About the inject method: How does it work?

Question

This part of code calculates the value of each arithmetic series up to and including the number that the user put in:

print "enter a number: "
num = gets.to_i
(1..num).inject(0) do |res, e|
  res += e
  p res
end

I think that (1..num) is the range, with num being the user input. I know that inject combines all elements of enum by applying a binary operation specified by a block or a symbol that names a method or operator.

I don't understand what each element in this line does:

(1..num).inject(0) do |res, e|

• What does |res, e| mean? It must be the block that defines what inject does, but what does for instance res and e stand for? (e is probably element?)
• What does (0) stand for?
• What does the command do do?
• what is its connection in regard to (1..num) and inject(0)?
• Am I right to assume that p at the end just stands for puts or print?

Answer 1

inject takes an optional start value, and a block taking an intermediate value and element and returning a new intermediate value.

So:

What does (0) stand for?

The start value parameter to inject.

What does the command "do" do?

It is not a command; it marks the start of the block (terminated by end). .inject(0) do ... end is almost (except for some syntactic issues) the same as .inject(0) { ... }. Usually, do ... end is used for multi-line blocks and { ... } for single-line blocks, but it is not a rule.

What does |res, e| mean?

Those are the block parameters (intermediate value and current element), here probably called after "result" and "element", respectively.

Let's see on a simplified example: (1..3).inject(0) do |res, e| res + e end will set the intermediate result to 0. Then it will pass this intermediate result and the first element of the enumerable being injected: res is 0 and e is 1. The value of the block is the value of its last expression, which is 1 (result of 0 + 1). This 1 now becomes the new intermediate value, and 2 becomes the next current element. The value of the block is 3 (result of 1 + 2). In the next iteration, intermediate value is 3, and the current element also 3, resulting in 6 (3 + 3). The range will stop yielding elements now that we reached its upper boundary, and inject returns with the last intermediate result calculated, 6.

Also, the last question am I right to assume that "p" at the end just stands for puts or print?

Almost. p is its own beast. p x is roughly synonymous with puts x.inspect; x - i.e. it prints the value in a bit different format, and unlike puts which always returns nil, p returns the value unchanged. Thus, p res at the end of your block will not destroy the code by making it return nil, but transparently return res.