Continue statement - Mini Project: Build & Apply

Using the Continue Statement in C

📖 Scenario: Imagine you are sorting through a list of numbers and want to skip certain numbers based on a rule, like ignoring even numbers.

🎯 Goal: You will write a C program that uses the continue statement inside a loop to skip printing even numbers from a list.

📋 What You'll Learn

Create an array of integers with specific values

Create a variable to hold the length of the array

Use a for loop to go through the array

Use the continue statement to skip even numbers

Print only the odd numbers

💡 Why This Matters

🌍 Real World

Skipping unwanted data while processing lists or arrays is common in many programs, such as filtering out invalid inputs or ignoring certain cases.

💼 Career

Understanding how to control loops with statements like continue is essential for writing efficient and clear code in software development.

1

Create an integer array

Create an integer array called numbers with these exact values: 2, 5, 8, 11, 14, 17.

C

#include 

int main() {
    // Create the integer array called numbers
    // Your code here

    return 0;
}

Hint

Use the syntax int numbers[] = {values}; to create the array.

2

Create a variable for array length

Create an integer variable called length and set it to the number of elements in the numbers array using sizeof(numbers) / sizeof(numbers[0]).

C

#include 

int main() {
    int numbers[] = {2, 5, 8, 11, 14, 17};
    // Create the length variable for the array
    // Your code here

    return 0;
}

Hint

Use sizeof to find the total size of the array and divide by the size of one element.

3

Use a for loop with continue to skip even numbers

Write a for loop using int i = 0; i < length; i++ to go through numbers. Inside the loop, use an if statement to check if numbers[i] is even (use numbers[i] % 2 == 0). If it is even, use the continue statement to skip to the next loop iteration.

C

#include 

int main() {
    int numbers[] = {2, 5, 8, 11, 14, 17};
    int length = sizeof(numbers) / sizeof(numbers[0]);

    // Loop through the array and skip even numbers using continue
    // Your code here

    return 0;
}

Hint

Use continue; inside the if block to skip even numbers.

4

Print only the odd numbers

Inside the for loop, after the continue statement, add a printf statement to print the odd numbers with a space after each number. Use printf("%d ", numbers[i]);. After the loop, print a newline with printf("\n");.

C

#include 

int main() {
    int numbers[] = {2, 5, 8, 11, 14, 17};
    int length = sizeof(numbers) / sizeof(numbers[0]);

    for (int i = 0; i < length; i++) {
        if (numbers[i] % 2 == 0) {
            continue;
        }
        // Print the odd number
        // Your code here
    }
    // Print a newline after the loop
    // Your code here

    return 0;
}

Hint

Use printf("%d ", numbers[i]); to print each odd number followed by a space.

Don't forget to print a newline after the 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