Question

My program reduce expression to one value.
I have a problem with changing “char sign” as a character of action. Could you show me some simple solution or idea how to do it?

I tried:

(tab[i]-'0') 'sign' (tab[i+1]-'0'); 

This is the full code:

#include <cstdlib>
#include <iostream>

using namespace std;

int main(int argc, char *argv[])
{

    char* tab = "12+";
    int b = sizeof (tab);
    char* tmp = new char[b] ;
    tmp [b-1] = '\0';

    int k = b/3;

    for(int i=0; i<k; i++){

            if(isdigit(tab[i]) && isdigit(tab[i+1]) ){

               if(tab[i+2]=='+' || tab[i+2]=='-' || tab[i+2]=='*'){
                  char sign = tab[i+2];

                  int n = (tab[i]-'0') + (tab[i+1]-'0');  //here is a problem, i want to replice + as a char sign which will be recognized

                  tmp[i] = n+'0';
               }
               else goto LAB;
            }

            else if (isdigit(tab[i]) && isdigit(tab[i+2])){


            }
            else if (isdigit(tab[i+1]) && isdigit(tab[i+2])){

            }


            else 
            LAB:
            tmp[i]= tab[i];

    }

    cout<<"Import "<<tmp[0]-'0'<<endl;        


    system("PAUSE");
    return EXIT_SUCCESS;
}

Answer 1

You cannot replace an operator, or function name with a symbol which is given by user, because compiler has to know which function should be called and there is no mechanism to convert a ‘+’ character to operator+(int, int) call embedded in a language. You have to write it by yourself.

The simplest solution is to write explicitly every operator you want to support:

int n;
switch(tab[i+2]){
    case '+':
        n = (tab[i]-'0') + (tab[i+1]-'0');
        break;
    case '-':
        n = (tab[i]-'0') - (tab[i+1]-'0');
        break;
    // etc...
}