In this HackerRank Ruby – Methods – Keyword Arguments problem solution Ruby allows you to (partially) mitigate this problem by passing a Hash as an argument or one of the arguments. For example, you have a method that takes a URI to download a file and another argument containing a Hash of other named options (proxy, timeout, active-connections etc.,)
def fetch_file(uri, options)
if options.has_key?(:proxy)
# do something
end
end
The main problem with this technique, however, is that you cannot easily set default value for arguments (Read more). Since this construct is highly useful, Ruby 2 introduced keyword arguments which allows you to write –
def foo(x, str: “foo”, num: 424242)
[x, str, num]
end
foo(13) #=> [13, ‘foo’, 424242]
foo(13, str: ‘bar’) #=> [13, ‘bar’, 424242]
Also, introducing the double splat (**) which similar to it’s counterpart collects all the extra named keywords as a hash parameter.
def foo(str: “foo”, num: 424242, **options)
[str, num, options]
end
foo #=> [‘foo’, 424242, {}]
foo(check: true) # => [‘foo’, 424242, {check: true}]
In this challenge, your task is to write a method convert_temp that helps in temperature conversion. The method will receive a number as an input (temperature), a named parameter input_scale (scale for input temperature), and an optional parameter output_scale (output temperature scale, defaults to Celsius) and return the converted temperature. It should perform interconversion between Celsius, Fahrenheit and Kelvin scale.
For example,
> convert_temp(0, input_scale: ‘celsius’, output_scale: ‘kelvin’)
=> 273.15
> convert_temp(0, input_scale: ‘celsius’, output_scale: ‘fahrenheit’)
=> 32.0
Note that the input values are a lowercase version of corresponding scales.
Problem solution.
# Your code here
def convert_temp(temp, input_scale, output_scale:'celsius')
temp = temp.to_f
case input_scale[:input_scale]
when "celsius"
case input_scale[:output_scale]
when "celsius"
return temp
when "kelvin"
return temp + 273.15
when "fahrenheit"
return ((9*temp)/5)+32
end
when "fahrenheit"
case input_scale[:output_scale]
when "fahrenheit"
return temp
when "celsius"
return (5*(temp - 32))/9
when "kelvin"
holdcel = (5*(temp - 32))/9
return holdcel + 273.15
end
when "kelvin"
case input_scale[:output_scale]
when "kelvin"
return temp
when "celsius"
return temp - 273.15
when "fahrenheit"
holdcel = temp - 273.15
return (5*(holdcel - 32))/9
end
end
end
Second solution.
# Your code here
def convert_temp(temp, input_scale:'celsius', output_scale:'celsius')
if input_scale==output_scale
return temp
end
if input_scale=='celsius' and output_scale=='kelvin'
return temp + 273
end
if input_scale=='celsius' and output_scale=='fahrenheit'
return temp*1.8+32
end
if input_scale=='fahrenheit' and output_scale=='celsius'
return (temp-32)/1.8
end
if input_scale=='fahrenheit' and output_scale=='kelvin'
return (temp+459.67)*5.0/9
end
if input_scale=='kelvin' and output_scale=='celsius'
return temp-273.15
end
if input_scale=='kelvin' and output_scale=='fahrenheit'
return temp*9.0/5-459.67
end
end
