RPN calculator / post fix calculator - c++

Question

So, i have to make a postfix calculator or a RPN calculator, the question goes like
The task is to write a simplified calculator that works only on integers. Your code has to provide a function called evaluate, which takes one argument: a std::string and returns an integer. The calculator has to work in the following way:
it reads a string character by character,
if it reads a digit it puts it on its own internal stack,
if it reads a space it has to ignore it,
if it reads a character +,-, * or / it performs the operation on two topmost elements of the stack, deletes them and puts the result on the stack,
when it reaches the end of the std::string argument it returns the top of the stack.
Code:
using namespace std;
int evaluate(string);
bool isdigit(char c);
bool isOp(char c);
bool isdigit(char c)
{
if (c >= '0' && c <= '9')
{
return true;
}
return false;
}
int main()
{
string str;
cout << "\n Enter the input : ";
cin >> str;
evaluate(str);
}
int evaluate(string str)
{
stack<int> mystack;
stack<int> vals;
for (int i = 0; i < str.size(); i++)
{
char c = str[i];
if (isdigit(c))
{
vals.push(c);
}
else if (c == ' ')
{
c = ' ';
cout << str;
}
else
{
int value1, value2, result;
value2 = vals.top();
vals.pop();
value1 = vals.top();
vals.pop();
switch (str[i])
{
case '+':
result = value1 + value2;
mystack.push(result);
break;
case '-':
result = value1 - value2;
mystack.push(result);
break;
case '*':
result = value1 * value2;
mystack.push(result);
break;
case '/':
result = value1 / value2;
mystack.push(result);
break;
}
cout << "result is " << mystack.top();
}
}
}
i expect the actual answers, but i guess the program is not ignoring
the space and when i input the string without spaces theres still a
wrong output

Answer 1

Pay attention that the provided algorithm works only if it gets postfix expression, and not a infix expression.
First problem:
Now, have a look on the following line:
vals.push(c);
c is a char, and vals is an integers stack. When c presenting 1 in your code, the c++ compiler actually see '0' + 1. For example:
For the input 23+, you'll get the result: 101. Why?
'2' != 2 and '3' != 3. Actually the calculation is:
'0' + 2 + '0' + 3, which means 48 + 2 + 48 + 3 because '0' == 48 in ascii code.
To fix this little problem, all you have to do is to decrease the inserted value into the vals stack by '0':
vals.push(c - '0');
Now the result for the input 23+ is: 5.
Second problem:
You are using two stacks instead of one that actually needed. When you push the result value into the second stack (mystack), you actually can't get an access to it (or rather say make it more complicated to use it's value) whenever you will get another part of expression, for example:
23+5*
You can debug this case (use watch table/debug code feature of the IDE you are working with) and see that you are trying to access the 5 of the first result, and get nothing left in the first stack, because the result stored in the second one. Solution: use single stack.
case '+':
result = value1 + value2;
//mystack.push(result);
vals.push(result);
break;
General Improvements
function: isdigit
The first thing I would advice you to do, is to remove the if statement- you don't need it. The following code will do the trick:
bool isdigit(char c) {
return c >= '0' && c <= '9';
}
If this sentence will give you true, the function will return true, otherwise it will return false.
The second thing, is like the comments says, this function already exists in the standard c++ library. Just use:
#include <cctype>
And remove your implementation to this function. The same function with the same name already exists: http://www.cplusplus.com/reference/locale/isdigit/
Important!
This answer was given due to extra time in my life and a boring situation I'm in, but more importantly because I think that it can help to understand how to debug code and to solve future (and more complicated) cases. This kind of questions are Debug questions type. You don't actually have to look here for the answer, but to debug you code and to see what's wrong. Please use the tools you got here wisely, and Good Luck!
EDIT:
For more c++ style solution, in higher difficult level, you can see the following solution (consider it like welcome saying from c++):
int evaluate(string str) {
// [] () -> {} ====> READ ABOUT: Lambda expressions
stack<int> vals;
/*
* pop
* ---
* Returns the top value of a stack, and the pop it from the stack.
*/
auto pop = [] (stack<int> &s) -> int {
int res = s.top();
s.pop();
return res;
};
/*
* op
* --
* Returns a function that execute the selected operator on two integers params.
*/
auto op = [] (char op) -> std::function<int(int, int)> {
switch (op) {
case '+':
default : return [] (int a, int b) -> int { return a + b; };
case '-': return [] (int a, int b) -> int { return a - b; };
case '*': return [] (int a, int b) -> int { return a * b; };
case '/': return [] (int a, int b) -> int { return a / b; };
}
};
/*
* for_each is a loop implementation in c++ as part of the standard library (std).
* It's get the first iterator place (str.begin()), end iterator place (str.end()), and function to execute on
* each value in the collection (between start and end iterators).
*/
std::for_each(str.begin(), str.end(), [&vals, pop, op] (char c) {
if (isdigit(c)) vals.push(c - '0');
else if (c != ' ') vals.push(op(c)(pop(vals), pop(vals)));
// op(c) -> returns a function according to the operator
// op(c)(n1, n2) -> use the returned operator on n1 and n2
// pop(vals) -> function that returns the top value of the stack, and then pop it from the stack.
// op(c)(pop(vals), pop(vals)) -> apply the selected op on the first two values in the stack
});
cout << "The result is: " << vals.top() << endl;
}