问题
Edit: So, seems I do have a problem somewhere in my coding. Whenever I run the program and input a variable, it always returns the same answer.."The content at location 76 is 0.
Okay guys, I posted on here a few days ago about a question but it was just a compilation error, so if this looks familiar, thats why. I will reiterate, I'm new to programming, I'm not the best, so I'm going for simplicity. Also this is an SML program. Anyway, this IS a homework assignment and I'm wanting a good grade on this. So I was looking for input and making sure this program will do what I'm hoping they are looking for. Anyway, here are the instructions: Write SML (Simpletron Machine language) programs to accomplish each of the following task:
A) Use a sentinel-controlled loop to read positive number s and compute and print their sum. Terminate input when a neg number is entered. B) Use a counter-controlled loop to read seven numbers, some positive and some negative, and compute + print the avg. C) Read a series of numbers, and determine and print the largest number. The first number read indicates how many numbers should be processed.
Without further a due, here is my program. All together.
Program 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;
}
Program 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;
}
Program 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;
}
回答1:
You have magic numbers throughout your code. You should do things like:
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;
回答2:
int memory[100] = {0};will define memory as an array ofintand all of its elements initialized to 0. You wont need theforloop then.- You can use shorthand notation. e.g.
accum += memory[operand];
in place ofaccum = accum + memory[operand];. Its more readable this way.
回答3:
mem[09] = 4009;
Which octal number is 09 exactly?? :)
Edit: For program A (and the rest for that matter) you may find it informative to print out the 'operand' and 'operation' variables before you hit the switch statement. Output:
Operand: 10 Operation: 10
Operand: 9 Operation: 20
Operand: 8 Operation: 30
Operand: 9 Operation: 21
Operand: 9 Operation: 11
Operand: 9 Operation: 10
Operand: 0 Operation: 0
Operand: 0 Operation: 0
... *LOTS of lines omitted....
Operand: 0 Operation: 0
Operand: -97 Operation: -9498072
Operand: 0 Operation: 0
Operand: 88 Operation: 20898776
Operand: 12 Operation: 0
Operand: 8 Operation: 22856
Operand: 69 Operation: 20898775
The code runs through the switch() almost 1300 times before finally asking for input from the user. It seems to me that program A goes off the rails when the operand = 0, which happens when you hit this line: memory [05] = 4300;
回答4:
I would suggest that try made the programs more modular, through the usage of functions. I'm assuming you've been introduced to them.
The initialization (clearing memory to zero values) and interpreter are prime candidates to be made into functions. These functions are the same (i.e. common) for all your examples. In both cases the only function parameter is the memory.
I think that combined with @R Samuel Klatchko's suggestions would make the programs easier to read for yourself and the marker.
Finally for 'extra credit' I would suggest that you validate the user's input (from cin) before storing in memory. Such as making sure it is a positive integer in the appropriate range. This input & validation could be made into its own function to avoid nesting too deep in your program.
I don't know, but I suspect that Program A incorrectly has a fixed address being branched to on instructions 10 and 11.
回答5:
This is now over 2 yrs old and thus no more use to your assignment, but others may end up here, so here goes...
The clue to this question is in your first paragraph, which ends:
"Write SML (Simpletron Machine language) programs to accomplish each of the following task: [sic]"
The key words are "Write SML programs..." - you have attempted to write C++ programs, which isn't likely to gain you many assignment credits.
I believe this assignment originates in Deitel & Deitel's book 'C: How to program'. In their book they present a very simple machine simulator, Simpletron, as an exercise. SML is the machine language of the this hypothetical computer. Having introduced the machine and its language, the exercise starts by requiring the student to write a few simple programs in SML. Later on the student is asked to construct the Simpletron in C.
来源：https://stackoverflow.com/questions/2412132/looking-for-some-input-on-my-c-program-simpletron-machine-language