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 a variable, then checks a condition. If true, it runs the loop body, updates the variable, and repeats. If false, it stops.

Execution Sample

C

for (int i = 0; i < 5; i++) {
    printf("%d\n", i);
}

Prints numbers 0 to 4, one per line, using a for loop.

Execution Table

Iteration	i value	Condition (i < 5)	Action	Output
1	0	True	Print 0	0
2	1	True	Print 1	1
3	2	True	Print 2	2
4	3	True	Print 3	3
5	4	True	Print 4	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 equals 5?

Because the condition i < 5 becomes false at i = 5, so the loop exits as shown in execution_table row 6.

When is the variable i updated in the loop?

i is updated after the loop body executes, before the next condition check, as seen between each iteration in variable_tracker.

Does the loop body run when the condition is false?

No, the loop body only runs when the condition is true. When i = 5, condition is false, so the loop body does not run (execution_table row 6).

Visual Quiz - 3 Questions

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

A 3

B 1

C 2

D 4

💡 Hint

Check the 'i value' column in execution_table row 3.

At which iteration does the loop condition become false?

A 6

B 4

C 5

D 1

💡 Hint

Look at the 'Condition (i < 5)' column in execution_table row 6.

If we change the loop to i < 3, how many times will the loop run?

A 5 times

B 3 times

C 2 times

D 4 times

💡 Hint

Refer to the pattern in variable_tracker and execution_table for how condition controls iterations.

Concept Snapshot

for loop syntax in C:
for (initialization; condition; update) {
  // code to repeat
}

Loop runs while condition is true.
Variable updates after each iteration.
Stops when condition is false.

Full Transcript

A for loop in C starts by setting a variable, here i=0. It checks if i is less than 5. If yes, it runs the code inside the loop, printing i. Then it adds 1 to i. This repeats until i reaches 5, when the condition fails and the loop stops. The variable i changes from 0 up to 5, but the loop body runs only when i is 0 to 4. This is shown step-by-step in the execution table and variable tracker.

Practice

1. What does a for loop in C primarily do?

easy

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

B. Defines a new variable

C. Stops the program execution

D. Declares a function

Solution

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

   A for loop is designed to repeat a block of 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 purpose.

3. Final Answer:

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

4. Quick Check:

   For loop = repeat code [OK]

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

Common Mistakes:

• Confusing for loop with variable declaration
• Thinking for loop stops program
• Mixing for loop with function declaration

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

easy

A. for i = 0 to 5 { }

B. for (int i = 0; i++ < 5;) { }

C. for (i < 5; i = 0; i++) { }

D. for (int i = 0; i < 5; i++) { }

Solution

1. Step 1: Review standard for loop syntax in C

   The correct syntax is: for (initialization; condition; update) { body }

2. Step 2: Check each option

   for (int i = 0; i < 5; i++) { } matches the correct syntax with initialization, condition, and update in proper order.

3. Final Answer:

   for (int i = 0; i < 5; i++) { } -> Option D

4. Quick Check:

   Syntax = for(init; cond; update) [OK]

Hint: For loop syntax: for(init; condition; update) [OK]

Common Mistakes:

• Using 'for i = 0 to 5' which is not C syntax
• Swapping condition and initialization
• Missing semicolons inside for parentheses

3. What is the output of this code?

for(int i = 1; i <= 3; i++) {
    printf("%d ", i);
}

medium

A. 3 2 1

B. 0 1 2 3

C. 1 2 3

D. 1 2 3 4

Solution

1. Step 1: Analyze the loop range

   The loop starts at i=1 and runs while i is less than or equal to 3, incrementing i by 1 each time.

2. Step 2: Determine printed values

   It prints i each time: 1, then 2, then 3, then stops.

3. Final Answer:

   1 2 3 -> Option C

4. Quick Check:

   Loop prints 1 to 3 [OK]

Hint: Check loop start, end, and increments carefully [OK]

Common Mistakes:

• Starting loop at 0 instead of 1
• Including an extra number beyond 3
• Printing in reverse order

4. Identify the error in this for loop:

for(int i = 0; i < 5; i)
    printf("%d ", i);

medium

A. Missing semicolon after printf statement

B. Missing increment operator in for loop update

C. Incorrect loop condition syntax

D. Variable i not declared

Solution

1. Step 1: Examine the for loop update part

   The update section is 'i' without increment operator, so i never changes.

2. Step 2: Understand the effect of missing increment

   Without 'i++' or similar, the loop runs infinitely or behaves incorrectly.

3. Final Answer:

   Missing increment operator in for loop update -> Option B

4. Quick Check:

   Update must change loop variable [OK]

Hint: For loop update must change variable (e.g., i++) [OK]

Common Mistakes:

• Forgetting to increment loop variable
• Confusing semicolon placement
• Assuming variable auto-increments

5. How can you use a for loop to print only even numbers from 2 to 10 inclusive in C?

hard

A. for(int i = 2; i <= 10; i += 2) { printf("%d ", i); }

B. for(int i = 0; i <= 10; i++) { if(i % 2 == 0) printf("%d ", i); }

C. for(int i = 2; i < 10; i++) { if(i % 2 == 0) printf("%d ", i); }

D. for(int i = 0; i <= 10; i += 3) { printf("%d ", i); }

Solution

1. Step 1: Understand the goal

   We want to print even numbers from 2 to 10 inclusive.

2. Step 2: Check each option

   for(int i = 2; i <= 10; i += 2) { printf("%d ", i); } starts at 2 and increments by 2, printing 2,4,6,8,10 exactly.

3. Final Answer:

   for(int i = 2; i <= 10; i += 2) { printf("%d ", i); } -> Option A

4. Quick Check:

   Increment by 2 prints evens directly [OK]

Hint: Increment by 2 to print evens directly [OK]

Common Mistakes:

• Using wrong increment step
• Off-by-one errors in loop condition
• Printing odd numbers by mistake