CODEWITHSUNDEEP
c++arrayssearchindexingreturn
soft_student
soft_student

Reputation: 13

Why is "2" always being returned from my function? Trying to find the index of a searched number within an array

I am trying to return the index of array if it matches a user defined number. I cant figure out why, no matter what I enter, the function returns "2".

    #include <iostream>
    using namespace std;
    int search(int x, int *list);
    int main()
    {
        int x;
        int list[10] = { 20, 1, 18, 3, 16, 5, 14, 7, 12, 9};

        cout << "Enter a number to search the list for: ";
        cin >> x;
        cout << endl;

        cout << search(x, &list[10]) << endl;

        system("PAUSE");
        return 0;
    }

    int search(int x, int *list)
    {
        bool match = false;
        for (int i = 0; i < 10; i++)
        {
            if (list[i] == x)
            {
                match = true;
            }
            if (match == true)
            {
                return i;
            }
            else if (match == false && i == 9)
            {
                return -1;
            }
        }
    }

Upvotes: 0

Views: 34

Answers (1)

John3136
John3136

Reputation: 29266

Because you have invoked undefined behavior by starting the search at index 10: search(x, &list[10]) should be search(x, list).

...and what's with that whole else if in the loop. Why not just return -1 at the end of the function. And why have a match variable - if a match is found just return i.

for (int i = 0; i < 10; i++)
{
    if (list[i] == x)
    {
        return i;
    }
}
return -1;

Upvotes: 1

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