如何在C ++中解决线性搜索问题？

Question

The output of the given code should be the index number of element that is in array but i am not getting the expected output.

Here is my code:

int main() {
    int n, a, b;
    int arr[100];
    cout << "Enter the size of array";
    cin >> n;
    cout << "Enter the number to find";
    cin >> a;
    cout << "Enter elements in array";
    for (int i = 0; i < n; i++)
        cin >> arr[n];
    for (int i = 0; i < n; i++) {
        if (arr[n] == a) {
            cout << n;
        } else
            n++;
    }
    return 0;
}

Answer 1

Here:

if(arr[n]==a)

You need to use i :

if(arr[i]==a)

The complete else -case is superfluous; remove it. Also, you have to make sure that n <= 100 , or else you have a security hole (buffer overflow). Use std::size_t instead of int for array indices.

Answer 2

You using 'n; as array index to take input and match search element. It should be 'i' instead of 'n'.

#include <iostream>
using namespace std;

int main()
{
 int n,a,b;
 int arr[100];
 cout<<"Enter the size of array";
 cin>>n;
 cout<<"Enter the number to find";
 cin>>a;
 cout<<"Enter elements in array";
 for(int i=0;i<n;i++)
    cin>>arr[i];

 for(int i=0;i<n;i++){
    if(arr[i]==a)
    {
        cout<<i<< "";
    }
 }
 return 0;
}

Answer 3

as @Erlkoenig said:

"You need to use i:

if(arr[i]==a) The complete else-case is superfluous; remove it. Also, you have to make sure that n <= 100, or else you have a security hole (buffer overflow). Use std::size_t instead of int for array indices."

Also n++ will propably give you a segmentation fault (depending no the starting n and how quickly you will find the "number to find". I'm pretty sure that if the number is not in the array you will get a sgementation fault.