Accessing and setting values in Ruby - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Ruby code?

Consider the following Ruby code snippet. What will be printed when it runs?

Ruby

hash = {a: 1, b: 2}
hash[:c] = 3
puts hash[:b]

A 2

B 3

C KeyError

D nil

💡 Hint

Look at how the hash is accessed and what key is used in puts.

The hash initially has keys :a and :b with values 1 and 2. We add :c with value 3. Then we print the value for key :b, which is 2.

❓ Predict Output

intermediate

What does this code output after setting a nested value?

Look at this Ruby code that sets a nested value in a hash. What will it print?

Ruby

data = {user: {name: "Alice", age: 30}}
data[:user][:age] = 31
puts data[:user][:age]

A 30

B NoMethodError

C nil

D 31

💡 Hint

Check how the nested hash value is updated before printing.

The nested hash under :user has :age initially 30. It is updated to 31, so printing it outputs 31.

❓ Predict Output

advanced

What is the output of this Ruby code?

Consider the following Ruby code snippet. What will be printed when it runs?

Ruby

hash = {a: 1, b: 2}
puts hash[:c].to_s.length

A 0

B TypeError

C NoMethodError

D 1

💡 Hint

Think about what happens when accessing a missing key in a hash and calling to_s on nil.

Accessing hash[:c] returns nil since :c is missing. Calling to_s on nil returns an empty string "". Its length is 0.

❓ Predict Output

advanced

What is the final value of the variable after setting values?

What will be the value of the variable 'arr' after running this Ruby code?

Ruby

arr = [1, 2, 3]
arr[1] = 5
arr[3] = 7
p arr

A [1, 5, 3]

B [1, 5, 3, 7]

C [1, 5, 7]

D [1, 2, 3, 7]

💡 Hint

Remember how Ruby arrays handle assignment beyond current length.

Assigning arr[1] = 5 changes second element to 5. Assigning arr[3] = 7 adds a new element at index 3. The array becomes [1, 5, 3, 7].

🧠 Conceptual

expert

Which option produces the exact output '42'?

Given the following Ruby code, which option will print exactly '42'?

Ruby

value = {x: 40}
# Your code here
puts value[:x]

A value[:x] = 42.to_s

B value[:x] = '42'

C value[:x] += 2

D value[:x] = 40 + '2'

💡 Hint

Consider the type of value[:x] and how to get the number 42 printed.

Option C adds 2 to the integer 40, resulting in 42. Printing value[:x] outputs 42. Option C sets it to string '42' but printing will output '42' as string, which looks the same but is not the integer 42. Option C converts 42 to string, also prints '42' as string. Option C causes a TypeError because you cannot add integer and string.