VHDL中mod和rem运算符的区别？

2 回答

• 24

  一种看待不同的方法是在测试平台上运行快速模拟，例如使用如下过程：

  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，因此在 a mod n 中 n

  • rem有分红的迹象，因此 a 在 a rem n

  mod 运算符给出一个向下舍入的除法（浮动除法）的残差，所以 a = floor_div(a, n) * n + (a mod n) . 优点是 a mod n 是一个重复的锯齿图，当 a 甚至通过零增加时，这在某些计算中很重要 .

  rem 运算符给出了常数整数除法 a / n 的余数，它朝向0（截断除法）舍入，因此 a = (a / n) * n + (a rem n) .

  回复于 2024-04-24T15:06:37+08:00
• 0

  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
  

  回复于 2024-04-24T15:06:37+08:00