In this article, we will discuss some important concepts related to arrays and problems based on that. Before understanding this, you should have basic idea about Arrays.

**Type 1. Based on array declaration –**

These are few key points on array declaration:

- A single dimensional array can be declared as int a[10] or int a[] = {1, 2, 3, 4}. It means specifying the number of elements is optional in 1-D array.
- A two dimensional array can be declared as int a[2][4] or int a[][4] = {1, 2, 3, 4, 5, 6, 7, 8}. It means specifying the number of rows is optional but columns are mandatory.
- The declaration of int a[4] will give the values as garbage if printed. However, int a[4] = {1,1} will initialize remaining two elements as 0.

**Que – 1.** Predict output of following program

int main() { int i; int arr[5] = {1}; for (i = 0; i(A) 1 followed by four garbage values:

(B) 1 0 0 0 0

(C) 1 1 1 1 1

(D) 0 0 0 0 0

Solution:As discussed, if array is initialized with few elements, remaining elements will be initialized to 0. Therefore, 1 followed by 0, 0, 0, 0 will be printed.

Que - 2.Predict output of the following program:int main() { int a[][] = {{1,2},{3,4}}; int i, j; for (i = 0; i(A) 1 2 3 4

(B) Compiler Error in line ” int a[][] = {{1,2},{3,4}};”

(C) 4 garbage values

(D) 4 3 2 1

Solution:As discussed, specifying the number of columns in 2-D array is mandatory, so, it will give compile time error.

Type 2. Finding address of an element with given base address -

When an array is declared, a contiguous block of memory is assigned to it which helps in finding address of elements from base address.For a single dimensional array a[100], address of ith element can be found as:

addr(a[i]) = BA+ i*SIZEWhere BA represents base address (address of 0th element) and SIZE represents size of each element in the array.

For a two dimensional array x[3][3], the elements can be represented as:

As 2-D array is stored in row major order in C language, row 0 will be stored first followed by row 1 and row 2. For finding the address of x[2][2], we need to go to 2nd row (each row having 3 elements). After reaching 2nd row, it can be accessed as single dimensional array. Therefore, we need to go to 2nd element of the array. Assuming BA as 0 and size as 1, the address of x[2][2] will be 0 + (2 * 3 + 2) * 1 = 8.

For a given array with m rows and n columns, the address can be calculated as:

add(a[i][j]) = BA + (i*n + j) * SIZEWhere BA represents base address (address of 0th element), n represents number of columns in 2-D array and SIZE represents size of each element in the array.

Que - 3.Consider the following declaration of a ‘two-dimensional array in C:char a[100][100];Assuming that the main memory is byte-addressable and that the array is stored starting from memory address 0, the address of a[40][50] is: (GATE CS 2002)

(A) 4040

(B) 4050

(C) 5040

(C) 5050

Solution:Using the formula discussed,addr[40][50] = 0 + (40*100 + 50) * 1 = 4050

Que - 4.For a C program accessing X[i][j][k], the following intermediate code is generated by a compiler. Assume that the size of an integer is 32 bits and the size of a character is 8 bits. (GATE-CS-2014)t0 = i * 1024 t1= j * 32 t2 = k * 4 t3 =t1 + t0 t4 = t3 + t2 t5 = X[t4]Which one of the following statement about the source code of C program is correct?

(A) X is declared as “int X[32][32][8]”

(B) X is declared as “int X[4][1024][32]”

(C) X is declared as “char X[4][32][8]”

(D) X is declared as “char X[32][16][2]”

Solution:For a three dimensional array X[10][20][30], we have 10 two dimensional matrices of size [20]*[30]. Therefore, for a 3 D array X[M][N][O], the address of X[i][j][k] can be calculated as:BA + (i*N*O+j*O+k)*SIZEGiven different expressions, the final value of t5 can be calculated as:

t5 = X[t4] = X[t3+t2] = X[t1+t0+t2] = X[i*1024+j*32+k*4]By equating addresses,

(i*N*O+j*O+k)SIZE = i*1024+j*32+k*4 = (i*256+j*8+k)4Comparing the values of i, j and SIZE, we get

SIZE = 4, N*O = 256 and O = 8, hence, N = 32As size is 4, array will be integer. The option which matches value of N and O and array as integer is (A).

Type 3. Accessing array elements using pointers -

- In a single dimensional array a[100], the element a[i] can be accessed as a[i] or *(a+i) or *(i+a)
- Address of a[i] can be accessed as &a[i] or (a+i) or (i+a)
- In two dimensional array a[100][100], the element a[i][j] can be accessed as a[i][j] or *(*(a+i)+j) or *(a[i]+j)
- Address of a[i][j] can be accessed as &a[i][j] or a[i]+j or *(a+i)+j
- In two dimensional array, address of ith row can be accessed as a[i] or *(a+i)

**Que - 5.** Assume the following C variable declaration

int *A [10], B[10][10];

Of the following expressions

I. A[2] II. A[2][3] III. B[1] IV. B[2][3]

which will not give compile-time errors if used as left hand sides of assignment statements in a C program (GATE CS 2003)?

(A) I, II, and IV only

(B) II, III, and IV only

(C) II and IV only

(D) IV only

**Solution:** As given in the question, A is an array of 10 pointers and B is a two dimensional array. Considering this, we take an example as:

int *A[10], B[10][10]; int C[] ={1, 2, 3, 4, 5};

As A[2] represents an integer pointer, it can store the address of integer array as: A[2] = C; therefore, I is valid.

As A[2] represents base address of C, A[2][3] can be modified as: A[2][3] = *(C +3) = 0; it will change the value of C[3] to 0. Hence, II is also valid.

As B is 2D array, B[2][3] can be modified as: B[2][3] = 5; it will change the value of B[2][3] to 5. Hence, IV is also valid.

As B is 2D array, B[2] represent address of 2nd row which can’t be used at LHS of statement as it is invalid to modify the address. Hence III is invalid.

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