class variables

[v.nainar]

Hello!

What is wrong with the following code? Why is @@id incremented before
and not after the assignment? . ( Sorry if this sounds silly and I have
overlooked something obvious )


#!/usr/local/bin/ruby
class Foo
@@id=1
def initialize(s)
@s=s
@id=setId()
end
def setId()
@id=@@id
@@id+=1
end
end

foo1=Foo.new("one")
foo2=Foo.new("two")
foo3=Foo.new("three")
p foo1,foo2,foo3
Output ->
# <Foo:0xb7d6b4f0 @id=2, @s="one"> # why is @id not 1
# <Foo:0xb7d6b4c8 @id=3, @s="two"> # and 2
# <Foo:0xb7d6b4a0 @id=4, @s="three"> # and 3 ?

 

[Daniel Tse]

Hello!

What is wrong with the following code? Why is @@id incremented before
and not after the assignment? . ( Sorry if this sounds silly and I have
overlooked something obvious )


#!/usr/local/bin/ruby
class Foo
@@id=1
def initialize(s)
@s=s
@id=setId()
end

You are assigning to the return value of setId, which is the incremented
@@id: in Ruby, the implicit return value of a function is that of the
last evaluated expression.

 

[Kevin Brown]

Hello!

What is wrong with the following code? Why is @@id incremented before
and not after the assignment? . ( Sorry if this sounds silly and I have
overlooked something obvious )


#!/usr/local/bin/ruby
class Foo
@@id=1
def initialize(s)
@s=s
@id=setId()
end
def setId()
@id=@@id
@@id+=1
end
end

foo1=Foo.new("one")
foo2=Foo.new("two")
foo3=Foo.new("three")
p foo1,foo2,foo3
Output ->
# <Foo:0xb7d6b4f0 @id=2, @s="one"> # why is @id not 1
# <Foo:0xb7d6b4c8 @id=3, @s="two"> # and 2
# <Foo:0xb7d6b4a0 @id=4, @s="three"> # and 3 ?

When somethings at the end of a function, it gets returned implicitly. So,
when you call @id - setId() you're setting @id to the current id (1), then
incrementing, and then then function returns the incremented value,
overwriting your 1 with a 2. Throw a return @s at the end and you'll get the
behaviour you're looking for.

 

[v.nainar]

When somethings at the end of a function, it gets returned implicitly. So,
when you call @id - setId() you're setting @id to the current id (1), then
incrementing, and then then function returns the incremented value,
overwriting your 1 with a 2. Throw a return @s at the end and you'll get the
behaviour you're looking for.

Well it **was** silly of me .Thanks for the quick replies