Это только первая часть моей домашней работы, я исправил все другие ошибки компиляции, но я продолжаю получать эту ошибку, пять.
1>\takehome\main.cpp(39) : error C2065: 'j' : undeclared identifier
1>\takehome\main.cpp(44) : error C2065: 'j' : undeclared identifier
\takehome\main.cpp(45) : error C2065: 'j' : undeclared identifier
\takehome\main.cpp(76) : error C2065: 'j' : undeclared identifier
\takehome\main.cpp(80) : error C2065: 'j' : undeclared identifier
Я пробовал делать все с этим, но я, вероятно, что-то делаю неправильно... очевидно, я. Я мог бы использовать какую-то помощь, если вы не возражаете:). Кстати, на случай, если кто-то задавался вопросом, делая simpletron.
#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 position = 0; //making the starting position to be 0.
for ( int 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. These are random variables.
memory [0] = 2942;
memory [1] = 2342;
memory [2] = 3523;
memory [3] = 2031;
memory [4] = 5000;
memory [5] = 8080;
memory [6] = 3425;
j = 0; //Makes the variable j start at 0.
while ( true )
{
memory[ j ]%100 = operand; // Finds the op codes from the limit on the memory (100)
memory[ j ]%100 = operation;
//using a switch loop to set up the loops for the cases
switch ( operation ){
case 1: //reads a variable into a word from loc.
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); break;
case 12: // Program ends
exit(0); break;
}
j++;
}
return 0;
}