Nearest value in an array

Question

I am supposed write a function that is passed two parameters: a one-dimensional array of int values, and an integer value. The function finds the value in the array that is closest in value to the second parameter. My code works but not when I enter numbers 1-5 my output is 0. When I enter numbers above 5 then I start getting accurate results. I am not too sure why this is happening but here is my code so far:

#include <iostream>
#include <cmath>
using namespace std;

const int MAX = 5;

int searchNearest(int anArray[],int key)
{
int value = abs(key - anArray[0]);
int num = 0;

for(int x = 0;x < MAX; x++)
{
    if(value > abs(key - anArray[x]))
    {
        value = abs(key - anArray[x]);
        num = anArray[x];
    }
}

return num;

}

int main()
{
int A[MAX] = {3,7,12,8,10};

int search;

int nearest;

cout << "Enter a number to search: ";

cin >> search;

nearest = searchNearest(A,search);

cout << "The nearest number is: " << nearest << endl;

system("pause");
return 0;

}

sfstewman · Accepted Answer · 2013-05-22 23:55:37Z

5

In your original code, numbers 1-5 are closest to the first element of the array. Because of this, the code in the if statement is never executed, and when you return num, you return its initial value, which happens to be 0. To fix this, just initialize num differently:

int num = anArray[0];  // <-- used to be 0

answered May 22, 2013 at 23:55

sfstewman

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2 Comments

Nick Over a year ago

Ahh I knew I was overlooking something. Thanks for the help!

sfstewman Over a year ago

@Nick This is something that's pretty easy to diagnose with a debugger (or using printf or cout statements). It's also far less frustrating than staring at the code.

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