Why modulus operation work different in ruby than other languages?
up vote
1
down vote
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In Ruby, I get:
-5 % 3 # => 1
whereas other languages like PHP, Javascript, C++, and Java all produce the result -2. I don't understand this concept. I hope someone can explain this ruby's calculation method. It would be better if you could use an example of how it works.
ruby
edited Nov 21 at 5:50
sawa
128k27193297
asked Nov 19 at 16:06
Pradhumn Sharma
1116
add a comment |
up vote
1
down vote
favorite
In Ruby, I get:
-5 % 3 # => 1
whereas other languages like PHP, Javascript, C++, and Java all produce the result -2. I don't understand this concept. I hope someone can explain this ruby's calculation method. It would be better if you could use an example of how it works.
ruby
edited Nov 21 at 5:50
sawa
128k27193297
asked Nov 19 at 16:06
Pradhumn Sharma
1116
4
You want-5.remainder(3)which returns-2
– Stefan
Nov 19 at 16:51
1
AFAIK-5 % 3is implementation defined in C and C++, that can be either-2or1depending on how the CPU implements modulus with negatives and what it thinks "smallest" means:-2is smaller than1because-2 < 1but-2is bigger than1because|-2| > |1|.
– mu is too short
Nov 19 at 23:55
It is a matter of what you want as the set of representatives of the elements derived by modulo. It is whether you think{-2, -1, 0}more natural or{0, 1, 2}more natural. Ruby takes the latter, and so do I.
– sawa
Nov 21 at 5:56
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In Ruby, I get:
-5 % 3 # => 1
whereas other languages like PHP, Javascript, C++, and Java all produce the result -2. I don't understand this concept. I hope someone can explain this ruby's calculation method. It would be better if you could use an example of how it works.
ruby
edited Nov 21 at 5:50
sawa
128k27193297
asked Nov 19 at 16:06
Pradhumn Sharma
1116
In Ruby, I get:
-5 % 3 # => 1
whereas other languages like PHP, Javascript, C++, and Java all produce the result -2. I don't understand this concept. I hope someone can explain this ruby's calculation method. It would be better if you could use an example of how it works.
ruby
ruby
edited Nov 21 at 5:50
sawa
128k27193297
asked Nov 19 at 16:06
Pradhumn Sharma
1116
edited Nov 21 at 5:50
sawa
128k27193297
asked Nov 19 at 16:06
Pradhumn Sharma
1116
edited Nov 21 at 5:50
sawa
128k27193297
edited Nov 21 at 5:50
sawa
128k27193297
edited Nov 21 at 5:50
sawa
128k27193297
128k27193297
asked Nov 19 at 16:06
Pradhumn Sharma
1116
asked Nov 19 at 16:06
Pradhumn Sharma
1116
asked Nov 19 at 16:06
Pradhumn Sharma
1116
1116
4
You want-5.remainder(3)which returns-2
– Stefan
Nov 19 at 16:51
1
AFAIK-5 % 3is implementation defined in C and C++, that can be either-2or1depending on how the CPU implements modulus with negatives and what it thinks "smallest" means:-2is smaller than1because-2 < 1but-2is bigger than1because|-2| > |1|.
– mu is too short
Nov 19 at 23:55
It is a matter of what you want as the set of representatives of the elements derived by modulo. It is whether you think{-2, -1, 0}more natural or{0, 1, 2}more natural. Ruby takes the latter, and so do I.
– sawa
Nov 21 at 5:56
add a comment |
4
You want-5.remainder(3)which returns-2
– Stefan
Nov 19 at 16:51
1
AFAIK-5 % 3is implementation defined in C and C++, that can be either-2or1depending on how the CPU implements modulus with negatives and what it thinks "smallest" means:-2is smaller than1because-2 < 1but-2is bigger than1because|-2| > |1|.
– mu is too short
Nov 19 at 23:55
It is a matter of what you want as the set of representatives of the elements derived by modulo. It is whether you think{-2, -1, 0}more natural or{0, 1, 2}more natural. Ruby takes the latter, and so do I.
– sawa
Nov 21 at 5:56
4
4
You want
-5.remainder(3) which returns -2– Stefan
Nov 19 at 16:51
You want
-5.remainder(3) which returns -2– Stefan
Nov 19 at 16:51
1
1
AFAIK
-5 % 3 is implementation defined in C and C++, that can be either -2 or 1 depending on how the CPU implements modulus with negatives and what it thinks "smallest" means: -2 is smaller than 1 because -2 < 1 but -2 is bigger than 1 because |-2| > |1|.– mu is too short
Nov 19 at 23:55
AFAIK
-5 % 3 is implementation defined in C and C++, that can be either -2 or 1 depending on how the CPU implements modulus with negatives and what it thinks "smallest" means: -2 is smaller than 1 because -2 < 1 but -2 is bigger than 1 because |-2| > |1|.– mu is too short
Nov 19 at 23:55
It is a matter of what you want as the set of representatives of the elements derived by modulo. It is whether you think
{-2, -1, 0} more natural or {0, 1, 2} more natural. Ruby takes the latter, and so do I.– sawa
Nov 21 at 5:56
It is a matter of what you want as the set of representatives of the elements derived by modulo. It is whether you think
{-2, -1, 0} more natural or {0, 1, 2} more natural. Ruby takes the latter, and so do I.– sawa
Nov 21 at 5:56
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
It's in the docs: https://ruby-doc.org/core-2.5.0/Numeric.html#method-i-divmod
If
q, r = x.divmod(y), then
q = floor(x/y)
x = q*y + r
The quotient is rounded toward negative infinity
So q is -3 (-5 / 2 and round down, as per usual integer division rules). And r = x - q * y = -5 - -3 * 2 = 1
answered Nov 19 at 16:18
Sergio Tulentsev
177k29286301
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
It's in the docs: https://ruby-doc.org/core-2.5.0/Numeric.html#method-i-divmod
If
q, r = x.divmod(y), then
q = floor(x/y)
x = q*y + r
The quotient is rounded toward negative infinity
So q is -3 (-5 / 2 and round down, as per usual integer division rules). And r = x - q * y = -5 - -3 * 2 = 1
answered Nov 19 at 16:18
Sergio Tulentsev
177k29286301
add a comment |
up vote
2
down vote
It's in the docs: https://ruby-doc.org/core-2.5.0/Numeric.html#method-i-divmod
If
q, r = x.divmod(y), then
q = floor(x/y)
x = q*y + r
The quotient is rounded toward negative infinity
So q is -3 (-5 / 2 and round down, as per usual integer division rules). And r = x - q * y = -5 - -3 * 2 = 1
answered Nov 19 at 16:18
Sergio Tulentsev
177k29286301
add a comment |
up vote
2
down vote
up vote
2
down vote
It's in the docs: https://ruby-doc.org/core-2.5.0/Numeric.html#method-i-divmod
If
q, r = x.divmod(y), then
q = floor(x/y)
x = q*y + r
The quotient is rounded toward negative infinity
So q is -3 (-5 / 2 and round down, as per usual integer division rules). And r = x - q * y = -5 - -3 * 2 = 1
answered Nov 19 at 16:18
Sergio Tulentsev
177k29286301
It's in the docs: https://ruby-doc.org/core-2.5.0/Numeric.html#method-i-divmod
If
q, r = x.divmod(y), then
q = floor(x/y)
x = q*y + r
The quotient is rounded toward negative infinity
So q is -3 (-5 / 2 and round down, as per usual integer division rules). And r = x - q * y = -5 - -3 * 2 = 1
answered Nov 19 at 16:18
Sergio Tulentsev
177k29286301
answered Nov 19 at 16:18
Sergio Tulentsev
177k29286301
answered Nov 19 at 16:18
Sergio Tulentsev
177k29286301
answered Nov 19 at 16:18
Sergio Tulentsev
177k29286301
177k29286301
add a comment |
add a comment |
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4
You want
-5.remainder(3)which returns-2– Stefan
Nov 19 at 16:51
1
AFAIK
-5 % 3is implementation defined in C and C++, that can be either-2or1depending on how the CPU implements modulus with negatives and what it thinks "smallest" means:-2is smaller than1because-2 < 1but-2is bigger than1because|-2| > |1|.– mu is too short
Nov 19 at 23:55
It is a matter of what you want as the set of representatives of the elements derived by modulo. It is whether you think
{-2, -1, 0}more natural or{0, 1, 2}more natural. Ruby takes the latter, and so do I.– sawa
Nov 21 at 5:56