I'm trying to figure out why the following codes doesn't return the same result:
CODE 1
p0 = "hello"
a = []
b = p0
1.upto(5) do |i|
b.insert(2,"B")
a.push b
end
a => ["heBBBBBllo", "heBBBBBllo", "heBBBBBllo", "heBBBBBllo", "heBBBBBllo"]
CODE 2
p0 = "hello"
a = []
b = p0
1.upto(5) do |i|
b.insert(2,"B")
a.push b.inspect
end
a => ["\"heBllo\"", "\"heBBllo\"", "\"heBBBllo\"", "\"heBBBBllo\"", "\"heBBBBBllo\""]
What I need is the Code 2's result, but I don't need the escaped char like the inspect method does.
Honestly, I really don't understand why with the inspect method works, and why in the code 1 doesn't.
It seems like that in code 1, "b" is used as a pointer, and every time it's updated, all the "linked"-b are updated.
Any clue??
thank you in advance.
In code 1 you're pushing references to the same object. The array will contain multiple references to the same thing.
In code 2 you're pushing the inspect output at a distinct moment in time. The array will contain a history of inspect's returned strings.
p0 = "hello"
a = []
b = p0
1.upto(5) do |i|
b.insert(2,"B")
a.push b.clone
end
It seems like that in code 1, "b" is used as a pointer, and every time
it's updated, all the "linked"-b are updated.
Correct! In the first case, you're pushing b into the array five times, so the result is that a contains five pointers to b. In the second case you're pushing b.inspect - which is a different object from b.
The easiest way to fix your first example would be to call a.push b.dup instead, which creates a duplicate of b which won't be affected by future changes to b.
'b' is a string, which is an object. When you insert a letter, it modifies the object.
In CODE 1, that same object b is being pushed into the array multiple times, and modified multiple times.
However, b.inspect returns a new string. So in CODE 2, you push a new string into the array each iteration, and that new string is a snapshot of how 'b' looked at the time.
You could use a.push b.dup instead, which creates a copy of b without changing the format.
Related
Consider the following code
a="123456789"
t=[[1,4],[3,4],[4,5],[1,2]]
p t.map{|x,y|
a[x],a[y]=a[y],a[x]
#p a
a
}
I know ruby map method collects the last expression of the given block but when using the above code to swap the chars in a using the indexes in t won't succeeds.My intention was to collect the state of a after each swap in the index of t.But map always gives the array of a which is in the last state ie)["135264789", "135264789", "135264789", "135264789"].
The results shows that the map method have collected the final result of a after completing each indexes in t.But when printing the a after each swap prints correct value of a at each state.
Is this the correct behavior or am i missing something?
This is because the String#[]= method mutates the string.
Quick fix would be something like this:
a="123456789"
t=[[1,4],[3,4],[4,5],[1,2]]
p t.map{|x,y]
b = "#{a}" # IMPORTANT - this builds a new string
b[x],b[y]=b[y],b[x] # this mutates the new string
#p b
b
}
An alternative to "#{a}" would be to say a.clone, it does the same thing in this case.
The reason this works, is because instead of directly modifying a with a[x],a[y]=a[y],a[x], you're making a temporary copy of a and modifying that instead
edit - I misread the question - if you want to show the result of chaining each operation on the previous result, use dup/clone after the modification as Stefan said in his answer
Is my understanding correct?
Yes, I believe it is. I second what Max says, and I'll also elaborate a bit in case it helps.
Each b is a newly created object because it gets created inside the block, so it gets recreated with every new iteration. The a is created outside the block, so the same object (a) keeps getting referenced inside the block for each iteration.
You can better understand how this works by experimenting with #object_id. Try running this code:
a="123456789"
t=[[1,4],[3,4],[4,5],[1,2]]
p t.map { |x,y|
b = "#{a}" # IMPORTANT - this builds a new string
b[x],b[y]=b[y],b[x]
p "a.object_id = #{a.object_id}"
p "b.object_id = #{b.object_id}"
b
}
You will notice that a is the same object for each iteration of the #map method, while b is a new one.
This is an example of the concept of a closure. A closure is some sort of enclosed code structure that retains access to whatever state is available in the context in which it was created, while that context doesn't have access to its, the enclosed code's, state. Sort of like a "one way mirror": the enclosed code can see outside, but the outside can't see into the enclosed code.
In Ruby, closures are implemented as blocks: blocks are closures. So, everything that is visible to whatever context a block is created in (in this case, main) is also visible to that block, although the reverse is not true — for example, you can't reference b from outside the block. (Methods are not closures: if your block were a method, it wouldn't be able to see a unless you passed it in as an argument to your method.)
So, as Max says, when you make changes to a inside your block, you are actually changing (mutating) the same a that you defined up top each time.
Side topic
Now, if you are referencing individual characters in strings it's important to understand that the underlying structure of strings differs from that of arrays. Also, arrays behave differently when you mutate their elements from strings when you mutate their characters.
I'm mentioning this because I have this vague feeling that you are thinking of string character references as pretty much analogous to array element references. This is pretty much only true with respect to syntax.
You may find the results of running this code interesting:
a = '123456789'
p a.object_id
p a[0].object_id
p a[1].object_id
a[0] = '7'
p a.object_id
p a[0].object_id
p a[1].object_id
puts
a = '123456789'.chars
p a.object_id
p a[0].object_id
p a[1].object_id
a[0] = '7'
p a.object_id
p a[0].object_id
p a[1].object_id
In particular, a comparison of the four outputs of a[1].object_id should be instructive, because it shows where strings and arrays differ. If you reassign an element in an array, that element and only that element gets a new object id. If you reassign a character in a string, the string object itself remains the same, but every character in the string gets recreated.
Since you are returning a from map, the result will contain a four times. Those a's all refer to the same object.
You probably want to return a copy of a to preserve its current state:
a = '123456789'
t = [[1, 4], [3, 4], [4, 5], [1, 2]]
r = t.map { |x, y|
a[x], a[y] = a[y], a[x]
a.dup
}
r #=> ["153426789", "153246789", "153264789", "135264789"]
I cannot understand this ruby behavior, the code explains better what I mean:
class DoNotUnderstand
def initialize
#tiny_array = [3,4]
test
end
def messing(ary)
return [ary[0]+=700, ary[1]+=999]
end
def test
puts #tiny_array.join(",") # before => 3,4
messing(#tiny_array)
puts #tiny_array.join(",") # after => 703,1003
end
end
question = DoNotUnderstand.new
#tiny_array was [3,4] and became [703,1003]
if I don't use a class, that happens:
#tiny = [1,2]
def messing(ary)
return [ary[0]+693,ary[1]+999]
end
puts #tiny.join(",") # before => 1,2
messing(#tiny)
puts #tiny.join(",") # after => 1,2
the array simply remains [1,2]
why?
The class is a red herring, and completely irrelevant to the issue.
In the first case, where the array was modified, you defined messing as:
def messing(ary)
return [ary[0]+=700, ary[1]+=999]
end
Whereas in the second case, where the array was not modified, you defined messing as:
def messing(ary)
return [ary[0]+693,ary[1]+999]
end
In one case you used +=, and in the other, you used merely +.
ary[0] += 700 is exactly equivalent to ary[0] = ary[0] + 700. In other words you are changing the value stored in the 0th index of ary.
In the second case you merely add to the values stored in the array and return the result, but in the first case you not only return the result, you also store it back in the array.
For an explanation of why modifying ary modifies #tiny_array, see this answer to the question Is Ruby pass by reference or by value?.
You're second code example (the one from outside the class) is missing the two characters in the first that make it work the way it does. In the first example, the += operator is used, modifying the array in place:
return [ary[0]+=700, ary[1]+=999]
In your second example, the + operator is used, leaving the array as is:
return [ary[0]+693,ary[1]+999]
If you change it use the += operator, it works the same way as the first code snippet.
A snippet of my code below flips a coin and outputs a result of 10 total heads or tails.
(e.g. Heads Tails Heads Tails...)
I'd like to store this into a variable where I can put it into an array and use its strings.
%w[act] only outputs the string "act". How can I get that line of code to output my array of strings from the line act = coin.flip?
Updated and added full code
class Coin
def flip
flip = 1 + rand(2)
if flip == 2
then puts "Heads"
else
puts "Tails"
end
end
end
array = []
10.times do
coin = Coin.new
array << coin.flip
end
puts array
This:
10.times do
coin = Coin.new
act = coin.flip
end
doesn't produce an array. It simply creates ten coin flips and throws them all away, the result of that expression is, in fact, 10. If you want an array, you'll need to build one.
You could take Douglas's approach or try something a bit more idiomatic.
The Integer#times method returns an enumerator so you can use any of the Enumerable methods on it rather than directly handing it a block. In particular, you could use collect to build an array in one nice short piece of code:
a = 10.times.collect { Coin.new.flip }
That gives you 10 flips in the Array a and then you can puts a or puts a.join(', ') or whatever you want.
The %w[] won't work because that's for generating an Array of whitespace separated words:
%w[] Non-interpolated Array of words, separated by whitespace
So %w[a b c] is just a nicer way of saying ['a', 'b', 'c'] and the words within %w[] are treated as single quoted strings rather than variables or method calls to be evaluated.
Seems that there is some editing going on. You'll also want to modify your flip method to return the flip rather than print it:
def flip
flip = 1 + rand(2)
if flip == 2
"Heads"
else
"Tails"
end
end
Then you'll get your Heads and Rails in the array.
Put the act results into an array.
arr = []
10.times do
coin = Coin.new
arr << coin.flip
end
p arr # => [...]
I came around this strange feature(?) of arrays in Ruby and it would be very helpful if someone could explain to me why they work the way they do.
First lets give an example of how things usually work.
a = "Hello" #=> "Hello"
b = a #=> "Hello"
b += " Goodbye" #=> "Hello Goodbye"
b #=> "Hello Goodbye"
a #=> "Hello"
Ok cool, when you use = it creats a copy of the object (this time a string).
But when you use arrays this happens:
a = [1,2,3] #=> [1,2,3]
b = a #=> [1,2,3]
b[1] = 5 #=> [1,5,3]
b #=> [1,5,3]
a #=> [1,5,3]
Now thats just strange. Its the only object I've found that doesn't get copied when using = but instead just creates a refrance to the original object.
Can someone also explain (there must be a method) for copying an array without having it point back to the original object?
Actually, you should re-examine your premise.
The string assignment is really b = b + " Goodbye". The b + " Goodbye" operation returns an entirely new string, so the variable b is pointing to a new object after the assignment.
But when you assign to an individual array element, you are not creating an entirely new array, so a and b continue to point to the same object, which you just changed.
If you are looking for a rationale for the mutating vs functional behavior of arrays, it's simple. There is nothing to be gained by modifying the string. It is most likely necessary to allocate new memory anyway, so an entirely new string is created.
But an array can be arbitrarily large. Creating a new array in order to change just one element could be hugely expensive. And in any case, an array is like any other composite object. Changing an individual attribute does not necessarily affect any other attributes.
And to answer your question, you can always do:
b = a.dup
What happends there is that ruby is treating the Array object by reference and not by value.
So you can see it as this:
b= [1,2,3]
a= b
--'b' Points to---> [1,2,3] <--'a' points t---
So as you can see both point to the same reference, that means that if you change anything in a it will be reflected on b.
As for your question on the copying the object you could use the Object#clone method to do so.
Try your Array case with a String:
a = "Hello" #=> "Hello"
b = a #=> "Hello"
b[1] = "x" #=> "x"
b #=> "Hxllo"
a #=> "Hxllo"
Strings and Arrays work the same way in this regard.
The key difference in the two cases, as you wrote them, is this:
b += " Goodbye"
This is syntactic shorthand for
b = b + " Goodbye"
which is creating a new string from b + " Goodbye" and then assigning that to b. The way to modify an existing string, rather than creating a new one, is
b << " Goodbye"
And if you plug that into your sequence, you'll see that it modifies both a and b, since both variables refer to the same string object.
As for deep copying, there's a decent piece about it here:
http://ruby.about.com/od/advancedruby/a/deepcopy.htm
I have a Ruby method that searches an array of hashes and returns a subset of that array.
def last_actions(type = 'all')
actions = #actions
if type == 'run'
actions = actions.select {|a| a['type'] == "run" }
end
return actions
end
This works, except when there is only one action to return, in which case I don't think it is returning an array with one element, but just the element itself.
This becomes problematic later.
What's a good way to ensure it returns an array of 1 element in this case?
Thanks.
Note that there is a nice clean ruby idiom to ensure single objects and arrays are treated as arrays:
a = [1,2,3]
b = 4
[*a]
=> [1, 2, 3]
[*b]
=> [4]
select always returns an array (except if you break inside the block, which you don't). So whatever is going wrong in your code, this is not the reason.
Of course if #actions does not contain an array (or another type of Enumerable), the call to select will cause an exception and the method will not return anything at all. The solution in that case would be to make sure that #actions always returns an array. How to do that depends on where/how you set #actions.
Your observation is more than strange, however you could try this:
def last_actions(type = 'all')
actions = #actions.dup || []
actions.delete_if {|a| a['type'] != "run" } if type == 'run'
actions
end
You can also do it like this:
a = [1,2,3]
b = 4
Array(a)
=> [1, 2, 3]
Array(b)
=> [4]