int i = 0; i < 5; i++) /* code */

B. for int i = 0; i < 5; i++ { /* code */ }

C. for (int i = 0; i < 5; i++) { /* code */ }

D. for (int i = 0; i < 5; i++) [] { /* code */ }

Solution

1. Step 1: Recall the correct for loop syntax

   The for loop syntax requires parentheses around initialization, condition, and increment, followed by a block in braces.

2. Step 2: Check each option

   for (int i = 0; i < 5; i++) { /* code */ } matches the correct syntax exactly. Others miss parentheses, braces, or use wrong symbols.

3. Final Answer:

   for (int i = 0; i < 5; i++) { /* code */ } -> Option C

4. Quick Check:

   Correct for loop syntax = for (int i = 0; i < 5; i++) { /* code */ } [OK]

Hint: For loops need parentheses and braces [OK]

Common Mistakes:

• Omitting parentheses around conditions
• Missing braces for loop body
• Using wrong symbols like [] instead of {}

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

int count = 0;
for (int i = 1; i <= 3; i++) {
  count += i;
}
std::cout << count;

medium

A. Error

B. 6

C. 0

D. 3

Solution

1. Step 1: Trace the loop iterations

   The loop runs with i = 1, 2, 3. Each time, count adds i: 0+1=1, 1+2=3, 3+3=6.

2. Step 2: Determine final count value

   After the loop, count is 6, which is printed.

3. Final Answer:

   6 -> Option B

4. Quick Check:

   Sum 1+2+3 = 6 [OK]

Hint: Add loop variable each iteration to sum quickly [OK]

Common Mistakes:

• Confusing loop end condition
• Forgetting to add inside loop
• Assuming count resets inside loop

4. Identify the error in this loop code:

int i = 0;
while (i < 3) {
  std::cout << i << " ";
}

medium

A. Variable i should be declared inside the loop

B. Incorrect while loop syntax

C. Missing semicolon after std::cout

D. Missing increment of i inside the loop

Solution

1. Step 1: Analyze the while loop behavior

   The loop condition is i < 3, but i never changes inside the loop, causing an infinite loop.

2. Step 2: Identify the missing increment

   To progress, i must increase inside the loop, e.g., i++.

3. Final Answer:

   Missing increment of i inside the loop -> Option D

4. Quick Check:

   Loop needs variable update to avoid infinite loop [OK]

Hint: Always update loop variable inside while loops [OK]

Common Mistakes:

• Forgetting to increment loop variable
• Thinking syntax error causes infinite loop
• Declaring variable inside loop unnecessarily

5. Consider this code snippet:

int sum = 0;
for (int i = 1; i <= 5; i++) {
  if (i % 2 == 0) continue;
  sum += i;
}
std::cout << sum;

What is the output and why?

hard

A. 9, because only odd numbers 1, 3, 5 are added

B. 15, because all numbers 1 to 5 are added

C. 10, because even numbers are added

D. 0, because continue skips all additions

Solution

1. Step 1: Understand the loop and condition

   The loop runs i from 1 to 5. If i is even, continue skips adding it to sum.

2. Step 2: Calculate sum of odd numbers

   Odd numbers are 1, 3, 5. Their sum is 1+3+5 = 9.

3. Final Answer:

   9, because only odd numbers 1, 3, 5 are added -> Option A

4. Quick Check:

   Sum odd 1+3+5 = 9 [OK]

Hint: Continue skips even numbers, sum only odds [OK]

Common Mistakes:

• Adding even numbers by mistake
• Ignoring continue statement effect
• Summing all numbers instead of filtered ones