汇编中的随机数 [英] Random number in assembly
问题描述
我是汇编新手，想知道如何在 EMU8086 中编写程序，在每次运行时打印不同的随机数.不使用中断可以做到吗?
I am new to assembly and would like to know how to write a program in EMU8086 that prints a different random number in every run of it. Is it possible to do it without using interrupts?
推荐答案
如果您使用的是真实版本的 DOS(不是 EMU8086)@fuz 方法是您可以做到的方式，它不需要中断.您只需在 BIOS 数据区(BDA).该位置的值是一个 32 位值，表示自午夜以来的计时器滴答数.不幸的是，EMU8086 不支持这种方法.
If you were using a real version of DOS (not EMU8086) @fuz method is the way you can do it, and it doesn't require interrupts. You just read the lower 16-bits of the 32-bit value at memory address 0x46c (0x00040:0x006c) in the BIOS Data Area(BDA). The value at that location is a 32-bit value representing the number of timer ticks since midnight. Unfortunately EMU8086 doesn't support this method.
要在带有中断(系统调用)的 EMU8086 中获取随机数，您可以使用 Int1ah/ah=0h:
To get a random number in EMU8086 with interrupts (system call) you can use Int 1ah/ah=0h:
时间 - 获取系统时间
TIME - GET SYSTEM TIME
AH = 00h
Return:
CX:DX = number of clock ticks since midnight
AL = midnight flag, nonzero if midnight passed since time last read
然后您可以使用该值并将其打印出来.该值是半随机的.您可以直接打印出来，但最好将其传递到 伪随机数生成器(PRNG) 作为种子值.有关基本LCG，请参阅以下部分.尽管 EMU8086 有一个 宏/函数 这样做.这段代码可以产生一个 1 到 10 之间的半随机数并打印出来:
You can then use that value and print it out. The value is semi random. You can print it out directly but it is preferable to pass it into a Pseudo-random Number Generator (PRNG) as a seed value. See section below for a basic LCG. Printing an integer is a separate issue although EMU8086 has a macro/function to do that. This code could produce a semi-random number between 1 and 10 and print it:
org 100h
include emu8086.inc
xor ax,ax ; xor register to itself same as zeroing register
int 1ah ; Int 1ah/ah=0 get timer ticks since midnight in CX:DX
mov ax,dx ; Use lower 16 bits (in DX) for random value
xor dx,dx ; Compute randval(DX) mod 10 to get num
mov bx,10 ; between 0 and 9
div bx ; Divide dx:ax by bx
inc dx ; DX = modulo from division
; Add 1 to give us # between 1 and 10 (not 0 to 9)
mov ax,dx ; Move to AX to print
call PRINT_NUM_UNS ; Print value in AX as unsigned
ret
DEFINE_PRINT_NUM_UNS ; Needed to support EMU8086 PRINT_NUM_UNS function
END
每次运行这个程序时，它应该打印一个 1 到 10 之间的数字.从时钟滴答中获得随机值后，我们将其转换为 1 到 10 之间的数字.代码与这个伪代码类似1:
Each time you run this program it should print a number between 1 and 10. After we get the random value from the clock ticks we convert it to a number between 1 and 10. The code would have been similar to this pseudo code1:
unsigned char num = (get_rand_value() % 10) + 1
我们除以 10 并使用模数(模值将在 0 到 9 之间)并加 1 使其成为 1 到 10 之间的值.get_rand_value 实际上是 Int 1ah/AH=0 系统调用.
We divide by 10 and use the modulo (Modulo value will be between 0 and 9) and add 1 to make it a value between 1 and 10. get_rand_value is effectively the Int 1ah/AH=0 system call.
注意:时钟滴答是半随机源，转换为1到10的值的方法有模偏差.我将上面的代码作为一种快速而肮脏的方法提供，但应该足以让你开始你的任务.
Note: The clock ticks is a semi random source, and the method to convert to a value from 1 to 10 suffers from modulo bias. I present the code above as a quick and dirty method but should be enough to get you started on your assignment.
无需发布INT 指令，但我们仍然通过对中断处理程序的代码进行间接 FAR CALL 来使用中断向量表.我怀疑这是否是您在询问是否可以不受干扰地完成的问题时的想法.引擎盖下的 INT 指令推送当前的 FLAGS 寄存器(使用 PUSHF) 后跟相当于 FAR CALL.控制被转移到位于中断向量表 (IVT) 中的 FAR 地址 0x0000:[interrupt_num * 4].当中断例程完成时，它会发出一个 IRET 指令来撤销推送，在 FAR CALL 之后恢复标志并返回指令.修改后的代码可能如下所示:
It is possible to do this without issuing an INT instruction, but we are still using the interrupt vector table by doing an indirect FAR CALL to the code for the interrupt handler. I doubt this is what you had in mind when you asked the question whether it can be done without interrupts. An INT instruction under the hood pushes the current FLAGS register (using PUSHF) followed by the equivalent of a FAR CALL. Control is transferred to the FAR address at 0x0000:[interrupt_num * 4] which is in the interrupt vector table (IVT). When the interrupt routine finishes it will issue an IRET instruction which undoes the pushes, restores the flags and returns the instruction after the FAR CALL. The revised code could look like:
org 100h
include emu8086.inc
xor ax,ax ; xor register to itself same as zeroing register
mov es,ax ; Zero the ES register for use with FAR JMP below so that we
; can make a FAR CALL relative to bottom of Interrupt Vector Table
; in low memory (0x0000 to 0x03FF)
; Do a system call without the INT instruction
; This is advanced assembly and relies on the
; understanding of how INT/IRETD work. We fake a
; system call by pushing FLAGS and rather
; than use int 1ah we do a FAR CALL indirectly
; through the interrupt vector table in lower memory
pushf ; Push FLAGS
call far es:[1ah*4] ; Indirectly call Int 1ah/ah=0 through far pointer in IVT
; get timer ticks since midnight in CX:DX
mov ax,dx ; Use lower 16 bits (in DX) for random value
xor dx,dx ; Compute randval(DX) mod 10 to get num
mov bx,10 ; between 0 and 9
div bx
inc dx ; DX = modulo from division
; Add 1 to give us # between 1 and 10 (not 0 to 9)
mov ax,dx ; Move to AX to print
call PRINT_NUM_UNS ; Print value in AX as unsigned
ret
DEFINE_PRINT_NUM_UNS ; Macro from include file to make PRINT_NUM_UNS usable
END
相关问题?可能的问题.简单的 LCG PRNG
还有一个与此模糊相似的问题是在此问题的一天内发布.如果此分配与另一个相关，则需要注意的是，如果您尝试从系统计时器滴答中快速连续获取随机数，则会遇到问题.在我上面的回答中，我说:
Related Question? Possible Issues. Simple LCG PRNG
There is another question vaguely similar to this that was posted within a day of this one. If this assignment is related to the other one then it needs to be noted that you will encounter issues if you attempt to get random numbers in quick succession from the system timer tick. In my answer above I stated:
该值为半随机.您可以直接打印出来，但最好将它传递到伪随机数生成器 (PRNG) 作为种子值.
The value is semi random. You can print it out directly but it is preferable to pass it into a Pseudo Random Number Generator (PRNG) as a seed value.
计时器分辨率为每秒 18.2 次.这不是一个非常高的分辨率，并且很可能一个接一个地调用 Int 1ah/ah=0 将导致返回相同的数字，或者第二个调用有更高的机会返回更高的值首先.这可以通过创建一个 PRNG(比如一个简单的 LCG)并使用一次计时器值来解决播种.对于您需要的每个值 - 您查询 PRNG 以获取下一个值而不是系统时间.
The timer resolution is 18.2 times a second. That isn't a very high resolution and likely calling Int 1ah/ah=0 one after the other will result in the same number being returned or the second call having a higher chance of returning a higher value than the first. This can be resolved by creating a PRNG (like a simple LCG) and use the timer value once to seed it. For each value you need - you query the PRNG for the next value not the system time.
一个简单的基于LCG的PRNG可以在这个相关的Stackoverflow 答案.根据该答案，您可以创建一个 srandsystime 函数来为 PRNG 设置定时器滴答的种子，以及一个 rand() 函数，该函数返回 PRNG 的下一个值.下面的代码演示了设置种子一次，然后显示 1 到 10 之间的两个随机值:
A simple LCG based PRNG can be found in this related Stackoverflow Answer. Based on that answer you can create an srandsystime function to seed the PRNG with the timer ticks, and a rand() function that returns the next value from the PRNG. The code below demonstrates setting the seed once, and then displaying two random values between 1 and 10:
org 100h
include emu8086.inc
start:
call srandsystime ; Seed PRNG with system time, call once only
call rand ; Get a random number in AX
call rand2num1to10 ; Convert AX to num between 1 and 10
call PRINT_NUM_UNS ; Print value in AX as unsigned
PRINT ", " ; Print delimiter between numbers
call rand ; Get another random number in AX
call rand2num1to10 ; Convert AX to num between 1 and 10
call PRINT_NUM_UNS ; Print value in AX as unsigned
ret
; Return number between 1 and 10
;
; Inputs: AX = value to convert
; Return: (AX) value between 1 and 10
rand2num1to10:
push dx
push bx
xor dx,dx ; Compute randval(DX) mod 10 to get num
mov bx,10 ; between 0 and 9
div bx
inc dx ; DX = modulo from division
; Add 1 to give us # between 1 and 10 (not 0 to 9)
mov ax,dx
pop bx
pop dx
ret
; Set LCG PRNG seed to system timer ticks
;
; Inputs: AX = seed
; Modifies: AX
; Return: nothing
srandsystime:
push cx
push dx
xor ax, ax ; Int 1Ah/AH=0 to get system timer in CX:DX
int 1ah
mov [seed], dx ; seed = 16-bit value from DX
pop dx
pop cx
ret
; Updates seed for next iteration
; seed = (multiplier * seed + increment) mod 65536
; multiplier = 25173, increment = 13849
;
; Inputs: none
; Return: (AX) random value
rand:
push dx
mov ax, 25173 ; LCG Multiplier
mul word ptr [seed] ; DX:AX = LCG multiplier * seed
add ax, 13849 ; Add LCG increment value
mov [seed], ax ; Update seed
; AX = (multiplier * seed + increment) mod 65536
pop dx
ret
seed dw 11 ; Default initial seed of 11
DEFINE_PRINT_NUM_UNS; Macro from include file to make PRINT_NUM_UNS usable
END
脚注:
1要获得
lower到upper(含)范围内的随机数，您可以使用以下通用公式:
1To get a random number in the range
lowertoupper(inclusive) you can use this general formula:
rndvalue = (rand() % (upper-lower+1)) +lower;
rndvalue = (rand() % (upper-lower+1)) + lower;
不足之处:将 PRNG 中的随机值转换为 1 到 10 之间的数字仍然存在模偏差.
Deficiency: converting the random value from the PRNG to a number between 1 and 10 still suffers from modulo bias.
我使用 Watcom 的寄存器调用约定(第 12 页的说明) 通常在开发 16 位汇编程序时.这可以根据自己的需求量身定制.
I use Watcom's register calling convention (Page 12 for a description) when developing 16-bit assembly routines in general. That can be tailored to one's own needs.
这个特定的 LCG PRNG 在模式重复之前有大约 65536 的周期.对于大多数简单的任务，这应该足够了.
This particular LCG PRNG has a period of about 65536 before the pattern repeats. This should be enough for most simple tasks.
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