find values in user entered array

Question

i am trying to find any user entered value in a array which the user has previously entered the values of. I made the following for finding which values were entered in array bu

Answer 1

Your problem is that an array of integers has no clear definition between assigned and unassigned. You need a way to "flag" these integers. You could do this with a parity array of booleans, that would be flagged based on std::cin.fail(), or use int* and use NULL to flag. There's a lot of ways to do this.

std::vector<int*> a(10);
for(int i=0;i<10;i++){
  int input;
  cout<<"enter value : ";
  cin >> input;
  if(!cin.fail()) {
    a[i] = new int(input);
  }else{
    cin.clear(std::ios_base::failbit);
  }
}

Now you're able to test the vector a for NULL pointers.

Answer 2

Working solution

You are solving a finding problem here. You have no any suggestions where the entered value may
probably be, so you should try each one step by step. This is a common problem.

The first way to solve it is to use a loop. It isn't a good way for modern C++. But you should probably
try it for a practice.

bool has(int val, int const* arr, size_t size) {
    size_t idx = 0;
    // Iterates each element of array and checks
    // whether the one is equal to the `val`. In
    // case of meeting of `val`, the loop stops.
    while (idx < size && arr[idx] != val) ++idx;
    return idx != size;
}

The way below is more convinient. Actually, the more general form of has function is already has in C++ standart library in <algorithm> header. It is called find. It do exactly the same, but much better. Actually, there are a lot of functions solving a common problems in <algorithm> header. You have to use it anywhere you can.

bool has_(int val, int const* arr, size_t size) {
    int const* end = arr + size;
    // Now `has_` don't iterate each element and
    // checks it. It finds the `val` in range
    // between the first element of array and
    // the last.
    return std::find(arr, end, val) != end;
}

I suggest you to read the subsection "Prefer algorithm calls to handwritten loops." in section "STL: Containers" in the book "C++ Coding Standarts" by Herb Sutter and Andrei Alexandrescu to gain an intuition about why to use the <algorithm> header.

Also, you may find the reference to the <algorithm> header here.

Mistakes in your solution

Lets consider your code and discuss why you end up with error. Actually, you just made a typo.
It is the one of the reasons to use the <algorithm> header instead of handwritten loops like yours.

#include<iostream>
#include<conio.h>

using namespace std;

void main()
{
    int a[10], found;

    for (int i = 0; i<10; i++)
    {
        cout << "enter value : ";
        cin >> a[i];
    }

    cout << "Enter Searching Value :";
    cin >> found;

    for (int i = 0; i<10; i++)
    {
        // Look at here: you compare entered value ten times
        // with the value after the last element of array.
        if (found == a[10])
        {
            // In case that you found an entered value in array
            // you just continue the loop. You should probably
            // break it at this point. This may be achieved by
            // using the `brake` operator or the `while` loop.
            cout << "Value Found";
            _getch();
        }
        else if (found != a[10])
            cout << "Value Not Found";

    }

    _getch();

}

Answer 3

I think this answer will solve your problem.. :)

#include<iostream>
#include<conio.h>

using namespace std;

void main ()
{
    int a[10];
    for(int i=0;i<10;i++)
    {
        cout<<"enter value : ";
        cin>>a[i];
    }
    int random;
    cin>>random;
    int flag=0;
    for(int i=0;i<10;i++)
    {
       if(a[i]==random)
       {
         flag=1;
         cout<<"Found at a["<<i<<"]"<<endl;
         break;
       }
    }
    if(flag==0)
    cout<<"element not found"<<endl;
    return 0;
}