arrays

This is a discussion on arrays within the C++ Programming forums, part of the General Programming Boards category; Hi there. I am learning and i have to have a program by sunday night. I am stuck. I have ...

  1. #1
    Registered User
    Join Date
    Feb 2002
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    2

    Unhappy arrays

    Hi there. I am learning and i have to have a program by sunday night. I am stuck. I have to input 20 numbers, if I put in a number that is already in the array then it ignores it. each input is stored to the next element in the array
    i can make my array, and it will take inputs 20 times, what i cannot figure out is how to compare all elements to see if my input matches one of them.


    Help

  2. #2
    Code Goddess Prelude's Avatar
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    9,796
    Use a loop the iterate through the array. Something along the lines of the following:
    Code:
    for ( int i = 0; i < LEN; i++ ) {
      if ( array[i] == input )
        flag = TRUE;
    }
    if ( flag == TRUE )
    // Ignore
    There are a number of ways to structure this.

    -Prelude
    My best code is written with the delete key.

  3. #3
    ....
    Join Date
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    For each number you should check if it is already in the array. In pseudo-code:

    Code:
    input_value = get_input_value ()
    already_in_array = FALSE
    for index = 0 to nr_of_input_values
        if input_value = array [index]
        then already_in_array = TRUE
    If already_in_array is TRUE, then ignore the input value.

    Note that nr_of_input_values is a variable and not a constant. If you have 5 elements in the array, it isn't very efficient to check 20 elements.

    Perhaps the algorithm can be done in other (more efficient) ways, but this one is usable.

  4. #4
    Registered User
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    I understand what you are saying, but i cannot figure out how to get it to work... let me show you what i have right now.

    #include <iostream.h>

    const int MAX = 20;

    int Count(int n, int Nbr[MAX])
    {
    cout << "Number " << n + 1 << ": ";
    cin >> Nbr[n];

    return Nbr[n];
    }

    int main()
    {
    int Number[MAX];

    cout << "Please type 20 integers.\n";



    for( int i = 0; i < MAX; i++ )
    {
    Count(i, Number);

    }


    return 0;


    }

  5. #5
    Just because ygfperson's Avatar
    Join Date
    Jan 2002
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    2,493
    you can use a goto:
    Code:
    int Count(int n, int Nbr[MAX]) 
    { 
    cout << "Number " << n + 1 << ": "; 
    
    e:
    cin >> Nbr[n];
    
    for (int i=0;i<n;i++)
     if (Nbr[i]==Nbr[n]) { 
      cout << "That number has already been typed in" << endl; goto e;
     }
    
    
    return Nbr[n]; 
    }
    if you're finiky about gotos, then do it this way:
    Code:
    int Count(int n, int Nbr[MAX])
    {
    int flag_already_typed;
    cout << "Number" << n+1 << ": ";
    do {
      flag_already_typed=0;
      cin >> Nbr[n];
      for (int i=0;i<n;i++) 
       if (Nbr[i]==Nbr[n]) {
        cout << endl << "that.. blah blah blah .. " << endl; flag_already_typed=1; 
       }
    } while (flag_already_typed);
    return Nbr[n];
    }
    whenever possible, put [ code] and [/code] around your source code
    Last edited by ygfperson; 02-24-2002 at 01:08 PM.

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