Arrays

Tools needed:

1.    PC running Windows Operating System (OS)

2.    DevC++ (installed)

Turn on the PC and log into it using your student account information.

Note:

The information and concepts discussed in this lab manual are in sufficient detail but are still not exhaustive. It is assumed that all these concepts are already taught to you in your ITC theory class.

1.    Arrays

When we need to store a lot of values of the same type, instead of declaring a lot of variables, it is convenient to be able to use a single name but access different values by indexing. Such data structures are called arrays.

Declaration:

An array is declared as follows:

datatype nameofarray[sizeofarray];

Note the square brackets [ ] are part of the syntax.

For example:

int age[10]; //to store 10 integer values

float marks[100]; //to store marks of up to 100 students

Indexing:

Individual values within an array are called elements. An array element is accessed through indexing. Index values start from 0 and the maximum index value can be one less than the size of the array. For example for an array of size 10, legitimate index values vary from 0 - 9. The index is provided within square brackets, as follows:

int age[10];

cin >> age[0]; //get input into first index of the array

Indices may not be constants. Any expression that evaluates to an integer value can be used as an index. For example consider the code below:

int age[10]; //to store 10 age values

for(int i = 0; i < 10; i++)

   cin >> age[i]; //get input into ith index of the array

Similarly a loop can be used to output (display) the values in an array. Also note the expression used as index:

int age[5]; //to store 5 age values

for(int i = 1; i <= 5; i++)

   cout << age[i-1] << endl; //display value at (i-1)th index

Initialization:

An array can be initialized right when it is declared, as follows:

datatype nameofarray[sizeofarray] = {value1, value2, ...};

Note that ellipsis (...) is not part of the syntax. Instead it means that we need to supply enough values to initialize the required number of elements in an array as needed. Number of values being initialized can be less than the size of the array. The rest of the values get initialized to 0.

int age[5] = {19,20,20,21,23}; //all values init

int age[5] = {12,19,20}; //first 3 values init, rest init to 0

A detailed memory-map of the array age[ ] (first example above) is as given below:

Element #	Index	Byte address	Memory contents (4 bytes block)	Access
1	0	0x00FF2200	19	age[0]
2	1	0x00FF2204	20	age[1]
3	2	0x00FF2208	20	age[2]
4	3	0x00FF220C	21	age[3]
5	4	0x00FF2210	23	age[4]

Note byte address differs by 4 as each element in the array is an int which takes 4 bytes to store the value.

In the second case, the map would look like this:

Element #	Index	Byte address	Memory contents (4 bytes block)	Access
1	0	0x00FF2214	12	age[0]
2	1	0x00FF2218	19	age[1]
3	2	0x00FF221C	20	age[2]
4	3	0x00FF2220	0	age[3]
5	4	0x00FF2224	0	age[4]

When an array is not initialized, the elements can take any values. It is customary to call such values as garbage values, since we do not know for sure what they are. We can denote a value with a ? if we do not know what value it is at a specific point in the execution of the code.

For example, after the declaration of an array as follows:

int age[5]; //uninitialized array

the memory map would be:

Element #	Index	Byte address	Memory contents (4 bytes block)	Access
1	0	0x00FF2214	?	age[0]
2	1	0x00FF2218	?	age[1]
3	2	0x00FF221C	?	age[2]
4	3	0x00FF2220	?	age[3]
5	4	0x00FF2224	?	age[4]

The values provided to initialize the array must conform to the datatype of the array.

Address of an array:

The array name is treated as its starting (or base) address. Note that an array is contiguous in memory, i.e., it is allocated all the memory in a single chunk. An array that is declared to hold 10 int values will be allocated 4x10 = 40 bytes. The array name serves as the starting address. Subsequent array elements are stored in increasing memory addresses.

int age[10];

cout << age;

This will display the address of the array (first element) in hex; not the value of any particular element in the array.

Caution: Array bounds

In C++ the array bounds must be taken care of by the programmer. Changing values of elements at indices that are not within limits (0 to one less than the size-of-array) would overwrite (i.e. change) the adjacent memory bytes and may corrupt other variables! Even in the case of read, we should be careful not to exceed the array bounds, otherwise we’re liable to read other variables’ data, which could be a privacy breach.

Example 1: For an array of integer values, the program below asks the user to enter an integer value and then checks whether that value exists in the array or not. If it does, the index at which the value is stored in the array is printed.

   int arr[10] = {25, -10, 3, 4, 99, 100, 74, -890, 0, 1}, val;

   cout << "Enter an integer value: "; cin >> val;

  

   int index = -1;

   for(int i = 0; i < 10; i++)

   {

         if(arr[i] == val)

               index = i;

   }

  

   if(index >= 0)

         cout << "Value found at " << index << " index.";

   else

         cout << "Value not found.";

   cout << endl;

The algorithm used above for searching a value in an array is known as Linear Search.

Task # 1: Based on Example 1 above

     I)        Print the average value of the elements of the array arr.

    II)        Print the average value of the even elements of the array arr, the average of odd values, and the average of the average of even and average of odd elements.

   III)        Print the smallest value of the array arr.

  IV)        Challenge Task: Print the smallest value of the array arr, then the smallest of the rest, then the smallest of the rest, and so on until you have displayed the whole array in ascending order. Hint: you need to use nested loops -- for every iteration of an outer array that iterates 10 times you need to iterate an inner array 10 times.

Code:

#include <iostream>

using namespace std;

int main()

{

int arr[10] = {25, -10, 3, 4, 99, 100, 74, -890, 0, 1};

int avg,evenavg,evensum=0,oddsum=0,oddavg,small,allavg,sum=0,n=0,reve,rodd,m=0;

           

            for(int i = 0; i < 10; i++)

            {

                        sum=sum+arr[i];

           

            }

            avg=sum/10;

            cout<<"Average value of elements of array :"<<avg<<endl;

                        for(int i = 0; i < 10; i++)

            {

                        reve=arr[i]%2;

                        if (reve==0)

                        {

                                    n++;

                                    evensum=evensum+arr[i];

                        }

           

            }

            evenavg=evensum/n;

            cout<<"Even element average is: "<<evenavg<<endl;

                                    for(int i = 0; i < 10; i++)

            {

                        rodd=arr[i]%2;

                        if (rodd==!0)

                        {

                                    m++;

                                    oddsum=oddsum+arr[i];

                        }

           

            }

            oddavg=oddsum/m;

            cout<<"ODD element average is : "<<oddavg<<endl;

            allavg=(oddavg+evenavg)/2;

            cout<<"average of even an odd :"<<allavg<<endl;

            int min = arr[0];

    for (int j = 0; j < 10; j++)

    {

        if (min > arr[j])

            min = arr[j];

    }

     cout << "Smallest element : " << min;

      for(int k=0; k<10; k++)

    {

        for(int j=k+1; j<10; j++)

        {

            if(arr[j] < arr[k])

            {

               int temp = arr[k];

                arr[k] = arr[j];

                arr[j] = temp;

            }

        }

    }

     cout << endl;

    cout<<"Elements of array in ascending order:"<<endl;

    for(int i=0; i<10; i++)

    {

        cout<<arr[i]<<endl;

    }

           

            cout << endl;

}

Task # 2: For an array of integer values, the program below asks the user to enter the element # of the array to print. The program then determines if the entered element is less than the size of the array. If not, it keeps on asking user to enter element # equal to or less than the array size. Once such number is entered, it prints out the element at that location.

Code:

#include <iostream>

using namespace std;

int main()

{

int size,n,x;

           

            cout <<"input size of array :";

            cin>>size;

            int arr[size];

cout<<"Input the Index for Ouput";

            cin>>n;

            x=n;

            do

            {

                        cout <<"out of bound"<<endl;

                        cout<<"again input according to size";

                        cin>>n;

            } while (n>size);

           

                        cout<<"value found at :"<<n;

cout<<endl;

return 0;

}

Task # 3: Take an array of 10 elements. Write a code to find the occurance of a number in an array entered by user. e.g.

1

5

4

1

1

3

2

5

1

5

Occurance of 1: 4

Code:

#include <iostream>

using namespace std;

int main()

{

int arr[10]={1,5,4,1,1,3,2,5,1,5};

            int n,oc=0;

            cout<<"input number to check occcurence: ";

            cin>>n;

                        for(int i = 0; i < 10; i++)

            {

            if (arr[i]==n)

                        oc++;

            }

            cout<<"Occurence of "<<n<<" in array is :"<<oc;

            return 0;

}

           

Task # 4: Take an array of 9 elements. Split it into middle and store the elements in two different arrays. e.g.
Initial array:

22

58

19

87

15

72

9

1

37

After splitting:

22	58	19	87	15
72	9	1	37

Code:

#include <iostream>

using namespace std;

int main()

{

int arr[9]={22,58,19,87,15,72,9,1,37};

int half[5],ndhalf[4];

           

           

            cout <<"1st Half"<<endl;

            for (int j=0; j<5;j++)

            {

  half[j]= arr[j];

  cout<<half[j]<<'\t';

}

cout<<endl;

 cout<<"2nd Half"<<endl;

for (int i=5; i<9; i++)

{

 ndhalf[i]=arr[i];

 cout<<ndhalf[i]<<'\t';

}

cout<<endl;

return 0;

}

Task # 5: Store an array of 10 integers and write a code that checks if the array is stored in ascending order or not.

Code:

#include <iostream>

using namespace std;

int main()

{

int arr[10];

            int j=0,n=0;

            for(int i = 0; i < 10; i++)

            {

                        cout<<"input "<<i <<" index element of array : ";

                        cin>>arr[i];

            }

            for(int i = 0; i < 10; i++)

            {

                        if (arr[i]>arr[i+1])

                        {

                                    n++;

                        }

                        else

                        {

                        j++;

                        }

            }

            if (n>0)

            {

                                    cout<<"array is not in ascending order";

            }

            else

            {

                                    cout<<"array is in ascending order";

            }

            return 0;

}