Continue statement in C++ - Step-by-Step Execution

Concept Flow - Continue statement

Start loop iteration

↓

Check condition

Yes↓

Check if continue condition

Yes|No↓

Skip rest of loop

↓

Next iteration or exit

The loop checks a condition each iteration. If the continue condition is true, it skips the rest of the loop body and moves to the next iteration.

Execution Sample

C++

for (int i = 1; i <= 5; i++) {
    if (i == 3) continue;
    std::cout << i << " ";
}

Print numbers 1 to 5 but skip printing 3 using continue.

Execution Table

Iteration	i	Condition i <= 5	Check if i == 3	Action	Output
1	1	True	False	Print 1	1
2	2	True	False	Print 2	2
3	3	True	True	Continue (skip print)
4	4	True	False	Print 4	4
5	5	True	False	Print 5	5
6	6	False	-	Exit loop

💡 i reaches 6, condition i <= 5 is False, loop ends

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	Final
i	1	2	3	4	5	6	6

Key Moments - 2 Insights

Why does the number 3 not get printed even though the loop runs when i is 3?

At iteration 3 in the execution_table, the condition i == 3 is True, so the continue statement skips the print action and moves to the next iteration.

Does the continue statement stop the entire loop?

No, continue only skips the rest of the current iteration. The loop continues with the next iteration as shown in rows 3 and 4 of the execution_table.

Visual Quiz - 3 Questions

Look at the execution_table, what is the output after iteration 2?

A "1 2 3 "

B "2 "

C "1 2 "

D ""

💡 Hint

Check the Output column after iteration 2 in the execution_table.

At which iteration does the continue statement cause the loop to skip printing?

A Iteration 3

B Iteration 1

C Iteration 5

D Iteration 6

💡 Hint

Look at the 'Check if i == 3' and 'Action' columns in the execution_table.

If the continue statement was removed, what would be the output after iteration 5?

A "3 "

B "1 2 3 4 5 "

C "1 2 4 5 "

D ""

💡 Hint

Without continue, the print happens every iteration where i <= 5.

Concept Snapshot

continue statement in C++:
- Used inside loops to skip the rest of current iteration
- Control jumps to next iteration immediately
- Syntax: if(condition) continue;
- Useful to skip specific cases without breaking the loop
- Loop continues until its main condition fails

Full Transcript

This example shows a for loop from 1 to 5. Each time, it checks if i equals 3. If yes, continue skips printing 3 and moves to next iteration. Otherwise, it prints the number. The loop ends when i becomes 6 because the condition i <= 5 is false. The continue statement only skips the current iteration's remaining code, not the whole loop.

Practice

1. What does the continue statement do inside a loop in C++?

easy

A. It stops the entire loop immediately.

B. It skips the rest of the current loop iteration and moves to the next iteration.

C. It exits the current function.

D. It restarts the program from the beginning.

Solution

1. Step 1: Understand the role of continue in loops

   The continue statement tells the program to skip the remaining code in the current loop cycle.

2. Step 2: Identify what happens after continue

   After skipping, the loop moves to the next iteration immediately without executing the rest of the loop body.

3. Final Answer:

   It skips the rest of the current loop iteration and moves to the next iteration. -> Option B

4. Quick Check:

   Continue skips current iteration [OK]

Hint: Continue skips current loop step, not whole loop [OK]

Common Mistakes:

• Thinking continue stops the loop completely
• Confusing continue with break
• Assuming continue exits the function

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

easy

A. continue;

B. continue()

C. continue{}

D. continue[]

Solution

1. Step 1: Recall the syntax of continue in C++

   The continue statement is a keyword and used alone followed by a semicolon.

2. Step 2: Check each option's syntax

   Only continue; is valid. The others use parentheses or braces which are incorrect.

3. Final Answer:

   continue; -> Option A

4. Quick Check:

   Continue ends with semicolon [OK]

Hint: Continue is a keyword, ends with semicolon [OK]

Common Mistakes:

• Adding parentheses after continue
• Using braces or brackets with continue
• Omitting the semicolon

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

for (int i = 1; i <= 5; i++) {
    if (i == 3) continue;
    std::cout << i << " ";
}

medium

A. 1 2 3 4 5

B. 1 2 3 4

C. 1 2 4 5

D. 2 3 4 5

Solution

1. Step 1: Analyze the loop and condition

   The loop runs from 1 to 5. When i == 3, the continue skips printing.

2. Step 2: Determine printed values

   Numbers 1 and 2 print normally. At 3, continue skips printing. Then 4 and 5 print.

3. Final Answer:

   1 2 4 5 -> Option C

4. Quick Check:

   Skip 3, print others [OK]

Hint: Continue skips printing 3, others print [OK]

Common Mistakes:

• Including 3 in output
• Stopping loop at 3
• Printing numbers after 3 incorrectly

4. Identify the error in the following C++ code snippet:

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

medium

A. No error in the code.

B. Incorrect loop condition.

C. Wrong variable type for i.

D. Missing semicolon after continue statement.

Solution

1. Step 1: Check syntax of continue statement

   The continue statement must end with a semicolon.

2. Step 2: Identify missing semicolon

   In the code, continue is not followed by a semicolon, causing a syntax error.

3. Final Answer:

   Missing semicolon after continue statement. -> Option D

4. Quick Check:

   Continue needs semicolon [OK]

Hint: Always put semicolon after continue [OK]

Common Mistakes:

• Forgetting semicolon after continue
• Assuming continue needs parentheses
• Misreading loop condition as error

5. Consider the following code that sums numbers from 1 to 10 but skips multiples of 3 using continue. What is the final sum?

int sum = 0;
for (int i = 1; i <= 10; i++) {
    if (i % 3 == 0) continue;
    sum += i;
}
std::cout << sum;

hard

A. 52

B. 55

C. 45

D. 50

Solution

1. Step 1: Identify numbers skipped by continue

   Numbers divisible by 3 (3, 6, 9) are skipped and not added to sum.

2. Step 2: Calculate sum of numbers 1 to 10 excluding 3, 6, 9

   Sum 1+2+4+5+7+8+10 = 37 + 15 = 52

3. Final Answer:

   52 -> Option A

4. Quick Check:

   Sum excluding multiples of 3 = 52 [OK]

Hint: Skip multiples of 3, sum others [OK]

Common Mistakes:

• Including multiples of 3 in sum
• Adding all numbers 1 to 10
• Misunderstanding continue effect