Ruby interview questions

Write a single line of Ruby code that prints the Fibonacci sequence of any length as an array.

(Hint: use inject)

Answer:

There are multiple ways to do this, but one possible answer is:

(1..20).inject( [0, 1] ) { | fib | fib << fib.last(2).inject(:+) }

As you go up the sequence fib, you sum, or inject(:+), the last two elements in the array and add the result to the end of fib.

In Ruby code, you quite often see the trick of using an expression like array.map(&:method_name) as a shorthand form of array.map { |element| element.method_name }. How exactly does it work?

Answer:

When a parameter is passed with & in front of it (indicating that is it to be used as a block), Ruby will call to_proc on it in an attempt to make it usable as a block. Symbol#to_proc quite handily returns a Proc that will invoke the method of the corresponding name on whatever is passed to it, thus enabling our little shorthand trick to work.

Given:

x = "hello"

Explain the difference between:

x += " world"

and

x.concat " world"

Answer:

The += operator re-initializes the variable with a new value, so a += b is equivalent to a = a + b.

Therefore, while it may seem that += is mutating the value, it’s actually creating a new object and pointing the the old variable to that new object.

This is perhaps easier to understand if written as follows:

foo = "foo"
foo2 = foo
foo.concat "bar"
puts foo
=> "foobar"
puts foo2
=> "foobar"
foo += "baz"
puts foo
=> "foobarbaz"
puts foo2
=> "foobar"

(Examining the object_id of foo and foo2 will also demonstrate that new objects are being created.)

The difference has implications for performance and also has different mutation behavior than one might expect.

Explain each of the following operators and how and when they should be used: ==, ===, eql?, equal?.

Answer:

== – Checks if the value of two operands are equal (often overridden to provide a class-specific definition of equality).

=== – Specifically used to test equality within the when clause of a case statement (also often overridden to provide meaningful class-specific semantics in case statements).

eql? – Checks if the value and type of two operands are the same (as opposed to the == operator which compares values but ignores types). For example, 1 == 1.0 evaluates to true, whereas 1.eql?(1.0) evaluates to false.

equal? – Compares the identity of two objects; i.e., returns true iff both operands have the same object id (i.e., if they both refer to the same object). Note that this will return false when comparing two identical copies of the same object.

(Thanks to Ruby Gotchas for this question.)

What is the difference between calling super and calling super()?

Answer:

A call to super invokes the parent method with the same arguments that were passed to the child method. An error will therefore occur if the arguments passed to the child method don’t match what the parent is expecting.

A call to super() invokes the parent method without any arguments, as presumably expected. As always, being explicit in your code is a good thing.

(Thanks to Ruby Gotchas for this question.)

Is the line of code below valid Ruby code? If so, what does it do? Explain your answer.

-> (a) {p a}["Hello world"]

Answer:

Yes, it’s valid. Here’s how to understand what it does:

The -> operator creates a new Proc, which is one of Ruby’s function types. (The -> is often called the “stabby proc”. It’s also called the “stabby lambda”, as it creates a new Proc instance that is a lambda. All lambdas are Procs, but not all Procs are lambdas. There are some slight differences between the two.)

This particular Proc takes one parameter (namely, a). When the Proc is called, Ruby executes the block p a, which is the equivalent of puts(a.inspect) (a subtle, but useful, difference which is why p is sometimes better than puts for debugging). So this Proc simply prints out the string that is passed to it.

You can call a Proc by using either the call method on Proc, or by using the square bracket syntax, so this line of code also invokes the Proc and passes it the string “Hello World”.

So putting that all together, this line of code (a) creates a Proc that takes a single parameter a which it prints out and (b) invokes that Proc and passes it the string “Hello world”. So, in short, this line of code prints “Hello World”.

Consider the following code:

VAL = 'Global'
module Foo
  VAL = 'Foo Local'
  class Bar
    def value1
      VAL
    end
  end
end
class Foo::Bar
  def value2
    VAL
  end
end

What will be the value of each of the following:

Foo::Bar.new.value1
Foo::Bar.new.value2

Explain your answer.

Answer:

Foo::Bar.new.value1 will be equal to 'Foo Local' and Foo::Bar.new.value2 will be equal to 'Global'.

Here’s why:

The module keyword (as well as the class and def keywords) will create a new lexical scope for all of its contents. The above module Foo therefore creates the scope Foo in which the VAL constant equal to 'Foo Local' is defined. Inside Foo, we declare class Bar, which creates another new lexical scope (named Foo::Bar) which also has access to its parent scope (i.e., Foo) and all of its constants.

However, when we then declare Foo::Bar (i.e., using ::), we are actually creating yet another lexical scope, which is also named Foo::Bar (how’s that for confusing!). However, this lexical scope has no parent (i.e., it is entirely independent of the lexcial scope Foo created earlier) and therefore does not have any access to the contents of the ‘Foo’ scope. Therefore, inside class Foo::Bar, we only have access to the VAL constant declared at the beginning of the script (i.e., outside of any module) with the value 'Global'.

Consider the following two methods:

def times_two(arg1);
  puts arg1 * 2;
end
def sum(arg1, arg2);
  puts arg1 + arg2;
end

What will be the result of each of the following lines of code:

times_two 5
times_two(5)
times_two (5)
sum 1, 2
sum(1, 2)
sum (1, 2)

Answer:

The first three lines of code will all print out 10, as expected.

The next two lines of code will both print out 3, as expected.

However, the last line of code (i.e., sum (1,2)) will result in the following:

syntax error, unexpected ',', expecting ')'
sum (1, 2)
       ^

The problem is the space between the method name and the open parenthesis. Because of the space, the Ruby parser thinks that (1, 2) is an expression that represents a single argument, but (1, 2) is not a valid Ruby expression, hence the error.

Note that the problem does not occur with single argument methods (as shown with our timesTwo method above), since the single value is a valid expression (e.g., (5) is a valid expression which simply evaluates to 5).

Write a function that sorts the keys in a hash by the length of the key as a string. For instance, the hash:

{ abc: 'hello', 'another_key' => 123, 4567 => 'third' }

should result in:

["abc", "4567", "another_key"]

Answer:

As is always true in programming, there are in fact multiple ways to accomplish this.

The most straightforward answer would be of the form:

hsh.keys.map(&:to_s).sort_by(&:length)

or:

hsh.keys.collect(&:to_s).sort_by { |key| key.length }

Alternatively, Ruby’s Enumerable mixin provides many methods to operate on collections. The key here is to turn the hash keys into a collection, convert them all to strings, then sort the array.

def key_sort hsh
	hsh.keys.collect(&:to_s).sort { |a, b| a.length <=> b.length }
end

An equivalent call of the collect method is done with the usual block syntax of:

collect { |x| x.to_s }

Which of the expressions listed below will result in "false"?

true    ? "true" : "false"
false   ? "true" : "false"
nil     ? "true" : "false"
1       ? "true" : "false"
0       ? "true" : "false"
"false" ? "true" : "false"
""      ? "true" : "false"
[]      ? "true" : "false"

Answer:

In Ruby, the only values that evaluate to false are false and nil. Everything else – even zero (0) and an empty array ([]) – evaluates to true.

This comes as a real surprise to programmers who have previously been working in other languages like JavaScript.

(Thanks to Ruby Gotchas for this question.)

What will val1 and val2 equal after the code below is executed? Explain your answer.

val1 = true and false  # hint: output of this statement in IRB is NOT value of val1!
val2 = true && false

Answer:

Although these two statements might appear to be equivalent, they are not, due to the order of operations. Specifically, the and and or operators have lower precedence than the = operator, whereas the && and || operators have higher precedence than the = operator, based on order of operations.

To help clarify this, here’s the same code, but employing parentheses to clarify the default order of operations:

(val1 = true) and false    # results in val1 being equal to true
val2 = (true && false)     # results in val2 being equal to false

This is, incidentally, a great example of why using parentheses to clearly specify your intent is generally a good practice, in any language. But whether or not you use parentheses, it’s important to be aware of these order of operations rules and to thereby ensure that you are properly determining when to employ and / or vs. && / ||.

Update the code, so it can run successfully:

band = "Blink" + 182

Answer:

We cannot concatenate a String and an Integer. However, we can convert the Integer to a String first and then concatenate the values as two Strings.

band = "Blink" + 182.to_s

Integers have useful built-in methods too. Convert the number 5 to the string "5".

Answer:

5.to_s

to_s stands for "to string"

Concatenate the following strings:

first = "Beautiful "
second = "face tattoo"

Answer:

first + second
# OR
first.+(second)
# OR
first.concat(second)

The +() and concat() methods do the same thing (they're not exactly the same, but very similar). Notice the similarity with the previous examples: there is a string, followed by a dot, followed by the method names with another string as an argument.

What does the following expression evaluate to?

"i am not shouting".upcase()

Answer:

"I AM NOT SHOUTING"

This shows that Ruby has nifty methods that are built in to the String class. The syntax for using the built in methods is the value, followed by dot, followed by the method name.

What does the following expression evaluate to?

2 ** 3

Answer:

8

** is used for exponents and 2 ** 3 is 2 to the power 3 or 2 * 2 * 2.

What does the following expression print?

something = "cats"
crazy = something
puts crazy

Answer:

This will print "cats" because the variables something and crazy both point to the same value.

Replace the "l" in the following string with "m":

word = "lace"

Answer:

word[0] = "m"
puts word
=> "mace"

Get the last element from the "Quizbucket" string.

Answer:

"Quizbucket"[-1]

Get the first through 3rd elements from the "Jaydakiss" string.

Answer:

"Jaydakiss"[0..2]