Calculator

[Hesham0]

I've created this calculator, which I'm using arrays in to accept any amount of numbers. For example, you can enter " 5 - 6 - 3 - 5 " then subtract them all. However, what I really want to do is using different operators in the same operation. For example, " 5 + 3 * 9 " which I have no clue how to do it.
How can I do it?

Thanks.

#include <stdio.h>
#include <iostream>
#include <cmath>
#include <new>
#include <climits>

using namespace std;
int main(int argc, char **argv)
{	
        double x[50] = { };
	double gm3=0, drb=1;
	char yesawy = '\0';

	for(int i=0; i<50; i++)
	{
			
		cout << "Enter Number or \"=\"" << endl;
		cin >> x[i];
		
		if(cin.good() && cin.peek() == '\n')
		{
			gm3 += x[i];
			drb *= x[i];
			continue;
		}
		else
		{
			cin.clear();
			cin >> yesawy;
			if(yesawy == '=')
			{
				break;
			}
			else
			{
				
				cin.ignore(INT_MAX, '\n');
				cout << "invalid char" << endl;
				i--;
			}
		}
	}	
		char op;
		start:
		cout << endl << "Enter Operator(+ - * /)" << endl;
		cin >> op;

		switch(op)
		{
			case '+':
				cout << gm3 << endl; break;
			case '-':
			{
				double z;
				z = x[0];
				for(int i=1; i<50; i++)
				{
					z = z - x[i]; 
				}
				cout << z;
				break;
			}
			case '*':
				cout << drb << endl; break;
			case '/':
			{
				double z;
				z = x[0];
				for(int i=1; i<50; i++)
				{
					z /= x[i]; 
				}
				cout << z;
				cout << "TT";
				break;
				
			}
			default :
			{
				cout << "Invalid Operator";
				cin.ignore(INT_MAX, '\n');
				goto start;
			}
		}
	}

[ats15]

Here is how I would approach this problem:
1. Store all numbers in an array (x), and all operations in a different array(op). The reason is that you would need to apply * and / before + or -. Note that the length of op is equal to length of x +1
2. Loop over the operator array. If op[i]=='*' then replace op[i] with '+', x[i+1] is going to be replaced by x[i]*x[i+1] and replace x[i] with 0. Similarly, if you have op[i]=='/' then x[i+1] is replaced by x[i]/x[i+1]
3. At this point your op array will have only + or -. You start with result=x[0], and loop over the op array. If op[i]=='+', you add x[i+1] to the result, else you subtract it from the result

[Hesham0]

That's kinda confusing, could you give me a scripted example please?

[andywestken]

Edit:

Ideally you need to extend your code so the user can enter a whole equation in one go (using getline()) and then split it up into numbers and operators (storing these in your array(s))

And replace your goto + label with a do-while loop!!

And then...

You might want to read up about the shunting yard algorithm

Shunting-yard algorithm
http://en.wikipedia.org/wiki/Shunting-yard_algorithm

In computer science, the shunting-yard algorithm is a method for parsing mathematical expressions specified in infix notation. ...

Andy

[ChrisMoore2]

Hesham0, that explanation makes perfect sense and is very elegant. It sounds like you want someone to write code for you.

[chipp]

Ideally you need to extend your code so the user can enter a whole equation in one go (using getline()) and then split it up into numbers and operators (storing these in your array(s))

how about using scanf()?

[andywestken]

Note that the length of op is equal to length of x +1

Thinking of 1 / 4 + 2 * 7, aren't there more operands than operators?

Andy

PS Regarding:

how about using scanf()?

Only if you're forced to use C, otherwise why wouldn't you use the std::getline(std::istream&, std::string&) ??