Continue statement - Step-by-Step Execution

Concept Flow - Continue statement

Start loop iteration

↓

Check condition

Yes↓

Check continue condition

↓

Skip rest

↓

Go to next

↓

Next iteration or exit

The loop starts each iteration by checking the main condition. If the continue condition is met, it skips the rest of the loop body and jumps to the next iteration.

Execution Sample

C

for (int i = 1; i <= 5; i++) {
    if (i == 3) continue;
    printf("%d ", i);
}

This code prints numbers from 1 to 5 but skips printing 3 using the continue statement.

Execution Table

Iteration	i	Condition (i <= 5)	Continue check (i == 3)	Action	Output
1	1	True	False	Print 1	1
2	2	True	False	Print 2	2
3	3	True	True	Skip print, continue to next
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 reaches i=3?

At iteration 3 in the execution_table, the continue check is True, so the loop skips the print statement and moves to the next iteration.

Does the continue statement stop the whole loop?

No, as shown in the execution_table, continue only skips the current iteration's remaining code and proceeds with the next iteration until the loop condition fails.

Visual Quiz - 3 Questions

Look at the execution_table, what is the output when i equals 3?

A "3 " is printed

B Nothing is printed

C "Continue" is printed

D Loop exits

💡 Hint

Check the row where Iteration is 3 and see the Action and Output columns.

At which iteration does the loop condition become false?

A 6

B 3

C 5

D 1

💡 Hint

Look at the last row in the execution_table where Condition (i <= 5) is False.

If the continue statement was removed, what would be the output at iteration 3?

A Nothing would be printed

B Loop would exit early

C "3 " would be printed

D Error occurs

💡 Hint

Refer to the Action and Output columns for iteration 3 and imagine no skipping happens.

Concept Snapshot

continue statement in C:
- Used inside loops to skip the rest of current iteration
- Control jumps immediately to next iteration check
- Syntax: if(condition) continue;
- Does NOT exit the loop, just skips current iteration
- Useful to ignore specific cases without breaking loop

Full Transcript

This example shows how the continue statement works in a C for loop. The loop runs from i=1 to i=5. When i equals 3, the continue condition is true, so the loop skips printing 3 and moves to the next iteration. For other values, 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 and does not stop the whole loop.

Practice

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

easy

A. Skips the rest of the current loop iteration and moves to the next iteration

B. Stops the entire loop immediately

C. Exits the current function

D. Restarts the program from the beginning

Solution

1. Step 1: Understand the role of continue

   The continue statement does not stop the loop but skips the remaining code in the current iteration.

2. Step 2: Effect on loop execution

   After continue, the loop immediately proceeds to the next iteration, ignoring any code below it in the current cycle.

3. Final Answer:

   Skips the rest of the current loop iteration and moves to the next iteration -> Option A

4. Quick Check:

   continue skips current iteration [OK]

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

Common Mistakes:

• Confusing continue with break
• Thinking continue exits the function
• Assuming continue restarts the program

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 correct statement syntax

   In C, continue is a statement and must end with a semicolon.

2. Step 2: Identify valid syntax

   Only continue; is valid; parentheses, braces, or brackets are incorrect.

3. Final Answer:

   continue; -> Option A

4. Quick Check:

   Statement ends with semicolon [OK]

Hint: Statements end with semicolon; continue is no exception [OK]

Common Mistakes:

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

3. What is the output of the following C code?

for (int i = 1; i <= 5; i++) {
    if (i == 3) continue;
    printf("%d ", i);
}

medium

A. 1 2 3 4 5

B. 1 2 4 5

C. 1 2 3 4

D. 2 3 4 5

Solution

1. Step 1: Analyze loop iterations

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

2. Step 2: Determine printed values

   Values 1, 2, 4, and 5 are printed with spaces. The number 3 is skipped.

3. Final Answer:

   1 2 4 5 -> Option B

4. Quick Check:

   continue skips printing 3 [OK]

Hint: Skip printing when continue triggers, others print normally [OK]

Common Mistakes:

• Including the skipped number in output
• Confusing continue with break
• Ignoring the loop condition

4. Identify the error in this code snippet using continue:

int i = 0;
while (i < 5) {
    if (i == 2)
        continue;
    printf("%d ", i);
    i++;
}

medium

A. Syntax error: continue cannot be used in while loops

B. The printf statement is unreachable

C. The loop will run infinitely because 'i' is not incremented when i == 2

D. The variable 'i' is not declared properly

Solution

1. Step 1: Understand loop flow with continue

   When i equals 2, continue skips the rest of the loop body, so i++ is not executed.

2. Step 2: Consequence of skipping i++

   Since i is not incremented at i == 2, the loop condition remains true forever, causing an infinite loop.

3. Final Answer:

   The loop will run infinitely because 'i' is not incremented when i == 2 -> Option C

4. Quick Check:

   continue skips i++ causing infinite loop [OK]

Hint: Always increment loop variable before continue to avoid infinite loops [OK]

Common Mistakes:

• Thinking continue causes syntax error
• Assuming printf is unreachable
• Ignoring loop variable increment

5. How can you modify this code to print only odd numbers from 1 to 10 using continue?

for (int i = 1; i <= 10; i++) {
    printf("%d ", i);
}

hard

A. Add if (i == 2) continue; before printf

B. Add if (i % 2 != 0) continue; before printf

C. Replace printf with if (i % 2 != 0) printf("%d ", i);

D. Add if (i % 2 == 0) continue; before printf

Solution

1. Step 1: Identify condition to skip even numbers

   Even numbers satisfy i % 2 == 0. Using continue here skips printing even numbers.

2. Step 2: Place continue before printf

   Adding if (i % 2 == 0) continue; before printf skips printing even numbers, so only odd numbers print.

3. Final Answer:

   Add if (i % 2 == 0) continue; before printf -> Option D

4. Quick Check:

   continue skips even numbers, prints odd [OK]

Hint: Skip even numbers with continue to print only odds [OK]

Common Mistakes:

• Skipping odd numbers instead of even
• Using continue after printf
• Incorrect condition for even numbers