Related
The pre/post increment/decrement operator (++ and --) are pretty standard programing language syntax (for procedural and object-oriented languages, at least).
Why doesn't Ruby support them? I understand you could accomplish the same thing with += and -=, but it just seems oddly arbitrary to exclude something like that, especially since it's so concise and conventional.
Example:
i = 0 #=> 0
i += 1 #=> 1
i #=> 1
i++ #=> expect 2, but as far as I can tell,
#=> irb ignores the second + and waits for a second number to add to i
I understand Fixnum is immutable, but if += can just instanciate a new Fixnum and set it, why not do the same for ++?
Is consistency in assignments containing the = character the only reason for this, or am I missing something?
Here is how Matz(Yukihiro Matsumoto) explains it in an old thread:
Hi,
In message "[ruby-talk:02706] X++?"
on 00/05/10, Aleksi Niemelä <aleksi.niemela#cinnober.com> writes:
|I got an idea from http://www.pragprog.com:8080/rubyfaq/rubyfaq-5.html#ss5.3
|and thought to try. I didn't manage to make "auto(in|de)crement" working so
|could somebody help here? Does this contain some errors or is the idea
|wrong?
(1) ++ and -- are NOT reserved operator in Ruby.
(2) C's increment/decrement operators are in fact hidden assignment.
They affect variables, not objects. You cannot accomplish
assignment via method. Ruby uses +=/-= operator instead.
(3) self cannot be a target of assignment. In addition, altering
the value of integer 1 might cause severe confusion throughout
the program.
matz.
One reason is that up to now every assignment operator (i.e. an operator which changes a variable) has a = in it. If you add ++ and --, that's no longer the case.
Another reason is that the behavior of ++ and -- often confuse people. Case in point: The return value of i++ in your example would actually be 1, not 2 (the new value of i would be 2, however).
It's not conventional in OO languages. In fact, there is no ++ in Smalltalk, the language that coined the term "object-oriented programming" (and the language Ruby is most strongly influenced by). What you mean is that it's conventional in C and languages closely imitating C. Ruby does have a somewhat C-like syntax, but it isn't slavish in adhering to C traditions.
As for why it isn't in Ruby: Matz didn't want it. That's really the ultimate reason.
The reason no such thing exists in Smalltalk is because it's part of the language's overriding philosophy that assigning a variable is fundamentally a different kind of thing than sending a message to an object — it's on a different level. This thinking probably influenced Matz in designing Ruby.
It wouldn't be impossible to include it in Ruby — you could easily write a preprocessor that transforms all ++ into +=1. but evidently Matz didn't like the idea of an operator that did a "hidden assignment." It also seems a little strange to have an operator with a hidden integer operand inside of it. No other operator in the language works that way.
I think there's another reason: ++ in Ruby wouldn't be remotely useful as in C and its direct successors.
The reason being, the for keyword: while it's essential in C, it's mostly superfluous in Ruby. Most of the iteration in Ruby is done through Enumerable methods, such as each and map when iterating through some data structure, and Fixnum#times method, when you need to loop an exact number of times.
Actually, as far as I have seen, most of the time +=1 is used by people freshly migrated to Ruby from C-style languages.
In short, it's really questionable if methods ++ and -- would be used at all.
You can define a .+ self-increment operator:
class Variable
def initialize value = nil
#value = value
end
attr_accessor :value
def method_missing *args, &blk
#value.send(*args, &blk)
end
def to_s
#value.to_s
end
# pre-increment ".+" when x not present
def +(x = nil)
x ? #value + x : #value += 1
end
def -(x = nil)
x ? #value - x : #value -= 1
end
end
i = Variable.new 5
puts i #=> 5
# normal use of +
puts i + 4 #=> 9
puts i #=> 5
# incrementing
puts i.+ #=> 6
puts i #=> 6
More information on "class Variable" is available in "Class Variable to increment Fixnum objects".
I think Matz' reasoning for not liking them is that it actually replaces the variable with a new one.
ex:
a = SomeClass.new
def a.go
'hello'
end
# at this point, you can call a.go
# but if you did an a++
# that really means a = a + 1
# so you can no longer call a.go
# as you have lost your original
Now if somebody could convince him that it should just call #succ! or what not, that would make more sense, and avoid the problem. You can suggest it on ruby core.
And in the words of David Black from his book "The Well-Grounded Rubyist":
Some objects in Ruby are stored in variables as immediate values. These include
integers, symbols (which look like :this), and the special objects true, false, and
nil. When you assign one of these values to a variable (x = 1), the variable holds
the value itself, rather than a reference to it.
In practical terms, this doesn’t matter (and it will often be left as implied, rather than
spelled out repeatedly, in discussions of references and related topics in this book).
Ruby handles the dereferencing of object references automatically; you don’t have to
do any extra work to send a message to an object that contains, say, a reference to
a string, as opposed to an object that contains an immediate integer value.
But the immediate-value representation rule has a couple of interesting ramifications,
especially when it comes to integers. For one thing, any object that’s represented
as an immediate value is always exactly the same object, no matter how many
variables it’s assigned to. There’s only one object 100, only one object false, and
so on.
The immediate, unique nature of integer-bound variables is behind Ruby’s lack of
pre- and post-increment operators—which is to say, you can’t do this in Ruby:
x = 1
x++ # No such operator
The reason is that due to the immediate presence of 1 in x, x++ would be like 1++,
which means you’d be changing the number 1 to the number 2—and that makes
no sense.
Some objects in Ruby are stored in variables as immediate values. These include integers, symbols (which look like :this), and the special objects true, false, and nil. When you assign one of these values to a variable (x = 1), the variable holds the value itself, rather than a reference to it.
Any object that’s represented as an immediate value is always exactly the same object, no matter how many variables it’s assigned to. There’s only one object 100, only one object false, and so on.
The immediate, unique nature of integer-bound variables is behind Ruby’s lack of pre-and post-increment operators—which is to say, you can’t do this in Ruby:
x=1
x++ # No such operator
The reason is that due to the immediate presence of 1 in x, x++ would be like 1++, which means you’d be changing the number 1 to the number 2—and that makes no sense.
Couldn't this be achieved by adding a new method to the fixnum or Integer class?
$ ruby -e 'numb=1;puts numb.next'
returns 2
"Destructive" methods seem to be appended with ! to warn possible users, so adding a new method called next! would pretty much do what was requested ie.
$ ruby -e 'numb=1; numb.next!; puts numb'
returns 2 (since numb has been incremented)
Of course, the next! method would have to check that the object was an integer variable and not a real number, but this should be available.
I want to be able to write number.incr, like so:
num = 1; num.incr; num
#=> 2
The error I'm seeing states:
Can't change the value of self
If that's true, how do bang! methods work?
You cannot change the value of self
An object is a class pointer and a set of instance methods (note that this link is an old version of Ruby, because its dramatically simpler, and thus better for explanatory purposes).
"Pointing" at an object means you have a variable which stores the object's location in memory. Then to do anything with the object, you first go to the location in memory (we might say "follow the pointer") to get the object, and then do the thing (e.g. invoke a method, set an ivar).
All Ruby code everywhere is executing in the context of some object. This is where your instance variables get saved, it's where Ruby looks for methods that don't have a receiver (e.g. $stdout is the receiver in $stdout.puts "hi", and the current object is the receiver in puts "hi"). Sometimes you need to do something with the current object. The way to work with objects is through variables, but what variable points at the current object? There isn't one. To fill this need, the keyword self is provided.
self acts like a variable in that it points at the location of the current object. But it is not like a variable, because you can't assign it new value. If you could, the code after that point would suddenly be operating on a different object, which is confusing and has no benefits over just using a variable.
Also remember that the object is tracked by variables which store memory addresses. What is self = 2 supposed to mean? Does it only mean that the current code operates as if it were invoked 2? Or does it mean that all variables pointing at the old object now have their values updated to point at the new one? It isn't really clear, but the former unnecessarily introduces an identity crisis, and the latter is prohibitively expensive and introduce situations where it's unclear what is correct (I'll go into that a bit more below).
You cannot mutate Fixnums
Some objects are special at the C level in Ruby (false, true, nil, fixnums, and symbols).
Variables pointing at them don't actually store a memory location. Instead, the address itself stores the type and identity of the object. Wherever it matters, Ruby checks to see if it's a special object (e.g. when looking up an instance variable), and then extracts the value from it.
So there isn't a spot in memory where the object 123 is stored. Which means self contains the idea of Fixnum 123 rather than a memory address like usual. As with variables, it will get checked for and handled specially when necessary.
Because of this, you cannot mutate the object itself (though it appears they keep a special global variable to allow you to set instance variables on things like Symbols).
Why are they doing all of this? To improve performance, I assume. A number stored in a register is just a series of bits (typically 32 or 64), which means there are hardware instructions for things like addition and multiplication. That is to say the ALU, is wired to perform these operations in a single clock cycle, rather than writing the algorithms with software, which would take many orders of magnitude longer. By storing them like this, they avoid the cost of storing and looking the object in memory, and they gain the advantage that they can directly add the two pointers using hardware. Note, however, that there are still some additional costs in Ruby, that you don't have in C (e.g. checking for overflow and converting result to Bignum).
Bang methods
You can put a bang at the end of any method. It doesn't require the object to change, it's just that people usually try to warn you when you're doing something that could have unexpected side-effects.
class C
def initialize(val)
#val = val # => 12
end # => :initialize
def bang_method!
"My val is: #{#val}" # => "My val is: 12"
end # => :bang_method!
end # => :bang_method!
c = C.new 12 # => #<C:0x007fdac48a7428 #val=12>
c.bang_method! # => "My val is: 12"
c # => #<C:0x007fdac48a7428 #val=12>
Also, there are no bang methods on integers, It wouldn't fit with the paradigm
Fixnum.instance_methods.grep(/!$/) # => [:!]
# Okay, there's one, but it's actually a boolean negation
1.! # => false
# And it's not a Fixnum method, it's an inherited boolean operator
1.method(:!).owner # => BasicObject
# In really, you call it this way, the interpreter translates it
!1 # => false
Alternatives
Make a wrapper object: I'm not going to advocate this one, but it's the closest to what you're trying to do. Basically create your own class, which is mutable, and then make it look like an integer. There's a great blog post walking through this at http://blog.rubybestpractices.com/posts/rklemme/019-Complete_Numeric_Class.html it will get you 95% of the way there
Don't depend directly on the value of a Fixnum: I can't give better advice than this without knowing what you're trying to do / why you feel this is a need.
Also, you should show your code when you ask questions like this. I misunderstood how you were approaching it for a long time.
It's simply impossible to change self to another object. self is the receiver of the message send. There can be only one.
If that's true, how do bang! methods work?
The bang (!) is simply part of the method name. It has absolutely no special meaning whatsoever. It is a convention among Ruby programmers to name surprising variants of less surprising methods with a bang, but that's just that: a convention.
As the title says I'm curious about the difference between "call-by-reference" and "call-by-value-return". I've read about it in some literature, and tried to find additional information on the internet, but I've only found comparison of "call-by-value" and "call-by-reference".
I do understand the difference at memory level, but not at the "conceptual" level, between the two.
The called subroutine will have it's own copy of the actual parameter value to work with, but will, when it ends executing, copy the new local value (bound to the formal parameter) back to the actual parameter of the caller.
When is call-by-value-return actually to prefer above "call-by-reference"? Any example scenario? All I can see is that it takes extra memory and execution time due to the copying of values in the memory-cells.
As a side question, is "call-by-value-return" implemented in 'modern' languages?
Call-by-value-return, from Wikipedia:
This variant has gained attention in multiprocessing contexts and Remote procedure call: if a parameter to a function call is a reference that might be accessible by another thread of execution, its contents may be copied to a new reference that is not; when the function call returns, the updated contents of this new reference are copied back to the original reference ("restored").
So, in more practical terms, it's entirely possible that a variable is in some undesired state in the middle of the execution of a function. With parallel processing this is a problem, since you can attempt to access the variable while it has this value. Copying it to a temporary value avoids this problem.
As an example:
policeCount = 0
everyTimeSomeoneApproachesOrLeaves()
calculatePoliceCount(policeCount)
calculatePoliceCount(count)
count = 0
for each police official
count++
goAboutMyDay()
if policeCount == 0
doSomethingIllegal()
else
doSomethingElse()
Assume everyTimeSomeoneApproachesOrLeaves and goAboutMyDay are executed in parallel.
So if you pass by reference, you could end up getting policeCount right after it was set to 0 in calculatePoliceCount, even if there are police officials around, then you'd end up doing something illegal and probably going to jail, or at least coughing up some money for a bribe. If you pass by value return, this won't happen.
Supported languages?
In my search, I found that Ada and Fortran support this. I don't know of others.
Suppose you have a call by reference function (in C++):
void foobar(int &x, int &y) {
while (y-->0) {
x++;
}
}
and you call it thusly:
int z = 5;
foobar(z, z);
It will never terminate, because x and y are the same reference, each time you decrement y, that is subsequently undone by the increment of x (since they are both really z under the hood).
By contrast using call-by-value-return (in rusty Fortran):
subroutine foobar(x,y):
integer, intent(inout) :: x,y
do while y > 0:
y = y - 1
x = x + 1
end do
end subroutine foobar
If you call this routine with the same variable:
integer, z = 5
call foobar(z,z)
it will still terminate, and at the end z will be changed have a value of either 10 or 0, depending on which result is applied first (I don't remember if a particular order is required and I can't find any quick answers to the question online).
Kindly go to the following link , the program in there can give u an practical idea regarding these two .
Difference between call-by-reference and call-by-value
Beginner Ruby question. What is the simplest way to change this code, but leaving the block completely intact, that eliminates the side effect?
x = lambda { |v| x = 2 ; v}
x.call(3)
#=> 3
x
#=> 2
This is the simplest example I could contrive to illustrate my issue, so "remove the assignment" or "don't assign the Proc to x" is not what I'm looking for.
I want to set local variables in a Proc (or lambda) that can be assigned without affecting the original enclosing scope. I could dynamically create a class or module to wrap the block, but that seems overkill for such a basic thing.
Equivalent Python to what I'm trying to do:
def x(v):
x = 2 # this is a local variable, what a concept
return v
Sometimes it is the desired behavior:
total = 0
(1..10).each{|x| total += x}
puts total
But sometimes it's accidental and you don't want to mess with an outside variable which happens to have the same name. In that case, follow the list of parameters with a semicolon and a list of the block-local variables:
x = lambda{|v; x| x = 2; v}
p x.call(3) #3
p x #<Proc:0x83f7570#test1.rb:2 (lambda)>
The reason for this is that the lambda is bound to its defining scope (NOT its calling scope), and is a full closure, which, among other things, includes the local variable x. What you really want here is an unbound proc to pass around and call without any particular binding. This isn't something that Ruby does very easily at all, by design (it's possible with eval, but that's less of a block and more just of a string statement). Procs, lambdas, and blocks are all bound to their defining scope. Lexical scope is only established on classes, modules, and methods; not on blocks/procs/lambdas/anything else.
It's worth noting that Python doesn't even permit assignment in lambdas in the first place.
The pre/post increment/decrement operator (++ and --) are pretty standard programing language syntax (for procedural and object-oriented languages, at least).
Why doesn't Ruby support them? I understand you could accomplish the same thing with += and -=, but it just seems oddly arbitrary to exclude something like that, especially since it's so concise and conventional.
Example:
i = 0 #=> 0
i += 1 #=> 1
i #=> 1
i++ #=> expect 2, but as far as I can tell,
#=> irb ignores the second + and waits for a second number to add to i
I understand Fixnum is immutable, but if += can just instanciate a new Fixnum and set it, why not do the same for ++?
Is consistency in assignments containing the = character the only reason for this, or am I missing something?
Here is how Matz(Yukihiro Matsumoto) explains it in an old thread:
Hi,
In message "[ruby-talk:02706] X++?"
on 00/05/10, Aleksi Niemelä <aleksi.niemela#cinnober.com> writes:
|I got an idea from http://www.pragprog.com:8080/rubyfaq/rubyfaq-5.html#ss5.3
|and thought to try. I didn't manage to make "auto(in|de)crement" working so
|could somebody help here? Does this contain some errors or is the idea
|wrong?
(1) ++ and -- are NOT reserved operator in Ruby.
(2) C's increment/decrement operators are in fact hidden assignment.
They affect variables, not objects. You cannot accomplish
assignment via method. Ruby uses +=/-= operator instead.
(3) self cannot be a target of assignment. In addition, altering
the value of integer 1 might cause severe confusion throughout
the program.
matz.
One reason is that up to now every assignment operator (i.e. an operator which changes a variable) has a = in it. If you add ++ and --, that's no longer the case.
Another reason is that the behavior of ++ and -- often confuse people. Case in point: The return value of i++ in your example would actually be 1, not 2 (the new value of i would be 2, however).
It's not conventional in OO languages. In fact, there is no ++ in Smalltalk, the language that coined the term "object-oriented programming" (and the language Ruby is most strongly influenced by). What you mean is that it's conventional in C and languages closely imitating C. Ruby does have a somewhat C-like syntax, but it isn't slavish in adhering to C traditions.
As for why it isn't in Ruby: Matz didn't want it. That's really the ultimate reason.
The reason no such thing exists in Smalltalk is because it's part of the language's overriding philosophy that assigning a variable is fundamentally a different kind of thing than sending a message to an object — it's on a different level. This thinking probably influenced Matz in designing Ruby.
It wouldn't be impossible to include it in Ruby — you could easily write a preprocessor that transforms all ++ into +=1. but evidently Matz didn't like the idea of an operator that did a "hidden assignment." It also seems a little strange to have an operator with a hidden integer operand inside of it. No other operator in the language works that way.
I think there's another reason: ++ in Ruby wouldn't be remotely useful as in C and its direct successors.
The reason being, the for keyword: while it's essential in C, it's mostly superfluous in Ruby. Most of the iteration in Ruby is done through Enumerable methods, such as each and map when iterating through some data structure, and Fixnum#times method, when you need to loop an exact number of times.
Actually, as far as I have seen, most of the time +=1 is used by people freshly migrated to Ruby from C-style languages.
In short, it's really questionable if methods ++ and -- would be used at all.
You can define a .+ self-increment operator:
class Variable
def initialize value = nil
#value = value
end
attr_accessor :value
def method_missing *args, &blk
#value.send(*args, &blk)
end
def to_s
#value.to_s
end
# pre-increment ".+" when x not present
def +(x = nil)
x ? #value + x : #value += 1
end
def -(x = nil)
x ? #value - x : #value -= 1
end
end
i = Variable.new 5
puts i #=> 5
# normal use of +
puts i + 4 #=> 9
puts i #=> 5
# incrementing
puts i.+ #=> 6
puts i #=> 6
More information on "class Variable" is available in "Class Variable to increment Fixnum objects".
I think Matz' reasoning for not liking them is that it actually replaces the variable with a new one.
ex:
a = SomeClass.new
def a.go
'hello'
end
# at this point, you can call a.go
# but if you did an a++
# that really means a = a + 1
# so you can no longer call a.go
# as you have lost your original
Now if somebody could convince him that it should just call #succ! or what not, that would make more sense, and avoid the problem. You can suggest it on ruby core.
And in the words of David Black from his book "The Well-Grounded Rubyist":
Some objects in Ruby are stored in variables as immediate values. These include
integers, symbols (which look like :this), and the special objects true, false, and
nil. When you assign one of these values to a variable (x = 1), the variable holds
the value itself, rather than a reference to it.
In practical terms, this doesn’t matter (and it will often be left as implied, rather than
spelled out repeatedly, in discussions of references and related topics in this book).
Ruby handles the dereferencing of object references automatically; you don’t have to
do any extra work to send a message to an object that contains, say, a reference to
a string, as opposed to an object that contains an immediate integer value.
But the immediate-value representation rule has a couple of interesting ramifications,
especially when it comes to integers. For one thing, any object that’s represented
as an immediate value is always exactly the same object, no matter how many
variables it’s assigned to. There’s only one object 100, only one object false, and
so on.
The immediate, unique nature of integer-bound variables is behind Ruby’s lack of
pre- and post-increment operators—which is to say, you can’t do this in Ruby:
x = 1
x++ # No such operator
The reason is that due to the immediate presence of 1 in x, x++ would be like 1++,
which means you’d be changing the number 1 to the number 2—and that makes
no sense.
Some objects in Ruby are stored in variables as immediate values. These include integers, symbols (which look like :this), and the special objects true, false, and nil. When you assign one of these values to a variable (x = 1), the variable holds the value itself, rather than a reference to it.
Any object that’s represented as an immediate value is always exactly the same object, no matter how many variables it’s assigned to. There’s only one object 100, only one object false, and so on.
The immediate, unique nature of integer-bound variables is behind Ruby’s lack of pre-and post-increment operators—which is to say, you can’t do this in Ruby:
x=1
x++ # No such operator
The reason is that due to the immediate presence of 1 in x, x++ would be like 1++, which means you’d be changing the number 1 to the number 2—and that makes no sense.
Couldn't this be achieved by adding a new method to the fixnum or Integer class?
$ ruby -e 'numb=1;puts numb.next'
returns 2
"Destructive" methods seem to be appended with ! to warn possible users, so adding a new method called next! would pretty much do what was requested ie.
$ ruby -e 'numb=1; numb.next!; puts numb'
returns 2 (since numb has been incremented)
Of course, the next! method would have to check that the object was an integer variable and not a real number, but this should be available.