Feature #8842

Integer#[] with range

Added by Yusuke Endoh 10 months ago. Updated 5 months ago.

[ruby-core:56902]
Status:Assigned
Priority:Normal
Assignee:Yusuke Endoh
Category:-
Target version:current: 2.2.0

Description

=begin
I propose to extend Integer#[] accepting a range.

0b01001101[2, 4] == 0b0011
0bHGFEDCBA[2, 4] == 0bFEDC

== Use case

I believe that everyone has written a code like this:

if (n >> 2) & 0xf == 0x3
  ...
end

because this is a very common idiom in C.
But it is less readable, writable, extendable and optimizable.

if n[2, 4] == 0x3
  ...
end

is much better in the all aspects.

== Corner cases

The current Integer# handle an integer as "a bit array with infinity length";
it returns 0 for any negative index and an (extended) sign bit for any index greater than MSB.
We also can use this standard to define the spec for a range argument.
For example:

15[-1, 42] #=> 30  (equivalent to (15 << 1) && (2 ** 42 - 1))
15[3, 42]  #=>  1  (equivalent to (15 >> 3) && (2 ** 42 - 1))
15[3..Float::INFINITY]  #=> 1  (equivalent to 15 >> 3)
15[-3..Float::INFINITY] #=> 2  (equivalent to 1 << 3)

-1[0..Float::INFINITY]   #=> -1
-1[1..Float::INFINITY]   #=> -1
-1[-1..Float::INFINITY]  #=> -2

1[-Float::INFINITY..0] #=> failed to allocate memory
2[-Float::INFINITY..0] #=> 0

Only tricky case that I thought of is a range (beg..end) whose "end" is smaller than "beg".
I think it should be handled as (beg..Float::INFINITY).

15[-3..-4] #=> 2  (equivalent to 1 << 3)
-1[0..-1] #=> -1
-1[0..-2] #=> -1

What do you think?
=end

integer-with-range.pdf (248 KB) Yusuke Endoh, 08/31/2013 11:47 AM

History

#1 Updated by Yusuke Endoh 10 months ago

#2 Updated by Yusuke Endoh 10 months ago

  • Tracker changed from Bug to Feature

#3 Updated by Yukihiro Matsumoto 10 months ago

  • Assignee changed from Yukihiro Matsumoto to Yusuke Endoh

Accepted.

Matz.

#4 Updated by Boris Stitnicky 10 months ago

I take it that the use of '&&' operator in the first 2 corner cases is a typo. But, pardon my ignorance,
in the 4th corner case, why 15[-3..Float::INFINITY] #=> 2 (equivalent to 1 << 3)?
1 << 3 is 8, and even if I assume a typo, 15 << 3 is 120. Is there something I misunderstood?

#5 Updated by Yusuke Endoh 10 months ago

  • Target version changed from next minor to 2.1.0

I take it that the use of '&&' operator in the first 2 corner cases is a typo.

Yes, sorry.

in the 4th corner case, why 15[-3..Float::INFINITY] #=> 2 (equivalent to 1 << 3)?
1 << 3 is 8, and even if I assume a typo, 15 << 3 is 120. Is there something I misunderstood?

No, you are right.

15[-3..Float::INFINITY] #=> 120 (equivalent to 15 << 3)

is correct.

Also, the attached slide has a bug: n[2..6] should be n[2..5] or n[2...6].

I'm sorry. Haste is from the devil. I'll make a patch more carefully.

Yusuke Endoh mame@tsg.ne.jp

#6 Updated by Hiroshi SHIBATA 5 months ago

  • Target version changed from 2.1.0 to current: 2.2.0

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