VHDL中mod和rem运算符的区别？

我在VHDL编程中遇到了这些语句，无法理解mod和rem两个操作符之间的区别

    9 mod 5
    (-9) mod 5
    9 mod (-5)
    9 rem 5
    (-9) rem 5
    9 rem (-5)

---

例如，在测试台上运行快速模拟可以看出不同之处 使用如下流程的示例：

process is
begin
  report "  9  mod   5  = " & integer'image(9 mod 5);
  report "  9  rem   5  = " & integer'image(9 rem 5);
  report "  9  mod (-5) = " & integer'image(9 mod (-5));
  report "  9  rem (-5) = " & integer'image(9 rem (-5));
  report "(-9) mod   5  = " & integer'image((-9) mod 5);
  report "(-9) rem   5  = " & integer'image((-9) rem 5);
  report "(-9) mod (-5) = " & integer'image((-9) mod (-5));
  report "(-9) rem (-5) = " & integer'image((-9) rem (-5));
  wait;
end process;

结果表明：

# ** Note:   9  mod   5  =  4
# ** Note:   9  rem   5  =  4
# ** Note:   9  mod (-5) = -1
# ** Note:   9  rem (-5) =  4
# ** Note: (-9) mod   5  =  1
# ** Note: (-9) rem   5  = -4
# ** Note: (-9) mod (-5) = -4
# ** Note: (-9) rem (-5) = -4

有详细的描述，包括规则：

• mod有除数符号，因此

  n

  在

  a mod n

• rem有红利的符号，因此

  a

  in

  a rem n

mod

运算符给出向下舍入的除法的余数（floorred除法），因此

a=floor\u div（a，n）*n+（a mod n）

。优点是当

a

增加甚至超过零时，

a mod n

是一个重复的锯齿图，这在某些计算中很重要

rem

运算符为向0（截断除法）舍入的常规整数除法

a/n

提供余数，因此

a=（a/n）*n+（a rem n）

For equal sign:
9/5=-9/-5=1.8 gets 1 
9 mod 5 = 9 rem 5
-9 mod -5 = -9 rem -5
-----------------------------------------
For unequal signs:
9/-5 = -9/5 = -1.8
In "mod" operator : -1.8 gets -2
In "rem" operator : -1.8 gets -1
----------------------------------------
example1: (9,-5)
9 = (-5*-2)-1  then:  (9 mod -5) = -1
9 = (-5*-1)+4  then:  (9 rem -5) = +4
----------------------------------------
example2: (-9,5)
-9 = (5*-2)+1  then:  (-9 mod 5) = +1
-9 = (5*-1)-4  then:  (-9 rem 5) = -4
----------------------------------------
example3: (-9,-5)
-9 = (-5*1)-4  then:  (-9 mod -5) = -4
-9 = (-5*1)-4  then:  (-9 rem -5) = -4
----------------------------------------
example4: (9,5)
9 = (5*1)+4  then:  (9 mod 5) = +4
9 = (5*1)+4  then:  (9 rem 5) = +4