c++ - 在我的 C++ 程序中寻找一些输入。 Simpletron，机器语言

Question

编辑:看来我的编码确实有问题。每当我运行程序并输入一个变量时，它总是返回相同的答案.."位置76处的内容为0。

好的伙计们，几天前我在这里发布了一个问题，但这只是一个编译错误，所以如果这看起来很眼熟，那就是原因。我会重申，我是编程的新手，我不是最好的，所以我只是为了简单起见。这也是一个 SML 程序。无论如何，这是一项家庭作业，我希望在这方面取得好成绩。所以我一直在寻找意见，并确保这个程序会做我希望他们正在寻找的事情。无论如何，这里是说明:编写 SML(Simpletron 机器语言)程序来完成以下每项任务:

A) 使用哨兵控制的循环读取正数 s 并计算和打印它们的总和。输入负数时终止输入。B) 使用计数器控制循环读取七个数字，一些是正数，一些是负数，然后计算并打印平均值。C) 读取一系列数字，确定并打印出最大的数字。读取的第一个数字表示应该处理多少个数字。

不用多说，这是我的程序。一起。

程序A

#include <iostream>
using namespace std;

int main()
{
 int memory[100]; //Making it 100, since simpletron contains a 100 word mem.

 int operation; //taking the rest of these variables straight out of the book seeing as how they were italisized.

 int operand;

 int accum = 0; // the special register is starting at 0

 int j;

 for (j = 0; j < 100; j++ ) //Simply stating that for int j is = to 0, j must be less than 100 because that is the memory limit, and for every pass-through, increment j.
  memory[j] = 0;


 // This is for part a, it will take in positive variables in a sent-controlled loop and compute + print their sum. Variables from example in text.
 memory [00] = 1010;

 memory [01] = 2009;

 memory [02] = 3008;

 memory [03] = 2109;

 memory [04] = 1109;

 memory [05] = 4300;

 memory [06] = 1009;

 j = 0; //Makes the variable j start at 0.

 while ( true )
 {

  operand = memory[ j ]%100; // Finds the op codes from the limit on the memory (100)
  operation = memory[ j ]/100;

  //using a switch loop to set up the loops for the cases
  switch ( operation ){
   case 1: //reads a variable into a word from loc. Enter in -1 to exit
    cout <<"\n Input a positive variable:  ";
    cin >> memory[ operand ]; break;

   case 2: // takes a word from location
    cout << "\n\nThe content at location " << operand   << "is " << memory[operand]; break;

   case 3:// loads
    accum = memory[ operand ]; break;

   case 4: //stores
    memory[ operand ] = accum; break;

   case 5: //adds
    accum = accum + memory[ operand ]; break;


   case 6: // subtracts
    accum = accum - memory[ operand ]; break;

   case 7: //divides
    accum = accum / (memory[ operand ]); break;

   case 8: // multiplies
    accum = accum*memory [ operand ]; break;

   case 9: // Branches to location
    j = -1; break;

   case 10: //branches if acc. is < 0
    if (accum < 0)
    j = 5; 
    break;

   case 11: //branches if acc = 0
    if (accum == 0)
     j = 5; 
    break;

   case 12: // Program ends
    exit(0); break;
 }
 j++;
 }
return 0;
}

程序B

//Part b finding the sum + avg.

int main()
{
 int mem[100];
 int operation;
 int operand;
 int accum = 0;
 int pos = 0;

 int j;

 for (j = 0; j < 100; j++ ) 
  memory[j] = 0;

 mem[22] = 7; // loop 7 times
 mem[25] = 1; // increment by 1

 mem[00] = 4306;

 mem[01] = 2303;

 mem[02] = 3402;

 mem[03] = 6410;

 mem[04] = 3412;

 mem[05] = 2111;

 mem[06] = 2002;

 mem[07] = 2312;

 mem[08] = 4210;

 mem[09] = 2109;

 mem[10] = 4001;

 mem[11] = 2015;

 mem[12] = 3212;

 mem[13] = 2116;

 mem[14] = 1101;

 mem[15] = 1116;

 mem[16] = 4300;

 j = 0;

 while ( true )
 {

  operand = memory[ j ]%100; // Finds the op codes from the limit on the memory (100)
  operation = memory[ j ]/100;

  //using a switch loop to set up the loops for the cases
  switch ( operation ){
   case 1: //reads a variable into a word from loc. Enter in -1 to exit
    cout <<"\n enter #:  ";
    cin >> memory[ operand ]; break;

   case 2: // takes a word from location
    cout << "\n\nThe content at location " << operand   << "is " << memory[operand]; break;

   case 3:// loads
    accum = memory[ operand ]; break;

   case 4: //stores
    memory[ operand ] = accum; break;

   case 5: //adds
    accum = accum + memory[ operand ]; break;


   case 6: // subtracts
    accum = accum - memory[ operand ]; break;

   case 7: //divides
    accum = accum / (memory[ operand ]); break;

   case 8: // multiplies
    accum = accum*memory [ operand ]; break;

   case 9: // Branches to location
    j = operand; break;

   case 10: //branches if acc. is < 0

    break;

   case 11: //branches if acc = 0
    if (accum == 0)
     j = operand; 
    break;

   case 12: // Program ends
    exit(0); break;
 }
 j++;
 }
return 0;
}

程序 C

///Part c
int main()
{
 int mem[100];
 int operation;
 int operand;
 int accum = 0;


 int j;

 for (j = 0; j < 100; j++ ) //Simply stating that for int j is = to 0, j must be less than 100 because that is the memory limit, and for every pass-through, increment j.
  memory[j] = 0;

 mem[23] = 1; //decrements 1 place in mem

 mem[0] = 1030; // Takes in # of values to be stored.

 mem[01] = 4123; // These 4 memory slots check for the largest variable then store
 mem[02] = 4134;
 mem[03] = 1011;
 mem[04] = 3204;

 mem[05] = 4005; // These 5 decrement the count+ store + branch.
 mem[06] = 4006;
 mem[07] = 4007;
 mem[08] = 4008;
 mem[09] = 4009;

 mem[10] = 4010;
 mem[11] = 4311; // exits

 j = 0; // this is the starting value..

 while ( true )
 {

  operand = memory[ j ]%100; // Finds the op codes from the limit on the memory (100)
  operation = memory[ j ]/100;

  //using a switch loop to set up the loops for the cases
  switch ( operation ){
   case 1: //reads a variable into a word from loc. Enter in -1 to exit
    cout <<"\n enter #:  ";
    cin >> memory[ operand ]; break;

   case 2: // takes a word from location
    cout << "\n\nThe content at location " << operand   << "is " << memory[operand]; break;

   case 3:// loads
    accum = memory[ operand ]; break;

   case 4: //stores
    memory[ operand ] = accum; break;

   case 5: //adds
    accum = accum + memory[ operand ]; break;


   case 6: // subtracts
    accum = accum - memory[ operand ]; break;

   case 7: //divides
    accum = accum / (memory[ operand ]); break;

   case 8: // multiplies
    accum = accum*memory [ operand ]; break;

   case 9: // Branches to location
    j = operand; break;

   case 10: //branches if acc. is < 0

    break;

   case 11: //branches if acc = 0
    if (accum == 0)
     j = operand; 
    break;

   case 12: // Program ends
    exit(0); break;
   case 13: // checks > than
    if (accum < mem[operand])
     accum = mem[operand];
    break;
  }
 j++;
 }
return 0;
}

Answer 1

您的整个代码中都有神奇的数字。你应该这样做:

const int OP_LOAD = 3;
const int OP_STORE = 4;
const int OP_ADD = 5;
...

const int OP_LOCATION_MULTIPLIER = 100;

mem[0] = OP_LOAD * OP_LOCATION_MULTIPLIER + ...;
mem[1] = OP_ADD * OP_LOCATION_MULTIPLIER + ...;

operand = memory[ j ] % OP_LOCATION_MULTIPLIER;
operation = memory[ j ] / OP_LOCATION_MULTIPLIER;