Lazy enumerators in Ruby - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of lazy enumerator with chained operations

What is the output of this Ruby code using lazy enumerators?

Ruby

result = (1..Float::INFINITY).lazy.select { |x| x % 3 == 0 }.map { |x| x * 2 }.first(5)
puts result.inspect

A [1, 2, 3, 4, 5]

B [3, 6, 9, 12, 15]

C [2, 4, 6, 8, 10]

D [6, 12, 18, 24, 30]

💡 Hint

Remember that lazy enumerators only compute values as needed and the select filters multiples of 3.

The code selects numbers divisible by 3 from an infinite range, then doubles them. The first five such numbers are 3,6,9,12,15, doubled to 6,12,18,24,30.

🧠 Conceptual

intermediate

Behavior of lazy enumerators with infinite sequences

Which statement best describes how Ruby's lazy enumerators handle infinite sequences?

A They cannot be used with infinite sequences.

B They compute elements only when needed, avoiding infinite loops.

C They compute all elements at once, causing memory overflow.

D They always return an array immediately.

💡 Hint

Think about how lazy evaluation delays computation.

Lazy enumerators compute elements on demand, so they can handle infinite sequences without running forever or using too much memory.

🔧 Debug

advanced

Identify the error in lazy enumerator chaining

What error does this Ruby code raise?

Ruby

result = (1..10).lazy.map { |x| x * 2 }.select { |x| x > 10 }
puts result

A RuntimeError: infinite loop detected

B TypeError: no implicit conversion of nil into Integer

C No error, prints a lazy enumerator object

D ArgumentError: wrong number of arguments (given 0, expected 1)

💡 Hint

Printing a lazy enumerator does not execute it.

The code creates a lazy enumerator and prints it. Printing a lazy enumerator shows its object representation without error.

❓ Predict Output

advanced

Output of lazy enumerator with take and map

What is the output of this Ruby code?

Ruby

result = (1..Float::INFINITY).lazy.map { |x| x * x }.take(4).force
puts result.inspect

A [1, 4, 9, 16]

B [1, 2, 3, 4]

C [1, 8, 27, 64]

D [2, 4, 6, 8]

💡 Hint

The map squares each number, take limits to first 4, force evaluates the lazy enumerator.

The code squares numbers starting from 1, takes the first 4 squares, and forces evaluation to get [1,4,9,16].

📝 Syntax

expert

Which option causes a syntax error with lazy enumerators?

Which of these Ruby code snippets will cause a syntax error?

A (1..10).lazy.select { |x| x.even? }.map ->(x) { x * 3 }

B (1..10).lazy.select { |x| x.even? }.map { |x| x * 3 }

C (1..10).lazy.select { |x| x.even? }.map { |x| x * 3 }.force

D (1..10).lazy.select { |x| x.even? }.map { |x| x * 3 }.first(3)

💡 Hint

Check the syntax of the block passed to map.

Option A uses incorrect syntax for passing a lambda to map; it should be map(&lambda) or map { ... }.