Stack with char array.How to execute "+" and "*" operations? - c++

Question

I need to sum up or multiplie(it depends on which operation i push in array) first two numbers in stack. If I push operation + and my first two numbers are one and two then my stack value where I pushed + must be 3 , but in my result i get some symbol.
#include<iostream>
using namespace std;
#define MAX 10
int sp=-1;
char stack[MAX];
void push(){
if(sp==MAX-1){
cout<<"Error: stack overflow"<<endl;
return;
}
else{
char x;
cout<<"Insert value in stack: ";
cin>>x;
if(x=='*'){
if(sp>=1){
stack[++sp]=stack[1]*stack[0];
return;
}
else {
stack[++sp]=x;
return;
}
}
else if(x=='+'){
if(sp>=1){
stack[++sp]=stack[0]+stack[1];
return;
}
else {
stack[++sp]=x;
return;
}
}
else stack[++sp]=x;
}
}
void pop(){
if(sp==-1){
cout<<"Error: Stack empty";
return;
}
else{
cout<<stack[sp]<<endl;
sp--;
}
}
void top(){
if(sp==-1){
cout<<"Error: Stack empty";
return;
}
else{
cout<<stack[sp]<<endl;
}
}
void isEmpty(){
if(sp==-1){
cout<<"Stack is empty"<<endl;
return;
}
else{
cout<<"Stack is not empty"<<endl;
}
}
int main(){
for(int i=0;i<8;i++){
push();
}
top();
return 0;
}
Original stack:
3 4 3 + 1 2 * *
Stack that I need to get:
3 4 3 7 1 2 12 12
Stack that I get:
3 4 3 \ 1 2 _ _ (let's say something like that)

Answer 1

So you wanted to implement some postfix calculator. Very good that you noticed that a stack is needed.
For calculators you normally parse the input, store some results on a stack and then operate on the stack elements.
In contrast to what timo said, a parse stack can never be implemented with a std::stack because several top stack elements must be matched against the production in the grammar. For that reason a std::vector is the ideal container.
Now to the problem in the program. OP tries to store integer values and operators like "+" and "*" and results of calculations in the same datatype char. That cannot work. char stores characters like '+' or '*' or '1' but not 1. Internally a character is also encoded as a number (for example using ASCII) and a '1' would be equal to 49. And, if you look at results you cannot store '12' (for all the pros. Yes of course I know) in a char. 12 is a number, not a character.
And if you mix these types and do calculations with that then you get wrong results.
The solutions is to use Tokens and implement them as a std::variant. The variant can have a char an an int. Then your parse stack should be a std:::vector of Token.
This the right approach.
Additionally. In all standard recursive descent parsers you do not operate always on elements 0 and 1. You would replace the used values with the newly calculated one.
There is always the same approach
Read next Token and shift it on a stack
Match the top elements of the stack against a production
Reduce. Replace used elements with result
Some example of a different parser: See here.
This is somehow complicated. You may want to read about formal languages and parser.