Elixir中的Modulo运算符

Question

How do I use the modulo operator in Elixir?

For example in Ruby you can do:

5 % 2 == 0

How does it differ from Ruby's modulo operator?

Answer 1

Use rem/2 see: https://hexdocs.pm/elixir/Kernel.html#rem/2 So in Elixir your example becomes:

rem(5,2) == 0

which returns false

BTW what you wrote using % in Elixir simply starts a comment to the end of the line.

Answer 2

For integers, use Kernel.rem/2 :

iex(1)> rem(5, 2)
1
iex(2)> rem(5, 2) == 0
false

From the docs:

Computes the remainder of an integer division.

rem/2 uses truncated division, which means that the result will always have the sign of the dividend .

Raises an ArithmeticError exception if one of the arguments is not an integer, or when the divisor is 0.

The main differences compared to Ruby seem to be:

1. rem only works with integers, but % changes its behavior completely depending on the datatype.
2. The sign is negative for negative dividends in Elixir (the same as Ruby's remainder ):

Ruby:

irb(main):001:0> -5 / 2
=> -3
irb(main):002:0> -5 % 2
=> 1
irb(main):003:0> -5.remainder(2)
=> -1

Elixir:

iex(1)> -5 / 2
-2.5
iex(2)> rem(-5, 2)
-1

Elixir's rem just uses Erlang's rem , so this related Erlang question may also be useful.

Answer 3

Use Integer.mod/2 , see: https://hexdocs.pm/elixir/Integer.html#mod/2

iex>Integer.mod(-5, 2)
1