编辑：所以，我的编码中似乎确实有问题。每当我运行程序并输入变量时，它总是返回相同的答案..“位置 76 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熟悉，那就是原因。我要重申，我是编程新手，我不是最好的，所以我会为了简单起见。这也是一个 SML 程序。无论如何，这是一个家庭作业，我想要一个好成绩。所以我一直在寻找意见，并确保这个程序能做我希望他们正在寻找的东西。无论如何，这里有说明： 编写 SML（Simpletron 机器语言）程序来完成以下每一项任务：
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，一些正数和一些负数，然后计算 + 打印平均值。C) 读取一系列数字，确定并打印最大的数字。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。全部一起。
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;
}
方案 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;
}