[SOLVED] reading from stdin

lord_didger · 01-14-2008, 02:45 AM

I want to make a calculator. It aim is to perform the simplest arithmetic operations. It expects a few lines similiar to this one '+ 8 6'. Each line is one task. The program reads the line, do the task and put the result to stdout.
Everything works but the program doesn't stop. I would like the programm read what is currently in buffer, did the tasks and end (not waiting).
How can I do it? The code is:

Code:

#include <iostream>
using namespace std;

int main()
{
  ios_base::sync_with_stdio(0);
  int a,b;
  char ch;
 
  while(!cin.eof())
  {
    cin >> ch >> a >> b;
    
    int wynik;
    switch(ch)
    {
      case '+':
        wynik = a+b;
        break;
      case '-':
        wynik = a-b;
        break;
      case '*':
        wynik = a*b;
        break;
      case '/':
        wynik = a/b;
        break;
      case '%':
        wynik = a%b;
        break;
    }
    
    cout << wynik << endl;
    cin.ignore();
  }
 
  system("PAUSE");
  return(0);
}

jailbait · 01-14-2008, 07:08 PM

What does system("PAUSE"); do? Try taking it out of the program.

--------------------
Steve Stites

lord_didger · 01-15-2008, 09:51 AM

Actually, 'system("PAUSE")' means nothing. I have written the program in Windows where this function is available. Nevertheless problem remains.

Uncle_Theodore · 01-15-2008, 10:06 AM

Your program will stop if you press Ctrl+D, signifying the cin.eof(), which is the stopping condition of your while loop.
I don't really understand the question. If you want your program to evaluate just one expression and then stop, why did you put in a while loop in the first place?

indienick · 01-15-2008, 12:51 PM

Agreeing with Uncle Theodore, if you want the program to end, take out the WHILE loop and replace with:

Code:

if (cin.eof())
  return(-1); 
  /* Or some other number that specifies 
     end of operation without failure */
else
{
  switch (ch)
  {
    ...
    ...
    ...
  }
}

lord_didger · 01-16-2008, 09:27 AM

I have already solved the problem.

The program's aim was to work as a calculator - it should read some data from stdin, execute the data and exit. When I type in command line 'program < data_file' I expect to get some information on stdout and the program to exit.

The main point is that I don't how much tasks there are on stdin

The code that lives up my expectations:

Code:

#include <iostream>
using namespace std;

int main()
{
  ios_base::sync_with_stdio(0);
  int a,b;
  char ch;
 
  while(cin >> ch >> a >> b)
  //while(!cin.eof())
  {
    //cin >> ch >> a >> b;
    int wynik;
    switch(ch)
    {
      case '+':
        wynik = a+b;
        break;
      case '-':
        wynik = a-b;
        break;
      case '*':
        wynik = a*b;
        break;
      case '/':
        wynik = a/b;
        break;
      case '%':
        wynik = a%b;
        break;
    }
    
    cout << wynik << endl;
    cin.ignore();
  }
 
  return(0);
}

and example of content of data_file

Quote:

+ 7 9
- 0 4
* 5 6
/ 8 3
% 5 2

and that is all.