Super simple newbie Q about methods

Ruby
1

If I write:

def add(i,j)
i = i + j
end

i = 1

puts i
add(i,3)
puts i

I get

1
1

What would I have to change to get

1
4

? I presume this is covered in the Pickaxe book but can’t find the
answer - a pointer to whereabouts it is covered would be great.

Sorry to be so dumb.

2

The ‘i’ in your method is local to the method. It is not the same ‘i’
as outside the method.

Instead, do

def add(i,j)
i = i + j
end

i = 1

puts i
i = add(i,3)
puts i

3

Alex DeCaria wrote:

The ‘i’ in your method is local to the method. It is not the same ‘i’
as outside the method.

Instead, do

def add(i,j)
i = i + j
end

i = 1

puts i
i = add(i,3)
puts i

Thanks for that, but I was looking for a way of passing by address. In
fact I thought that, because everything was an object in Ruby everything
was always passed by address. I would like to understand when it is not
the case.

4

Thanks for that, but I was looking for a way of passing by address. In
fact I thought that, because everything was an object in Ruby everything
was always passed by address. I would like to understand when it is not
the case.

Then you need to use instance variables such as ‘@i’, or class variables
such as ‘@@i’. Ruby doesn’t support the use of pointers (as far as I
know at least).

–Alex

5

Henry O. wrote:

Thanks for that, but I was looking for a way of passing by address. In
fact I thought that, because everything was an object in Ruby everything
was always passed by address. I would like to understand when it is not
the case.

Well…

def i, j
i[0] = i[0] + j
end

You will notice that the previous method requires i to be an Array.

i = [1]
puts i
add i, 2
puts i

Happy? :slight_smile:

6

Henry O. wrote:

The ‘i’ in your method is local to the method. It is not the same ‘i’
as outside the method.

Instead, do

def add(i,j)
i = i + j
end

i = 1

puts i
i = add(i,3)
puts i

Thanks for that, but I was looking for a way of passing by address. In
fact I thought that, because everything was an object in Ruby everything
was always passed by address.

The number 1 is an object. The variable i is not an object; it is a
placeholder which contains a reference to the object.

Object references are always passed by value. You can never get a
pointer to the placeholder.

This means that foo(i) cannot affect the value of i, because inside the
method it’s using a copy of that reference. If i refers to a mutable
object, then the object itself can alter its state, but the reference
does not change.

def a(x, y)
x << y
end

i = “hello”
a(i, “x”)
puts i # “hellox” - but it’s the same string object

i = 3
a(i, 1) # calculates and returns 3 << 1 (=6)
puts i # but i is still the same Fixnum object (=3)

7

Hello,

Then you need to use instance variables such as ‘@i’, or class variables
such as ‘@@i’. Ruby doesn’t support the use of pointers (as far as I
know at least).

I would say rather on the contrary. Everything is a pointer but you
can reassign a variable in a new pointing direction without affecting
the previously pointed object. See

def add(a,b)
a.replace(a+b)
end
=> nil

s = “string”
=> “string”

add(s,"_added")
=> “string_added”

s
=> “string_added”

Here, the add method does not reassign variable a but send a message
to the object where a is pointing, saying that it should change it’s
value.
Of course, that would not work for a Fixnum, as those are immutable in
ruby.

Cheers,