PPP2 Chapter 6 Section 6.3.1 Expression Calculator

[DragonOsman]

In the PPP2 book, there's for an expression calculator that Stroustrup has us make through trial and error and he's using it to teach us program design and how think through a problem. He mentioned that each problem has a different set of requirements and that he won't go into that here, but that he'll try to arm us with enough knowledge to be able to solve a problem on our own (or at least, that's how I understood it).

Anyway, I have the code, but first I'll post about a use-case for it. Please compare it with the output from my code and help me figure out what I did wrong.

Here's the paragraph from the use-case.

This isn’t bad, but then we try 1+2*3 and see that the result is 9 and not the 7 our
arithmetic teachers told us was the right answer. Similarly, 1 – 2*3 gives – 3 rather
than the – 5 we expected. We are doing the operations in the wrong order: 1+2*3
is calculated as (1+2)*3 rather than as the conventional 1+(2*3) . Similarly, 1 – 2*3
is calculated as (1 – 2)*3 rather than as the conventional 1 – (2*3) . Bummer! We
might consider the convention that “multiplication binds tighter than addition” as
a silly old convention, but hundreds of years of convention will not disappear just
to simplify our programming.

And here's my code:

// expression_calc_1.cpp : Defines the entry point for the console application.
// Osman Zakir
// 1 / 3 / 2017
// Bjarne Stroustrup: Programming: Principles and Practice Using C++ 2nd Edition
// Chapter 6 Section 6.3.1 
// First attempt at expression calculator program

#include "../../std_lib_facilities.h"

int main()
{
	cout << "Please enter expression (we can handle +, -, *, and /)\n";
	cout << "add an x to end expression (e.g., 1+2*3x): ";
	int lval = 0;
	int rval;
	cin >> lval;
	cin.ignore();
	if (!cin)
	{
		error("no first operand");
	}
	for (char op; cin >> op;)
	{
		cin.ignore();
		if (op != 'x')
		{
			cin >> rval;
			cin.ignore();
		}
		if (!cin)
		{
			error("no second operand");
		}
		switch (op) 
		{
		case '+':
			lval += rval;
			break;
		case '–':
			lval -= rval;
			break;
		case '*':
			lval *= rval;
			break;
		case '/':
			lval /= rval;
			break;
		default:
			cout << "Result: " << lval << '\n';
			keep_window_open();
			return 0;
		}
	}
	error("bad expression");
}

Here's an example output:

Please enter expression (we can handle +, -, *, and /)
add an x to end expression (e.g., 1+2*3x): 1+2*3
Result: 1
Please enter a character to exit
k
Press any key to continue . . .

An output of 9 would also have been wrong, but at least it'd have been like the book said it should be. But here it's a completely different result.

Edit: From my debug attempt, it seems I should put a space there. But it doesn't seem like Stroustrup is putting in a space. So what's going on?

I noticed that when I do it without spaces, while lval has the right value, op ends up with the value of what should've been the second operand instead of the operation character.

[cire]

Get rid of all those cin.ignore(); lines.

[Marcus Aseth]

You get the 1 into lval then cin.ignore() discards the '+', then you take a 2 into op...and from there, everything get's messed up.
Do as cire says xD

[DragonOsman]

Yeah, it works better that way. I put in those cin.ignore() calls because it clears out the input buffer, but I guess it's best for when there's an extra newline to clear out.

[cire]

I put in those cin.ignore() calls because it clears out the input buffer, but I guess it's best for when there's an extra newline to clear out.

That strategy is for when you're mixing unformatted and formatted input extraction. You aren't doing so here.