Replace input style (cin to array)

Posted on 2006-11-25
Last Modified: 2008-04-27
Hi everybody,
I got a code (calculate math expression) from a website and I want to use it in my program.

Part of the code:

    cout << "Don't forget add ';' to the tail of the expression:)" <<endl;  

           cin >> temp;
                           cin >> number[ntop++];
    cout << "And the answer is... " << number[0] <<endl;                    
    return 0;

In my program I already have an expression (char array) so I don't want to use the "cin" commands (especialy cin.putback) to deal with the expression..
I want to use my array or maybe put the string into cin streaming function(if possible) so I'll can use this code..
Any ideas..?
Thanks for your help.  
Question by:flexlight

Accepted Solution

UrosVidojevic earned 50 total points
ID: 18013113
Hi flexlight,

this is sample code that will solve your problem.

The idea is similar to that, you used in your code.
Program is goint through array of chars and checking
for arrays of successive digits (numbers) and puting
them in an array of numbers. At the end program prints
the first of them.

Althought, I don't understand why first of them
should be answer. :-)

Sample code:
#include <iostream>
#include <cstring>
using namespace std;

const int MAX_LEN = 1000, // Maximal char array length
        MAX_NUM = 500;  // Maxinal number of numbers in array      

int main() {
      char text[MAX_LEN];
      cout << "Char array: ";
      cin.getline(text, MAX_LEN);

      // text is array of characters that you allready have.
      // so this is part of code that is particulary
      // interesting for your problem.

      int i = 0, len = strlen(text), number[MAX_NUM], ntop = 0;
      while (true) {
            while (i < len && !isdigit(text[i]))
            if (i == len) break;
            long num = 0;
            while (i < len && isdigit(text[i])) {
                  num = 10*num + (text[i] - '0');
            number[ntop++] = num;

      // i is index of current character in array
      // len is length of array
      // number is array if numbers found in array
      // num is current number
      // ntop is index of current number

      cout << "And the answer is: " << number[0] << endl;

      // note that if there is no numbers in array
      // number[0] will be some random number.

      // this is is optional, just to show you
      // that program is correct.
      cout << "Other numbers: ";
      for (int i = 1; i < ntop; i++)
            cout << number [i] << " ";
      cout << endl;

Sample input\output 1:
Char array: f(123) + g(12) - sin(45)/tg(32) + 123*456
And the answer is: 123
Other numbers: 12 45 32 123 456

Sample input\output 2:
Char array: a(1+b(2+c(3)+2)+1)
And the answer is: 1
Other numbers: 2 3 2 1

If I was not clear enough,
I would be glad to help.
Regards, Uros.

Expert Comment

ID: 18721853
Looks like you would like to convert your char array to the array of numbers as in the small example.
Would this be helpful:
for(int i=0;...) {
  number[i] = atoi(yourCharArr[i]; //for int atof for double
Now you have numbers instead of characters if that was what you tried to achieve

Expert Comment

ID: 21427223
No comment has been added to this question in more than 21 days, so it is now classified as abandoned.

I will leave the following recommendation for this question in the Cleanup Zone:
  Accept: UrosVidojevic {http:#18013113}

Any objections should be posted here in the next 4 days. After that time, the question will be closed.

Sean Durkin
Experts Exchange Cleanup Volunteer

Expert Comment

ID: 21449012
Forced accept.

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