Question

我在VHDL编程中遇到过这些语句，无法理解两个运算符mod和rem

之间的区别

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

Answer 1

一种看待不同的方法是在测试台上运行快速模拟示例使用这样的过程：

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;

它显示结果：

# ** 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

Wikipedia - Modulo operation 有详尽的描述，包括规则：

mod有除数的符号，因此n

a mod n

rem有分红的迹象，因此a

a rem n

mod运算符为舍入（浮动除法）的除法提供余数，因此a = floor_div(a, n) * n + (a mod n)。优点是a mod n是a即使在零增加时重复的锯齿图，这在某些计算中很重要。

rem运算符给出了向0（截断除法）舍入的常规整数除a / n的余数，因此a = (a / n) * n + (a rem n)。

Answer 2

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

VHDL中mod和rem运算符的区别？

2 个答案: