VHDL中的mod和rem运算符之间的区别?

问题描述

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

I came across these statements in VHDL programming and could not understand the difference between the two operators mod and rem

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

推荐答案

查看不同之处的一种方法是在测试台上运行快速模拟，例如使用如下流程的示例:

A way to see the different is to run a quick simulation in a test bench, forexample using a process like this:

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;

显示结果为:

# ** 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

维基百科 - 模运算有详细的描述，包括规则:

Wikipedia - Modulo operationhas an elaborate description, including the rules:

• mod 有除数的符号，因此 n 在 a mod n
• rem 有被除数，因此 a 在 a rem n

• mod has sign of divisor, thus n in a mod n
• rem has sign of dividend, thus a in a rem n

mod 运算符给出向下舍入的除法的余数(下除法)，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 增加甚至通过零时，这在某些计算中很重要.

The mod operator gives the residue for a division that rounds down (floored division), so a = floor_div(a, n) * n + (a mod n). The advantage is that a mod n is a repeated sawtooth graph when a is increasing even through zero, which is important in some calculations.

rem 运算符给出向 0(截断除法)舍入的常规整数除法 a/n 的余数，所以 a = (a/n) * n + (a rem n).

The rem operator gives the remainder for the regular integer division a / n that rounds towards 0 (truncated division), so a = (a / n) * n + (a rem n).

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