Loop execution flow in Rust - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print numbers from 1 to 5 using a loop.

Rust

for i in 1..=[1] {
    println!("{}", i);
}

Drag options to blanks, or click blank then click option'

A10

B6

C0

D5

Attempts:

3 left

💡 Hint

Common Mistakes

Using 6 will print numbers up to 6, which is one more than needed.

Using 0 will cause the loop to not run as expected.

The range 1..=5 includes numbers from 1 to 5, so the loop prints 1 to 5.

2fill in blank

medium

Complete the code to skip printing the number 3 inside the loop.

Rust

for i in 1..=5 {
    if i == [1] {
        continue;
    }
    println!("{}", i);
}

Drag options to blanks, or click blank then click option'

A2

B4

C3

D5

Attempts:

3 left

💡 Hint

Common Mistakes

Using a number other than 3 will skip the wrong iteration.

Forgetting to use continue will print all numbers.

The continue statement skips the current loop iteration when i equals 3, so 3 is not printed.

3fill in blank

hard

Fix the error in the loop to stop printing numbers when i equals 4.

Rust

for i in 1..=5 {
    if i == [1] {
        break;
    }
    println!("{}", i);
}

Drag options to blanks, or click blank then click option'

A4

B5

C6

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Using 5 will print 4 before stopping, which is not desired.

Using 3 will stop too early.

The break statement stops the loop when i equals 4, so numbers 1 to 3 are printed.

4fill in blank

hard

Fill both blanks to create a loop that prints only even numbers from 1 to 10.

Rust

for i in 1..=[1] {
    if i % 2 [2] 0 {
        println!("{}", i);
    }
}

Drag options to blanks, or click blank then click option'

A10

B==

C!=

D5

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' will print odd numbers instead.

Using 5 as upper limit will miss numbers 6 to 10.

The loop runs from 1 to 10. The condition i % 2 == 0 checks if i is even, so only even numbers are printed.

5fill in blank

hard

Fill all three blanks to create a loop that collects numbers greater than 3 into a vector.

Rust

let mut result = Vec::new();
for num in 1..=[1] {
    if num [2] [3] {
        result.push(num);
    }
}
println!("{:?}", result);

Drag options to blanks, or click blank then click option'

A5

B>

C3

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' instead of '>' will only select number 3, not greater numbers.

Using 3 as upper limit will miss number 4 and 5.

The loop runs from 1 to 5. The condition num > 3 selects numbers greater than 3, which are pushed into the vector. The output will be [4, 5].

Practice

1. What does the break statement do inside a Rust loop?

easy

A. It restarts the loop from the beginning.

B. It skips the current iteration and continues with the next one.

C. It pauses the loop until a condition is met.

D. It stops the loop immediately and exits it.

Solution

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

   The break statement immediately stops the loop and exits it.

2. Step 2: Compare with other loop controls

   continue skips to the next iteration, but break ends the loop entirely.

3. Final Answer:

   It stops the loop immediately and exits it. -> Option D

4. Quick Check:

   break stops loop = C [OK]

Hint: Remember: break = stop loop, continue = skip iteration [OK]

Common Mistakes:

• Confusing break with continue
• Thinking break pauses instead of stops
• Assuming break restarts the loop

2. Which of the following is the correct syntax to skip the current iteration in a Rust for loop?

easy

A. skip;

B. continue;

C. break;

D. next;

Solution

1. Step 1: Identify the keyword to skip iteration

   In Rust, continue; skips the rest of the current loop iteration and moves to the next.

2. Step 2: Check other options for correctness

   break; stops the loop, skip; and next; are not valid Rust keywords for this purpose.

3. Final Answer:

   continue; -> Option B

4. Quick Check:

   Skip iteration = continue [OK]

Hint: Use continue to skip current loop iteration [OK]

Common Mistakes:

• Using break instead of continue
• Using non-existent keywords like skip or next
• Omitting the semicolon

3. What will be the output of this Rust code?

let mut count = 0;
loop {
    count += 1;
    if count == 3 {
        break;
    }
    println!("{}", count);
}

medium

A. 1 2 3

B. 0 1 2

C. 1 2

D. 3

Solution

1. Step 1: Trace the loop iterations and print statements

   Loop increments count from 0 to 1, prints 1; then 2, prints 2; when count is 3, break stops loop before printing.

2. Step 2: Determine printed values

   Only 1 and 2 are printed before break stops the loop.

3. Final Answer:

   1 2 -> Option C

4. Quick Check:

   Prints before break = 1 2 [OK]

Hint: Break stops before printing count 3 [OK]

Common Mistakes:

• Assuming count 3 is printed
• Starting count from 1 instead of 0
• Confusing break and continue effects

4. Identify the error in this Rust loop code:

for i in 1..5 {
    if i == 3 {
        continue
    }
    println!("{}", i);
}

medium

A. Missing semicolon after continue

B. Incorrect range syntax

C. println! macro is used incorrectly

D. continue cannot be used inside for loops

Solution

1. Step 1: Check syntax of continue statement

   In Rust, statements like continue require a semicolon to end the statement.

2. Step 2: Verify other parts of the code

   The range 1..5 and println! usage are correct; continue is valid inside for loops.

3. Final Answer:

   Missing semicolon after continue -> Option A

4. Quick Check:

   Statements need semicolon = A [OK]

Hint: Add semicolon after continue to fix syntax [OK]

Common Mistakes:

• Forgetting semicolon after continue
• Thinking continue is invalid in for loops
• Misreading range syntax

5. Given this Rust code, what will be the final value of sum?

let mut sum = 0;
for i in 1..6 {
    if i % 2 == 0 {
        continue;
    }
    if i == 5 {
        break;
    }
    sum += i;
}

hard

A. 4

B. 5

C. 6

D. 9

Solution

1. Step 1: Analyze loop iterations and conditions

   Loop runs i from 1 to 5 (exclusive 6). For even i, continue skips adding. For i == 5, break stops loop before adding.

2. Step 2: Calculate sum for valid i values

   i=1 (odd, add 1), i=2 (even, skip), i=3 (odd, add 3), i=4 (even, skip), i=5 (break before add). sum = 1 + 3 = 4.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   Sum odd before 5 = 4 [OK]

Hint: Add only odd numbers before 5, skip evens [OK]

Common Mistakes:

• Adding even numbers mistakenly
• Including 5 in sum despite break
• Confusing continue and break effects