Ruby's times method and for loop does the same thing, but what are the respective processes?-CodePudding

I have a paradigm where the purpose is to calculate "the total accumulated from 'variable from' to 'variable to'". I got two methods, but I don't know exactly what steps they go through.

The first is the times method. I don't understand what (to - from 1) produces and why (i from)? Also, this code throws syntax error, unexpected end, expecting end-of-input, I don't know where I am going wrong .

from = 10
to = 20
sum = 0
  (to - from  1).times to |i|
    sum = sum   (i   from)
  end
puts sum

Then there is the for loop. I can't understand what is the "total sum accumulated from 'variable from' to 'variable to'", what is the process of adding it?

from = 10
to = 20
sum = 0
for i in from..to
  sum = sum   i
end
puts sum

I'm bad at math, thank you all.

CodePudding user response：

Here are a few idiomatic ways of solving this problem in Ruby. They might be a little easier to follow than the solution you proposed.

sum = 0

10.upto(20) do |i|
  sum = sum   i
end

For the first iteration of loop, i will = 10, the second, i will = 11 and so "up to" 20. Each time it will add the value of i to sum. When it completes, sum will have a value of 165.

Another way is this:

10.upto(20).sum

The reason this works is that 10.upto(20) returns an Enumerator. You can call sum on an Enumerator.

CodePudding user response：

Using the IRB you can create an array that represents all the numbers from to to:

irb(main):043:0> from=10
=> 10
irb(main):044:0> to=20
=> 20
irb(main):047:0> a=(from..to).to_a
=> [10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]

You can get a sum of that array:

irb(main):048:0> a.sum
=> 165

And, actually, you don't need to make an array at all:

irb(main):049:0> (from..to).sum
=> 165

Side note. If you have three dots the last value of to is not included and that is a different sum obviously:

irb(main):053:0> (from...to).to_a
=> [10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
irb(main):054:0> (from...to).sum
=> 145

So upto should really be called uptoandincluding:

irb(main):055:0> from.upto(to).to_a
=> [10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]

You can also use reduce or the alias inject either with a block or symbol to an operation to reduct a list the the pairwise application of the operator.

Example (reduce and inject are the same, but I usually use inject with a block or reduce with the parenthesis):

# block form where n is repeatedly added to an accumulator
irb(main):058:0> (from..to).inject{|sum, n| sum n}
=> 165

# the operator   is repeated applied from the left to right 
irb(main):059:0> (from..to).reduce(: )
=> 165