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I was going through the exercises in Ruby Koans and I was struck by the following Ruby quirk that I found really unexplainable:
array = [:peanut, :butter, :and, :jelly]
array[0] #=> :peanut #OK!
array[0,1] #=> [:peanut] #OK!
array[0,2] #=> [:peanut, :butter] #OK!
array[0,0] #=> [] #OK!
array[2] #=> :and #OK!
array[2,2] #=> [:and, :jelly] #OK!
array[2,20] #=> [:and, :jelly] #OK!
array[4] #=> nil #OK!
array[4,0] #=> [] #HUH?? Why's that?
array[4,100] #=> [] #Still HUH, but consistent with previous one
array[5] #=> nil #consistent with array[4] #=> nil
array[5,0] #=> nil #WOW. Now I don't understand anything anymore...
So why is array[5,0] not equal to array[4,0]? Is there any reason why array slicing behaves this weird when you start at the (length+1)th position??
Slicing and indexing are two different operations, and inferring the behaviour of one from the other is where your problem lies.
The first argument in slice identifies not the element but the places between elements, defining spans (and not elements themselves):
:peanut :butter :and :jelly
0 1 2 3 4
4 is still within the array, just barely; if you request 0 elements, you get the empty end of the array. But there is no index 5, so you can't slice from there.
When you do index (like array[4]), you are pointing at elements themselves, so the indices only go from 0 to 3.
this has to do with the fact that slice returns an array, relevant source documentation from Array#slice:
* call-seq:
* array[index] -> obj or nil
* array[start, length] -> an_array or nil
* array[range] -> an_array or nil
* array.slice(index) -> obj or nil
* array.slice(start, length) -> an_array or nil
* array.slice(range) -> an_array or nil
which suggests to me that if you give the start that is out of bounds, it will return nil, thus in your example array[4,0] asks for the 4th element that exists, but asks to return an array of zero elements. While array[5,0] asks for an index out of bounds so it returns nil. This perhaps makes more sense if you remember that the slice method is returning a new array, not altering the original data structure.
EDIT:
After reviewing the comments I decided to edit this answer. Slice calls the following code snippet when the arg value is two:
if (argc == 2) {
if (SYMBOL_P(argv[0])) {
rb_raise(rb_eTypeError, "Symbol as array index");
}
beg = NUM2LONG(argv[0]);
len = NUM2LONG(argv[1]);
if (beg < 0) {
beg += RARRAY(ary)->len;
}
return rb_ary_subseq(ary, beg, len);
}
if you look in the array.c class where the rb_ary_subseq method is defined, you see that it is returning nil if the length is out of bounds, not the index:
if (beg > RARRAY_LEN(ary)) return Qnil;
In this case this is what is happening when 4 is passed in, it checks that there are 4 elements and thus does not trigger the nil return. It then goes on and returns an empty array if the second arg is set to zero. while if 5 is passed in, there are not 5 elements in the array, so it returns nil before the zero arg is evaluated. code here at line 944.
I believe this to be a bug, or at least unpredictable and not the 'Principle of Least Surprise'. When I get a few minutes I will a least submit a failing test patch to ruby core.
At least note that the behavior is consistent. From 5 on up everything acts the same; the weirdness only occurs at [4,N].
Maybe this pattern helps, or maybe I'm just tired and it doesn't help at all.
array[0,4] => [:peanut, :butter, :and, :jelly]
array[1,3] => [:butter, :and, :jelly]
array[2,2] => [:and, :jelly]
array[3,1] => [:jelly]
array[4,0] => []
At [4,0], we catch the end of the array. I'd actually find it rather odd, as far as beauty in patterns go, if the last one returned nil. Because of a context like this, 4 is an acceptable option for the first parameter so that the empty array can be returned. Once we hit 5 and up, though, the method likely exits immediately by nature of being totally and completely out of bounds.
This makes sense when you consider than an array slice can be a valid lvalue, not just an rvalue:
array = [:peanut, :butter, :and, :jelly]
# replace 0 elements starting at index 5 (insert at end or array):
array[4,0] = [:sandwich]
# replace 0 elements starting at index 0 (insert at head of array):
array[0,0] = [:make, :me, :a]
# array is [:make, :me, :a, :peanut, :butter, :and, :jelly, :sandwich]
# this is just like replacing existing elements:
array[3, 4] = [:grilled, :cheese]
# array is [:make, :me, :a, :grilled, :cheese, :sandwich]
This wouldn't be possible if array[4,0] returned nil instead of []. However, array[5,0] returns nil because it's out of bounds (inserting after the 4th element of a 4-element array is meaningful, but inserting after the 5th element of a 4 element array is not).
Read the slice syntax array[x,y] as "starting after x elements in array, select up to y elements". This is only meaningful if array has at least x elements.
This does make sense
You need to be able to assign to those slices, so they are defined in such a way that the beginning and the end of the string have working zero-length expressions.
array[4, 0] = :sandwich
array[0, 0] = :crunchy
=> [:crunchy, :peanut, :butter, :and, :jelly, :sandwich]
I found explanation by Gary Wright very helpful as well.
http://www.ruby-forum.com/topic/1393096#990065
The answer by Gary Wright is -
http://www.ruby-doc.org/core/classes/Array.html
The docs certainly could be more clear but the actual behavior is
self-consistent and useful.
Note: I'm assuming 1.9.X version of String.
It helps to consider the numbering in the following way:
-4 -3 -2 -1 <-- numbering for single argument indexing
0 1 2 3
+---+---+---+---+
| a | b | c | d |
+---+---+---+---+
0 1 2 3 4 <-- numbering for two argument indexing or start of range
-4 -3 -2 -1
The common (and understandable) mistake is too assume that the semantics
of the single argument index are the same as the semantics of the
first argument in the two argument scenario (or range). They are not
the same thing in practice and the documentation doesn't reflect this.
The error though is definitely in the documentation and not in the
implementation:
single argument: the index represents a single character position
within the string. The result is either the single character string
found at the index or nil because there is no character at the given
index.
s = ""
s[0] # nil because no character at that position
s = "abcd"
s[0] # "a"
s[-4] # "a"
s[-5] # nil, no characters before the first one
two integer arguments: the arguments identify a portion of the string to
extract or to replace. In particular, zero-width portions of the string
can also be identified so that text can be inserted before or after
existing characters including at the front or end of the string. In this
case, the first argument does not identify a character position but
instead identifies the space between characters as shown in the diagram
above. The second argument is the length, which can be 0.
s = "abcd" # each example below assumes s is reset to "abcd"
To insert text before 'a': s[0,0] = "X" # "Xabcd"
To insert text after 'd': s[4,0] = "Z" # "abcdZ"
To replace first two characters: s[0,2] = "AB" # "ABcd"
To replace last two characters: s[-2,2] = "CD" # "abCD"
To replace middle two characters: s[1..3] = "XX" # "aXXd"
The behavior of a range is pretty interesting. The starting point is the
same as the first argument when two arguments are provided (as described
above) but the end point of the range can be the 'character position' as
with single indexing or the "edge position" as with two integer
arguments. The difference is determined by whether the double-dot range
or triple-dot range is used:
s = "abcd"
s[1..1] # "b"
s[1..1] = "X" # "aXcd"
s[1...1] # ""
s[1...1] = "X" # "aXbcd", the range specifies a zero-width portion of
the string
s[1..3] # "bcd"
s[1..3] = "X" # "aX", positions 1, 2, and 3 are replaced.
s[1...3] # "bc"
s[1...3] = "X" # "aXd", positions 1, 2, but not quite 3 are replaced.
If you go back through these examples and insist and using the single
index semantics for the double or range indexing examples you'll just
get confused. You've got to use the alternate numbering I show in the
ascii diagram to model the actual behavior.
I agree that this seems like strange behavior, but even the official documentation on Array#slice demonstrates the same behavior as in your example, in the "special cases" below:
a = [ "a", "b", "c", "d", "e" ]
a[2] + a[0] + a[1] #=> "cab"
a[6] #=> nil
a[1, 2] #=> [ "b", "c" ]
a[1..3] #=> [ "b", "c", "d" ]
a[4..7] #=> [ "e" ]
a[6..10] #=> nil
a[-3, 3] #=> [ "c", "d", "e" ]
# special cases
a[5] #=> nil
a[5, 1] #=> []
a[5..10] #=> []
Unfortunately, even their description of Array#slice doesn't seem to offer any insight as to why it works this way:
Element Reference—Returns the element at index, or returns a subarray starting at start and continuing for length elements, or returns a subarray specified by range. Negative indices count backward from the end of the array (-1 is the last element). Returns nil if the index (or starting index) are out of range.
An explanation provided by Jim Weirich
One way to think about it is that index position 4 is at the very edge
of the array. When asking for a slice, you return as much of the
array that is left. So consider the array[2,10], array[3,10] and
array[4,10] ... each returns the remaining bits of the end of the
array: 2 elements, 1 element and 0 elements respectively. However,
position 5 is clearly outside the array and not at the edge, so
array[5,10] returns nil.
Consider the following array:
>> array=["a","b","c"]
=> ["a", "b", "c"]
You can insert an item to the begining (head) of the array by assigning it to a[0,0]. To put the element between "a" and "b", use a[1,0]. Basically, in the notation a[i,n], i represents an index and n a number of elements. When n=0, it defines a position between the elements of the array.
Now if you think about the end of the array, how can you append an item to its end using the notation described above? Simple, assign the value to a[3,0]. This is the tail of the array.
So, if you try to access the element at a[3,0], you will get []. In this case you are still in the range of the array. But if you try to access a[4,0], you'll get nil as return value, since you're not within the range of the array anymore.
Read more about it at http://mybrainstormings.wordpress.com/2012/09/10/arrays-in-ruby/ .
tl;dr: in the source code in array.c, different functions are called depending on whether you pass 1 or 2 arguments in to Array#slice resulting in the unexpected return values.
(First off, I'd like to point out that I don't code in C, but have been using Ruby for years. So if you're not familiar with C, but you take a few minutes to familiarize yourself with the basics of functions and variables it's really not that hard to follow the Ruby source code, as demonstrated below. This answer is based on Ruby v2.3, but is more or less the same back to v1.9.)
Scenario #1
array.length == 4; array.slice(4) #=> nil
If you look at the source code for Array#slice (rb_ary_aref), you see that when only one argument is passed in (lines 1277-1289), rb_ary_entry is called, passing in the index value (which can be positive or negative).
rb_ary_entry then calculates the position of the requested element from the beginning of the array (in other words, if a negative index is passed in, it computes the positive equivalent) and then calls rb_ary_elt to get the requested element.
As expected, rb_ary_elt returns nil when the length of the array len is less than or equal to the index (here called offset).
1189: if (offset < 0 || len <= offset) {
1190: return Qnil;
1191: }
Scenario #2
array.length == 4; array.slice(4, 0) #=> []
However when 2 arguments are passed in (i.e. the starting index beg, and length of the slice len), rb_ary_subseq is called.
In rb_ary_subseq, if the starting index beg is greater than the array length alen, nil is returned:
1208: long alen = RARRAY_LEN(ary);
1209:
1210: if (beg > alen) return Qnil;
Otherwise the length of the resulting slice len is calculated, and if it's determined to be zero, an empty array is returned:
1213: if (alen < len || alen < beg + len) {
1214: len = alen - beg;
1215: }
1216: klass = rb_obj_class(ary);
1217: if (len == 0) return ary_new(klass, 0);
So since the starting index of 4 is not greater than array.length, an empty array is returned instead of the nil value that one might expect.
Question answered?
If the actual question here isn't "What code causes this to happen?", but rather, "Why did Matz do it this way?", well you'll just have to buy him a cup of coffee at the next RubyConf and ask him.
ary = [1, 4, 6, 9]
(0...ary.size).bsearch { |i|
ary[i] - 1
} # => nil
1 - ary[i] # => 0
When the code is written in a form ary[i] - 1 which doesn't work as expected.
What I am trying to do is to find the index of the number 1 in the array.
But 1 - ary[i] can return the number's index correctly. Why doesn't ary[i] - 1 work?
Array#bsearch is meant to perform binary search to find an element that meets certain criteria. As per documentation, if you return numeric values from the block, the find-any mode type of search is used.
The search starts at center of the sorted array - and if block returns negative value, it continues search in first half, and if block returns positive value, it continues the search in second half of the array.
In your case when you use ary[i] - 1, the value returned by block is always positive and search continues recursively on second half of the array - and never finds the value 1.
Here is the code with some debug statements:
ary = [1, 4, 6, 9]
p (0...ary.size).bsearch { |i|
puts "Elem: #{ary[i]} Index: #{i}"
ary[i] - 1
}
Output:
Elem: 4 Index: 1
Elem: 6 Index: 2
Elem: 9 Index: 3
nil
[Finished in 0.4s]
Array#bsearch returns an element of the array, not the index of a matching element.
You might want to use Array#index instead.
If you want to find the index of the element instead of the element itself, you need to use Array#bsearch_index. Note: this method was introduced in Ruby 2.3, which at the time of this writing has not been released yet (it will be released on Christmas 2015).
The feature request for Array#bsearch_index contains a comment by Yusuke Endoh showing how to implement Array#bsearch_index (and in fact Array#bsearch as well) based on Range#bsearch:
class Array
def bsearch_index(&blk)
return enum_for(__method__) unless blk
(0...size).bsearch {|i| yield self[i] }
end
end
When either running Ruby 2.3 or using the above monkey patch, you can then do:
ary.bsearch_index(&1.method(:-))
in order to find the index of the 1 element in your array.
The reason why it doesn't work with
ary.bsearch_index {|el | el - 1 }
is simple: the block violates the contract of bsearch_index (and also bsearch since they are the same). The block needs to return a positive number for indices left of the one you are searching for, a negative number for indices right of the one you are searching for, and 0 for indices within the range you are searching for. Your block does the opposite.
a = [1,2,3,4]
a[4,43]
# => []
a[5,43]
# => nil
What is the reason for this anomaly?
Assuming you mean this:
a = [1,2,3,4,5]
a[5,43]
# => []
a[6,43]
# => nil
From http://bugs.ruby-lang.org/issues/4245:
This is per spec.
Think of it this way:
[5..-1]: give me all elements after the 5th one and up to the last
one. There are none, so []
[6..-1]: give me all elements after the 6th
one and up to the last one. Say what? There is no 6th one, so returns
nil as the given request is out of bounds.
I agree with spartacus and the bug reporter from the link by Dylan -- there are indeed inconsistencies and Array#slice could use a change. Although yes, it is the spec; from Array#slice
For start and range cases the starting index is just before an element. Additionally, an empty array is returned when the starting index for an element range is at the end of the array ... Returns nil if the index (or starting index) are out of range.
[ 0, 1, 2, 3 ]
^ ^ ^ ^ ^
0 1 2 3 4
Let's say the array a was constructed as [0,1,2,3] , and the numbers below the hats represent where one would "start" when using the a[start,length] form. So a[0,0] is [], and a[1,0] is [], since we have not "hopped" anywhere. Similarly, a[4,0] is [], and that's fine. Now a[4,x] where x is any nonzero number, should return nil, in my opinion, since a[4] is undefined, and in theory, a[4,x] is asking for an array that would look like [undefined, undefined...]
For ranges, however, it appears that an array is built from all indices that appear in the range, so the "hat" start visualization does not apply. So a[0..2] returns [a[0],a1,a[2]], or [0,1,2] in this case.
a[4..x] should return nil for any x, since there is no a[4].
So working through the above exercise and found this solution on GitHub.
def count_between arr, lower, upper
return 0 if arr.length == 0 || lower > upper
return arr.length if lower == upper
range = (lower..upper).to_a
arr.select { |value| range.include?(value) }.length
end
I understand what the first three lines mean and why they return the values they do. What I'd like to understand are the following lines of code.
Line 4 (below) is defining "range" as a variable and uses the lower...upper as the range variables (just discovered you don't need to put an integer value in a range. What does '.to_a' mean, can't seem to find it in the ruby docs, and what does it do?
range = (lower..upper).to_a
Line 5 (below) is using an Array#select method and its saying select this value if the value is included in this range and then give me the Array#length of all selected values, but I don't quite understand A. what |value| is doing and what it means. B. range.include?(value) means is this value included in this range I am assuming.
arr.select { |value| range.include?(value) }.length
Actually, I'd simplify to this:
def count_between arr, lower, upper
return 0 if lower > upper
arr.count{|v| (lower..upper).include?(v)}
end
to_a is documented here; it returns an Array containing each element in the Range. However, there's no reason to call to_a on the Range before calling include?.
There's also no reason to special-case the empty array.
Returning the length of the array when lower equals upper makes no sense.
value is the name given to the value the block is called with. I think a simple v is better for such a trivial case.
select calls the block for each value in arr and returns a new Array containing the elements for which the block returns true, so the length of that new Array is the number of matching values. However, count exists, and makes more sense to use, since the count is all we care about.
Update: As #steenslag points out in the comments, Comparable#between? can be used instead of creating a Range on which to call include?, and this eliminates the need to ensure that lower is less than or equal to upper:
def count_between arr, lower, upper
arr.count{|v| v.between?(lower, upper)}
end
to_a means convert to array
irb(main):001:0> (1..5).to_a
=> [1, 2, 3, 4, 5]
select method passes each element to the block and Returns a new array containing all elements of ary for which the given block returns a true value.. In your case it simply checks if the value is contained in the range array. range is an array not a range.
## if arr is [1,5] for eg:
irb(main):005:0> [1,5].select {|value| range.include?(value)}
=> [1, 5]
irb(main):006:0> [1,5].select {|value| range.include?(value)}.length
=> 2
so the elements of arr are contained in the |value| variable inside the block.
It's a block.
As the documentation says: select "Returns a new array containing all elements of ary for which the given block returns a true value."
So for each object in arr it is passed to the block in which you provide whatever code you want to that returns true or false, and the select statement uses this result to add the value to the the array that it returns. And after that, length is called on the array.
So you have an array, you filter the array to contain only the numbers that are in the range, and then you take the length - effectively counting the number of elements.
I was going through the exercises in Ruby Koans and I was struck by the following Ruby quirk that I found really unexplainable:
array = [:peanut, :butter, :and, :jelly]
array[0] #=> :peanut #OK!
array[0,1] #=> [:peanut] #OK!
array[0,2] #=> [:peanut, :butter] #OK!
array[0,0] #=> [] #OK!
array[2] #=> :and #OK!
array[2,2] #=> [:and, :jelly] #OK!
array[2,20] #=> [:and, :jelly] #OK!
array[4] #=> nil #OK!
array[4,0] #=> [] #HUH?? Why's that?
array[4,100] #=> [] #Still HUH, but consistent with previous one
array[5] #=> nil #consistent with array[4] #=> nil
array[5,0] #=> nil #WOW. Now I don't understand anything anymore...
So why is array[5,0] not equal to array[4,0]? Is there any reason why array slicing behaves this weird when you start at the (length+1)th position??
Slicing and indexing are two different operations, and inferring the behaviour of one from the other is where your problem lies.
The first argument in slice identifies not the element but the places between elements, defining spans (and not elements themselves):
:peanut :butter :and :jelly
0 1 2 3 4
4 is still within the array, just barely; if you request 0 elements, you get the empty end of the array. But there is no index 5, so you can't slice from there.
When you do index (like array[4]), you are pointing at elements themselves, so the indices only go from 0 to 3.
this has to do with the fact that slice returns an array, relevant source documentation from Array#slice:
* call-seq:
* array[index] -> obj or nil
* array[start, length] -> an_array or nil
* array[range] -> an_array or nil
* array.slice(index) -> obj or nil
* array.slice(start, length) -> an_array or nil
* array.slice(range) -> an_array or nil
which suggests to me that if you give the start that is out of bounds, it will return nil, thus in your example array[4,0] asks for the 4th element that exists, but asks to return an array of zero elements. While array[5,0] asks for an index out of bounds so it returns nil. This perhaps makes more sense if you remember that the slice method is returning a new array, not altering the original data structure.
EDIT:
After reviewing the comments I decided to edit this answer. Slice calls the following code snippet when the arg value is two:
if (argc == 2) {
if (SYMBOL_P(argv[0])) {
rb_raise(rb_eTypeError, "Symbol as array index");
}
beg = NUM2LONG(argv[0]);
len = NUM2LONG(argv[1]);
if (beg < 0) {
beg += RARRAY(ary)->len;
}
return rb_ary_subseq(ary, beg, len);
}
if you look in the array.c class where the rb_ary_subseq method is defined, you see that it is returning nil if the length is out of bounds, not the index:
if (beg > RARRAY_LEN(ary)) return Qnil;
In this case this is what is happening when 4 is passed in, it checks that there are 4 elements and thus does not trigger the nil return. It then goes on and returns an empty array if the second arg is set to zero. while if 5 is passed in, there are not 5 elements in the array, so it returns nil before the zero arg is evaluated. code here at line 944.
I believe this to be a bug, or at least unpredictable and not the 'Principle of Least Surprise'. When I get a few minutes I will a least submit a failing test patch to ruby core.
At least note that the behavior is consistent. From 5 on up everything acts the same; the weirdness only occurs at [4,N].
Maybe this pattern helps, or maybe I'm just tired and it doesn't help at all.
array[0,4] => [:peanut, :butter, :and, :jelly]
array[1,3] => [:butter, :and, :jelly]
array[2,2] => [:and, :jelly]
array[3,1] => [:jelly]
array[4,0] => []
At [4,0], we catch the end of the array. I'd actually find it rather odd, as far as beauty in patterns go, if the last one returned nil. Because of a context like this, 4 is an acceptable option for the first parameter so that the empty array can be returned. Once we hit 5 and up, though, the method likely exits immediately by nature of being totally and completely out of bounds.
This makes sense when you consider than an array slice can be a valid lvalue, not just an rvalue:
array = [:peanut, :butter, :and, :jelly]
# replace 0 elements starting at index 5 (insert at end or array):
array[4,0] = [:sandwich]
# replace 0 elements starting at index 0 (insert at head of array):
array[0,0] = [:make, :me, :a]
# array is [:make, :me, :a, :peanut, :butter, :and, :jelly, :sandwich]
# this is just like replacing existing elements:
array[3, 4] = [:grilled, :cheese]
# array is [:make, :me, :a, :grilled, :cheese, :sandwich]
This wouldn't be possible if array[4,0] returned nil instead of []. However, array[5,0] returns nil because it's out of bounds (inserting after the 4th element of a 4-element array is meaningful, but inserting after the 5th element of a 4 element array is not).
Read the slice syntax array[x,y] as "starting after x elements in array, select up to y elements". This is only meaningful if array has at least x elements.
This does make sense
You need to be able to assign to those slices, so they are defined in such a way that the beginning and the end of the string have working zero-length expressions.
array[4, 0] = :sandwich
array[0, 0] = :crunchy
=> [:crunchy, :peanut, :butter, :and, :jelly, :sandwich]
I found explanation by Gary Wright very helpful as well.
http://www.ruby-forum.com/topic/1393096#990065
The answer by Gary Wright is -
http://www.ruby-doc.org/core/classes/Array.html
The docs certainly could be more clear but the actual behavior is
self-consistent and useful.
Note: I'm assuming 1.9.X version of String.
It helps to consider the numbering in the following way:
-4 -3 -2 -1 <-- numbering for single argument indexing
0 1 2 3
+---+---+---+---+
| a | b | c | d |
+---+---+---+---+
0 1 2 3 4 <-- numbering for two argument indexing or start of range
-4 -3 -2 -1
The common (and understandable) mistake is too assume that the semantics
of the single argument index are the same as the semantics of the
first argument in the two argument scenario (or range). They are not
the same thing in practice and the documentation doesn't reflect this.
The error though is definitely in the documentation and not in the
implementation:
single argument: the index represents a single character position
within the string. The result is either the single character string
found at the index or nil because there is no character at the given
index.
s = ""
s[0] # nil because no character at that position
s = "abcd"
s[0] # "a"
s[-4] # "a"
s[-5] # nil, no characters before the first one
two integer arguments: the arguments identify a portion of the string to
extract or to replace. In particular, zero-width portions of the string
can also be identified so that text can be inserted before or after
existing characters including at the front or end of the string. In this
case, the first argument does not identify a character position but
instead identifies the space between characters as shown in the diagram
above. The second argument is the length, which can be 0.
s = "abcd" # each example below assumes s is reset to "abcd"
To insert text before 'a': s[0,0] = "X" # "Xabcd"
To insert text after 'd': s[4,0] = "Z" # "abcdZ"
To replace first two characters: s[0,2] = "AB" # "ABcd"
To replace last two characters: s[-2,2] = "CD" # "abCD"
To replace middle two characters: s[1..3] = "XX" # "aXXd"
The behavior of a range is pretty interesting. The starting point is the
same as the first argument when two arguments are provided (as described
above) but the end point of the range can be the 'character position' as
with single indexing or the "edge position" as with two integer
arguments. The difference is determined by whether the double-dot range
or triple-dot range is used:
s = "abcd"
s[1..1] # "b"
s[1..1] = "X" # "aXcd"
s[1...1] # ""
s[1...1] = "X" # "aXbcd", the range specifies a zero-width portion of
the string
s[1..3] # "bcd"
s[1..3] = "X" # "aX", positions 1, 2, and 3 are replaced.
s[1...3] # "bc"
s[1...3] = "X" # "aXd", positions 1, 2, but not quite 3 are replaced.
If you go back through these examples and insist and using the single
index semantics for the double or range indexing examples you'll just
get confused. You've got to use the alternate numbering I show in the
ascii diagram to model the actual behavior.
I agree that this seems like strange behavior, but even the official documentation on Array#slice demonstrates the same behavior as in your example, in the "special cases" below:
a = [ "a", "b", "c", "d", "e" ]
a[2] + a[0] + a[1] #=> "cab"
a[6] #=> nil
a[1, 2] #=> [ "b", "c" ]
a[1..3] #=> [ "b", "c", "d" ]
a[4..7] #=> [ "e" ]
a[6..10] #=> nil
a[-3, 3] #=> [ "c", "d", "e" ]
# special cases
a[5] #=> nil
a[5, 1] #=> []
a[5..10] #=> []
Unfortunately, even their description of Array#slice doesn't seem to offer any insight as to why it works this way:
Element Reference—Returns the element at index, or returns a subarray starting at start and continuing for length elements, or returns a subarray specified by range. Negative indices count backward from the end of the array (-1 is the last element). Returns nil if the index (or starting index) are out of range.
An explanation provided by Jim Weirich
One way to think about it is that index position 4 is at the very edge
of the array. When asking for a slice, you return as much of the
array that is left. So consider the array[2,10], array[3,10] and
array[4,10] ... each returns the remaining bits of the end of the
array: 2 elements, 1 element and 0 elements respectively. However,
position 5 is clearly outside the array and not at the edge, so
array[5,10] returns nil.
Consider the following array:
>> array=["a","b","c"]
=> ["a", "b", "c"]
You can insert an item to the begining (head) of the array by assigning it to a[0,0]. To put the element between "a" and "b", use a[1,0]. Basically, in the notation a[i,n], i represents an index and n a number of elements. When n=0, it defines a position between the elements of the array.
Now if you think about the end of the array, how can you append an item to its end using the notation described above? Simple, assign the value to a[3,0]. This is the tail of the array.
So, if you try to access the element at a[3,0], you will get []. In this case you are still in the range of the array. But if you try to access a[4,0], you'll get nil as return value, since you're not within the range of the array anymore.
Read more about it at http://mybrainstormings.wordpress.com/2012/09/10/arrays-in-ruby/ .
tl;dr: in the source code in array.c, different functions are called depending on whether you pass 1 or 2 arguments in to Array#slice resulting in the unexpected return values.
(First off, I'd like to point out that I don't code in C, but have been using Ruby for years. So if you're not familiar with C, but you take a few minutes to familiarize yourself with the basics of functions and variables it's really not that hard to follow the Ruby source code, as demonstrated below. This answer is based on Ruby v2.3, but is more or less the same back to v1.9.)
Scenario #1
array.length == 4; array.slice(4) #=> nil
If you look at the source code for Array#slice (rb_ary_aref), you see that when only one argument is passed in (lines 1277-1289), rb_ary_entry is called, passing in the index value (which can be positive or negative).
rb_ary_entry then calculates the position of the requested element from the beginning of the array (in other words, if a negative index is passed in, it computes the positive equivalent) and then calls rb_ary_elt to get the requested element.
As expected, rb_ary_elt returns nil when the length of the array len is less than or equal to the index (here called offset).
1189: if (offset < 0 || len <= offset) {
1190: return Qnil;
1191: }
Scenario #2
array.length == 4; array.slice(4, 0) #=> []
However when 2 arguments are passed in (i.e. the starting index beg, and length of the slice len), rb_ary_subseq is called.
In rb_ary_subseq, if the starting index beg is greater than the array length alen, nil is returned:
1208: long alen = RARRAY_LEN(ary);
1209:
1210: if (beg > alen) return Qnil;
Otherwise the length of the resulting slice len is calculated, and if it's determined to be zero, an empty array is returned:
1213: if (alen < len || alen < beg + len) {
1214: len = alen - beg;
1215: }
1216: klass = rb_obj_class(ary);
1217: if (len == 0) return ary_new(klass, 0);
So since the starting index of 4 is not greater than array.length, an empty array is returned instead of the nil value that one might expect.
Question answered?
If the actual question here isn't "What code causes this to happen?", but rather, "Why did Matz do it this way?", well you'll just have to buy him a cup of coffee at the next RubyConf and ask him.