16.5. Returning a Pointer to a Particular Element in an Array

Problem

You need to create a method that accepts a pointer to an array, searches that array for a particular element, and returns a pointer to the found element.

Solution

The FindInArray method, shown here, returns a pointer to an element found in an array:

public unsafe int* FindInArray(int* theArray, int arrayLength, int valueToFind)
{
    for (int counter = 0; counter < arrayLength; counter++)
    {
        if (theArray[counter] == valueToFind)
        {
            return (&theArray[counter]);
        }
    }

    return (null);
}

This method is strongly typed for arrays that contain integers. To modify this method to use another type, change the int* types to the pointer type of your choice. Note that if no elements are found in the array, a null pointer is returned.

The method that creates an array of integers and passes a pointer to this array into the FindInArray method is shown here:

public void TestFind( )
{
    unsafe
    {
        int[] numericArr = new int[3] {2,4,6};
        fixed(int* ptrArr = &numericArr[0])
        {
            int* foundItem = FindInArray(ptrArr, numericArr.Length, 4);
            if (foundItem != null)
            {
                Console.WriteLine(*foundItem);
            }
            else
            {
                Console.WriteLine("Not Found");
            }
        }
    }
}

Discussion

The FindInArray method accepts three parameters. The first parameter, theArray, is a pointer to the first element in the array that will be searched. The second parameter, arrayLength, is the length of the array, and the final parameter, valueToFind, is the value we wish to find in the array theArray.

The ...

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