Thread: simple ALU

I am trying to write a simple ALU. I think I am on the right track, but don't really know how I can improve on it. I also dont fully understand what to do with the carryin value? I found this site on google and tried a search but could not find anything related(sorry if it's already been covered). Any tips?

Code:

#include <iostream>
#include <cstdlib>
#include <string> 

using namespace std;

int main() {
/*4-bit ALU*/
		char A[5] = "0101";
		char B[5] = "0011";
		char zero[2] = "0";
		int a[5];
		int b[5];
		char Cin[2] = "0";
		int Result[5];
		
		/*Makes an int out of input A*/
		for(int i=0; i<4; i++){
			if(A[i] == zero[0]){
				a[i] = 0;
			}
			else{
				a[i] = 1;
			}
		}
		/*Makes an int out of input B*/
		for(int i=0; i<4; i++){
			if(B[i] == zero[0]){
				b[i] = 0;
			}
			else{
				b[i] = 1;
			}
		}
		cout << "4 bit ALU" << endl;
		cout << "--------------------";
		cout << "\nThe first input is:  ";
		for(int i=0; i<4; i++){
			cout << a[i];
			}
		cout << "\nThe second input is: ";
		for(int i=0; i<4; i++){
			cout << b[i];
			}
		cout << "\nThe carry-in is:        ";
		for(int i=0;i<1;i++){
			cout << Cin[i];
		}
		cout << "\n--------------------";
		/*AND*/	
		for(int i=0; i<4; i++){
			Result[i] = a[i]&b[i];
			}
		cout << "\nThe Result of AND is: ";
		for(int i=0;i<4;i++){
			cout << Result[i];
		}
		cout << "\n";
		/*OR*/
		for(int i=0; i<4; i++){
			Result[i] = a[i]|b[i];
			}
		cout << "The Result of OR is:  ";
		for(int i=0;i<4;i++){
			cout << Result[i];
		}
		cout << "\n";
		/*XOR*/
		for(int i=0; i<4; i++){
			Result[i] = a[i]^b[i];
			}
		cout << "The Result of XOR is: ";
		for(int i=0;i<4;i++){
			cout << Result[i];
		}
		
		cout << "\n";
/*32 bit ALU*/
		char A2[33] = "000011111010011001011001";
		char B2[33] = "001100101101111000100101";
		char zero2[2] = "0";
		int a2[33];
		int b2[33];
		int bb[33];
		char Cin2[2] = "0";
		int Result2[33];
		
		/*Makes an int out of input A*/
		for(int i=0; i<32; i++){
			if(A2[i] == zero2[0]){
				a2[i] = 0;
			}
			else{
				a2[i] = 1;
			}
		}
		/*Makes an int out of input B*/
		for(int i=0; i<32; i++){
			if(B2[i] == zero2[0]){
				b2[i] = 0;
			}
			else{
				b2[i] = 1;
			}
		}
		/*Makes an int out of input B INVERSE using twos compliment*/
		for(int i=0; i<32; i++){
			if(B2[i] == zero2[0]){
				bb[i] = 1;
			}
			else{
				bb[i] = 0;
			}
		}
		bb[31]++;
		cout << "32 bit ALU" << endl;
		cout << "--------------------";
		cout << "\nThe first input is:                 ";
		for(int i=0; i<32; i++){
			cout << a2[i];
			}
		cout << "\nThe second input is:                ";
		for(int i=0; i<32; i++){
			cout << b2[i];
			}
		cout << "\nThe INVERSE of the second input is: ";
		for(int i=0; i<32; i++){
			cout << bb[i];
			}
		cout << "\nThe carry-in is:                    ";
		for(int i=0;i<1;i++){
			cout << Cin2[i];
		}
		cout << "\n--------------------";
		/*AND*/	
		for(int i=0; i<32; i++){
			Result2[i] = a2[i]&b2[i];
			}
		cout << "\nThe Result of AND is: ";
		for(int i=0;i<32;i++){
			cout << Result2[i];
		}
		cout << "\n";
		/*OR*/
		for(int i=0; i<32; i++){
			Result2[i] = a2[i]|b2[i];
			}
		cout << "The Result of OR is:  ";
		for(int i=0;i<32;i++){
			cout << Result2[i];
		}
		cout << "\n";
		/*XOR*/
		for(int i=0; i<32; i++){
			Result2[i] = a2[i]^b2[i];
			}
		cout << "The Result of XOR is: ";
		for(int i=0;i<32;i++){
			cout << Result2[i];
		}
		cout << "\n";
		/*Sub*/
		for(int i=0; i<32; i++){
			Result2[i] = a2[i]|bb[i];
			}
		cout << "The Result of Sub is: ";
		for(int i=0;i<32;i++){
			cout << Result2[i];
		}
		cout << "\n";
	return 0;
}

Looks promising.

Make some utility functions to clean main() up.
Eg

Code:

void convertToInt ( int *result, const char *input, int len ) {
		/*Makes an int out of input */
		for(int i=0; i<len; i++){
			if(input[i] == '0' ){
				result[i] = 0;
			}
			else{
				result[i] = 1;
			}
		}
}

Eg.
01
11

For addition, you start with carry = 0

Then take the two right-most bits
r = 1 + 1 + carry

In a single bit, r = 0 and the carry is now 1

The next bits would be
r = 0 + 1 + carry
Again, r = 0 and carry is again 1

So our complete result (in 2 bits) is 00 (and carry = 1)