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 ?
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.
···
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.
···
On Wednesday 28 September 2005 00:03, v.nainar wrote:
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 ?
> 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 )
···
On Wed, Sep 28, 2005 at 03:52:54PM +0900, Kevin Brown wrote:
On Wednesday 28 September 2005 00:03, v.nainar wrote:
-------
> # <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.
Well it **was** silly of me .Thanks for the quick replies