Ruby Interview Questions
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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 && falseAlthough 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. && / ||. -
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"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.) -
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"]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 } -
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)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). -
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.value2Explain your 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'. -
Is the line of code below valid Ruby code? If so, what does it do? Explain your answer.
Yes, it’s valid. Here’s how to understand what it does:-> (a) {p a}["Hello world"]
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”. -
What is the difference between calling super and calling super()?
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.)
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Explain each of the following operators and how and when they should be used: ==, ===, eql?, equal?.
== 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). === is the pattern matching operator!
=== matches regular expressions
=== checks range membership
=== checks being instance of a class
=== calls lambda expressions
=== sometimes checks equality, but mostly it does not
Some examplescase value when /regexp/ # value matches this regexp when 4..10 # value is in range when MyClass # value is an instance of class when ->(value) { ... } # lambda expression returns true when a, b, c, d # value matches one of a through d with `===` when *array # value matches an element in array with `===` when x # values is equal to x unless x is one of the above endeql? 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.)
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Given:
x = "hello"Explain the difference between:
x += " world"and
The += operator re-initializes the variable with a new value, so a += b is equivalent to a = a + b.x.concat " world"
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. -
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?
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.
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Write a single line of Ruby code that prints the Fibonacci sequence of any length as an array.
(Hint: use inject)
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.
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What does the “$” character mean in Ruby?
$ identifies a global variable, as opposed to a local variable, @instance variable, or @@class variable. This is one of the pre-defined Ruby global variables. All globals are prefixed with $, and in this case $$ represents the current process ID. This was inherited from Perl.