Arrays - 2 - Free MCQ Practice Test with solutions, GATE CSE (CSE) Programming

Elements in an array are accessed randomly. In Linked lists, elements are accessed sequentially.

The program is described as:

In the given code, Consider the address of the array as 100.
The main function has an array arrr[] = {12, 30, 40, 15};
int n = sizeof(arr) / sizeof(arr[0]);
sizeof(arr) = 16(4 x 2) i.e four elements and each size is 2. 
sizeof(arr[0]) = Size of element arr[0] i.e first element size = 2.
So n becomes n=16/2
∴ n=4
And sum(arr, n)
Here arr is the address of the array. 
sum(100, 5)

The sum function calculates the summation of elements
Initially, the sum=0 sums each element and stores it in the sum variable.
sum=12+30+40+15
sum=97 and returns the sum variable to the main function i.e 97.
Hence the correct answer is 97.

Since there are 15 int elements and each int is of 4bytes, we get 15*4 = 60bytes.

• Array of size 6 is declared and is initialized by 1.
• arr[0] will contain 1 and rest are by default initialized as zero.
• for loop will print all that values starting from arr[0] to arr[5].

arr[0]=1
arr[1]=0
arr[2]=0
arr[3]=0
arr[4]=0
arr[5]=0

Final output print by the program is 1 0 0 0 0 0.

ArrayIndexOutOfBoundsException is a run-time exception and the compilation is error-free.

Column major order concept: 

• Column major order means the element in a column is stored adjacent to each other. This basically works in a 2d and 3d array.
• Now we are given the base address along with the size of each element and the construction of a 2d array(row and column number )
• First, we have to calculate the no. of rows here = (Row upper limit - Row lower limit) + 1 = (20 - 10) + 1 = 11.
• We need no. of the column if the ques will ask about Row major order.
• There is a formula available to calculate it directly(Rather than going into deep how the formula came, it's better to remember the formula.)

Formula:-
Address of A[I][J] is can be calculated with the formula is A[I][J] = B + ((J – LC) * M + (I – LR))*W.
where;

• I- Row subset of an element whose address is to be found ( 15 here)
• J = Column Subset of an element whose address is to be found,(30 here)
• B- Base address(10 here)
• W- storage size of one element  (2 bytes here)
• LR- the lower limit of row (u can take start index of the matrix)- (10 in this ques)
• LC-Lower limit of column(start column index of the matrix) - (25 in this ques)
• M- number of rows given in the matrix. (11 here)

 
Now solve the equation by solving the inner bracket first.
10+[5*11+5] *2  = 10+(60)*2 = 10+120= 130 will be the answer.

An array is defined as the collection of similar types of data items stored at contiguous memory locations. 
int array[3][4] = (0);
Here Array is an integer array. which is in a two-dimensional array of 3 rows and 4 columns. So each row has 4 elements and there are three rows.
Hence total number of elements = 3 x 4 =12
Each element of size = 2 bytes.
So size of array is = 12 x 2 bytes =24 bytes.
Hence the correct answer is 24 bytes.

Array indexing starts from 0.

*(arr[1] + 9) can be written as arr[1][9].
as C doesn't follow bound check and follow the row major ordering
arr[1][5] = arr[2][0]  // arr[1][4] will be first row of array and then arr[2][0] will be second row of array 
arr[1][6] = arr[2][1]
arr[1][7] = arr[2][2]
arr[1][8] = arr[2][3]
arr[1][9] = arr[2][4]
arr[2][4] = 10*i + j = 10*2+4 = 24

Option 4 is the answer.

• In the following declaration, the array name is num.
• The size of the array is four.
• The array is initialized with four elements 1,2,3,4.

Note that int arr[]; is declaration whereas int arr[] = new int[3]; is to instantiate an array.

Array contains elements only of the same type.

Where \0 is a null character 
After *p = '0';   \\s1[2] = '0'
100  101 102 103  104

printf(“%s”, s1) will print all starting from address location 100 until the null character.
Output: 1204 

An array is a collection of items stored at contiguous memory locations. The idea is to store multiple items of the same type together.

When an array is used as a value and array name represents the address of the first element. When an array is not used as a value its name represents the whole array.

Array name is a type of name or a type of any element name that is share by all elements of an array but its indexes are different. Array name handle as a constant pointer, it can never change during execution of a program. Array name is also used to reach its all element.

Therefore Option 3 is correct

Array a

for(I = 2; I < 6; ++i)            // initially i = 2 and 2< 6(true)
a [a [i] ] = a [i];         
Iteration 1:
i = 2;
a[a[2]] = a[2]
a[3] = [3]

Iteration 2:
When i = 3, 3 < 6 (true)
a[a[3]] = a[3]
a[3] = 3              // as in iteration 1, a[3] becomes was 3

Iteration 3:
When i = 4
a[a[4]] = a[4]
a[5] = 5

Iteration 4:
When i = 5, 5 < 6 (true)
a[a[5]] = a[5]
a[5] = 5                // as in iteration 3 a[5] becomes 5.

In next iteration i becomes and condition 6 < 6 false here.
Content of array A are as follows:

Finally, it prints: 1 2 3 3 5 5 7 8