I don't understand this method - ruby

Question

I'm a beginner in Ruby and I don't understand what this code is doing, could you explain it to me, please?
def a(n)
s = 0
for i in 0..n-1
s += i
end
s
end

Answer 1

def defines a method. Methods can be used to run the same code on different values. For example, lets say you wanted to get the square of a number:
def square(n)
n * n
end
Now I can do that with different values and I don't have to repeat n * n:
square(1) # => 1
square(2) # => 4
square(3) # => 9
= is an assignment.
s = 0 basically says, behind the name s, there is now a zero.
0..n-1 - constructs a range that holds all numbers between 0 and n - 1. For example:
puts (0..3).to_a
# 0
# 1
# 2
# 3
for assigns i each consecutive value of the range. It loops through all values. So first i is 0, then 1, then ... n - 1.
s += i is a shorthand for s = s + i. In other words, increments the existing value of s by i on each iteration.
The s at the end just says that the method (remember the thing we opened with def) will give you back the value of s. In other words - the sum we accumulated so far.
There is your programming lesson in 5 minutes.

Answer 2

This example isn't idiomatic Ruby code even if it is syntactically valid. Ruby hardly ever uses the for construct, iterators are more flexible. This might seem strange if you come from another language background where for is the backbone of many programs.
In any case, the program breaks down to this:
# Define a method called a which takes an argument n
def a(n)
# Assign 0 to the local variable s
s = 0
# For each value i in the range 0 through n minus one...
for i in 0..n-1
# ...add that value to s.
s += i
end
# The result of this method is s, the sum of those values.
s
end
The more Ruby way of expressing this is to use times:
def a(n)
s = 0
# Repeat this block n times, and in each iteration i will represent
# a value in the range 0 to n-1 in order.
n.times do |i|
s += i
end
s
end
That's just addressing the for issue. Already the code is more readable, mind you, where it's n.times do something. The do ... end block represents a chunk of code that's used for each iteration. Ruby blocks might be a little bewildering at first but understanding them is absolutely essential to being effective in Ruby.
Taking this one step further:
def a(n)
# For each element i in the range 0 to n-1...
(0..n-1).reduce |sum, i|
# ...add i to the sum and use that as the sum in the next round.
sum + i
end
end
The reduce method is one of the simple tools in Ruby that's quite potent if used effectively. It allows you to quickly spin through lists of things and compact them down to a single value, hence the name. It's also known as inject which is just an alias for the same thing.
You can also use short-hand for this:
def a(n)
# For each element in the range 0 to n-1, combine them with +
# and return that as the result of this method.
(0..n-1).reduce(&:+)
end
Where here &:+ is shorthand for { |a,b| a + b }, just as &:x would be short for { |a,b| a.x(b) }.

Answer 3

As you are a beginner in Ruby, let's start from the small slices.
0..n-1 => [0, n-1]. E.g. 0..3 => 0, 1, 2, 3 => [0, 3]
for i in 0.. n-1 => this is a for loop. i traverses [0, n-1].
s += i is same as s = s + i
So. Method a(n) initializes s = 0 then in the for loop i traverse [0, n - 1] and s = s + i
At the end of this method there is an s. Ruby omits key words return. so you can see it as return s
def a(n)
s = 0
for i in 0..n-1
s += i
end
s
end
is same as
def a(n)
s = 0
for i in 0..n-1
s = s + i
end
return s
end
a(4) = 0 + 1 + 2 + 3 = 6
Hope this is helpful.

Answer 4

The method a(n) calculates the sums of the first n natural numbers.
Example:
when n=4, then s = 0+1+2+3 = 6

Answer 5

Let's go symbol by symbol!
def a(n)
This is the start of a function definition, and you're defining the function a that takes a single parameter, n - all typical software stuff. Notably, you can define a function on other things, too:
foo = "foo"
def foo.bar
"bar"
end
foo.bar() # "bar"
"foo".bar # NoMethodError
Next line:
s = 0
In this line, you're both declaring the variable s, and setting it's initial value to 0. Also typical programming stuff.
Notably, the value of the entire expression; s = 0, is the value of s after the assignment:
s = 0
r = t = s += 1 # You can think: r = (t = (s += 1) )
# r and t are now 1
Next line:
for i in 0..n-1
This is starting a loop; specifically a for ... in ... loop. This one a little harder to unpack, but the entire statement is basically: "for each integer between 0 and n-1, assign that number to i and then do something". In fact, in Ruby, another way to write this line is:
(0..n-1).each do |i|
This line and your line are exactly the same.
For single line loops, you can use { and } instead of do and end:
(0..n-1).each{|i| s += i }
This line and your for loop are exactly the same.
(0..n-1) is a range. Ranges are super fun! You can use a lot of things to make up a range, particularly, time:
(Time.now..Time.new(2017, 1, 1)) # Now, until Jan 1st in 2017
You can also change the "step size", so that instead of every integer, it's, say, every 1/10:
(0..5).step(0.1).to_a # [0.0, 0.1, 0.2, ...]
Also, you can make the range exclude the last value:
(0..5).to_a # [0, 1, 2, 3, 4, 5]
(0...5).to_a # [0, 1, 2, 3, 4]
Next line!
s += i
Usually read aloud a "plus-equals". It's literally the same as: s = s + 1. AFAIK, almost every operator in Ruby can be paired up this way:
s = 5
s -= 2 # 3
s *= 4 # 12
s /= 2 # 6
s %= 4 # 2
# etc
Final lines (we'll take these as a group):
end
s
end
The "blocks" (groups of code) that are started by def and for need to be ended, that's what you're doing here.
But also!
Everything in Ruby has a value. Every expression has a value (including assignment, as you saw with line 2), and every block of code. The default value of a block is the value of the last expression in that block.
For your function, the last expression is simply s, and so the value of the expression is the value of s, after all is said and done. This is literally the same as:
return s
end
For the loop, it's weirder - it ends up being the evaluated range.
This example may make it clearer:
n = 5
s = 0
x = for i in (0..n-1)
s += i
end
# x is (0..4)
To recap, another way to write you function is:
def a(n)
s = 0
(0..n-1).each{ |i| s = s + i }
return s
end
Questions?