在Erlang中，5 rem 3.给出2，-5 rem 3.给出-2。如果我理解你的问题，你会希望-5 rem 3.给出1，因为-5 = -2 * 3 + 1.
这是你想要的吗？

mod(X,Y) when X > 0 -> X rem Y;
mod(X,Y) when X < 0 -> Y + X rem Y;
mod(0,Y) -> 0.

Erlang模运算符是rem

Eshell V5.6.4  (abort with ^G)
1> 97 rem 10.
7

我用了以下药剂：

defp mod(x,y) when x > 0, do: rem(x, y);
defp mod(x,y) when x < 0, do: rem(x, y) + y;
defp mod(0,_y), do: 0

根据this blog post，它是rem。

上面的Y + X雷姆Y似乎错了：（Y + X）rem Y或Y +（X rem Y）产生不正确的结果。例如：设Y=3。如果X= -4，第一种形式返回-1，如果X=-3，第二种形式返回3，它们都不在[0; 3[.
我用这个代替：

% Returns the positive remainder of the division of X by Y, in [0;Y[. 
% In Erlang, -5 rem 3 is -2, whereas this function will return 1,  
% since -5 =-2 * 3 + 1.

modulo(X,Y) when X > 0 ->   
   X rem Y;

modulo(X,Y) when X < 0 ->   
    K = (-X div Y)+1,
    PositiveX = X + K*Y,
    PositiveX rem Y;

modulo(0,_Y) -> 
    0.

Erlang余数不适用于负数，因此您必须为负数参数编写自己的函数。

mod(A, B) when A > 0 -> A rem B;
mod(A, B) when A < 0 -> mod(A+B, B); 
mod(0, _) -> 0.

% console:
3> my:mod(-13, 5).
2

The accepted answer is wrong.
rem behaves exactly like the % operator in modern C. It uses truncated division.
The accepted answer fails for X<0 and Y<0. Consider mod(-5,-3) :

C:                     -5 % -3 == -2
rem:                 -5 rem -3 == -2
Y + X rem Y:    -3 + -5 rem -3 == -5 !! wrong !!

The alternative implementations for the modulo operator use floored division and Euclidean division. The results for those are

flooring division:   -5 mod -3 == -2
euclidean division:  -5 mod -3 == 1

So

Y + X rem Y

doesn't reproduce any modulo operator for X < 0 and Y < 0.
And rem works as expected -- it's using truncated division.