Quiz-1

Declaration: char *strrchr(const char *s, int c);

It scans a string s in the reverse direction, looking for a specific character c.

extern int x; - is an external variable declaration.

double pow(double, double); - is a function prototype declaration.

Therefore, 1 and 3 are declarations. 2 is definition.

1. The array int num[26]; can store 26 elements. This statement is true.

2. The expression num[1] designates the very first element in the array. This statement is false, because it designates the second element of the array.

3. It is necessary to initialize the array at the time of declaration. This statement is false.

4. The declaration num[SIZE] is allowed if SIZE is a macro. This statement is true, because the MACRO just replaces the symbol SIZE with given value.

Hence the statements '1' and '4' are correct statements.

Step 1: int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 8, 7, 8, 9, 0}; The variable a is declared as an multidimensional integer array with size of 3 rows 4 columns.

Step 2: int *ptr; The *ptr is a integer pointer variable.

Step 3: ptr = &a[0][0]; Here we are assigning the base address of the array a to the pointer variable *ptr.

Step 4: fun(&ptr); Now, the &ptr contains the base address of array a.

Step 4: Inside the function fun(&ptr); The printf("%d\n", **p); prints the value '1'.

because the *p contains the base address or the first element memory address of the array a (ie. a[0])

**p contains the value of *p memory location (ie. a[0]=1).

Hence the output of the program is '1'

Step 1: int arr[] = {12, 14, 15, 23, 45}; The variable arr is declared as an integer array and initialized.

Step 2: printf("%u, %u\n", arr, &arr); Here,

The base address of the array is 65486.

=> arr, &arr is pointing to the base address of the array arr.

Hence the output of the program is 65486, 65486

Step 1: char ch; ch = 'A'; here variable ch is declared as an character type an initialized to 'A'.

Step 2: printf("The letter is"); It prints "The letter is".

Step 3: printf("%c", ch >= 'A' && ch <= 'Z' ? ch + 'a' - 'A':ch);

The ASCII value of 'A' is 65 and 'a' is 97.

Here

=> ('A' >= 'A' && 'A' <= 'Z') ? (A + 'a' - 'A'):('A')

=> (TRUE && TRUE) ? (65 + 97 - 65) : ('A')

=> (TRUE) ? (97): ('A')

In printf the format specifier is '%c'. Hence prints 97 as 'a'.

Step 4: printf("Now the letter is"); It prints "Now the letter is".

Step 5: printf("%c\n", ch >= 'A' && ch <= 'Z' ? ch : ch + 'a' - 'A');

Here => ('A' >= 'A' && 'A' <= 'Z') ? ('A') : (A + 'a' - 'A')

=> (TRUE && TRUE) ? ('A') :(65 + 97 - 65)

=> (TRUE) ? ('A') : (97)

It prints 'A'

Hence the output is

The letter is a
Now the letter is A

Step 1: int i=-3, j=2, k=0, m; here variable i, j, k, m are declared as an integer type and variable i, j, k are initialized to -3, 2, 0 respectively.

Step 2: m = ++i && ++j && ++k;
becomes m = -2 && 3 && 1;
becomes m = TRUE && TRUE; Hence this statement becomes TRUE. So it returns '1'(one). Hence m=1.

Step 3: printf("%d, %d, %d, %d\n", i, j, k, m); In the previous step the value of i,j,k are increemented by '1'(one).

Hence the output is "-2, 3, 1, 1".

Step 1: int i=4, j=-1, k=0, w, x, y, z; here variable i, j, k, w, x, y, z are declared as an integer type and the variable i, j, k are initialized to 4, -1, 0 respectively.

Step 2: w = i || j || k; becomes w = 4 || -1 || 0;. Hence it returns TRUE. So, w=1

Step 3: x = i && j && k; becomes x = 4 && -1 && 0; Hence it returns FALSE. So, x=0

Step 4: y = i || j &&k; becomes y = 4 || -1 && 0; Hence it returns TRUE. So, y=1

Step 5: z = i && j || k; becomes z = 4 && -1 || 0; Hence it returns TRUE. So, z=1.

Step 6: printf("%d, %d, %d, %d\n", w, x, y, z); Hence the output is "1, 0, 1, 1".

Output will be 3

Step 1: int a[5] = {5, 1, 15, 20, 25}; The variable arr is declared as an integer array with a size of 5 and it is initialized to

a[0] = 5, a[1] = 1, a[2] = 15, a[3] = 20, a[4] = 25 .

Step 2: int i, j, m; The variable i,j,m are declared as an integer type.

Step 3: i = ++a[1]; becomes i = ++1; Hence i = 2 and a[1] = 2

Step 4: j = a[1]++; becomes j = 2++; Hence j = 2 and a[1] = 3.

Step 5: m = a[i++]; becomes m = a[2]; Hence m = 15 and i is incremented by 1(i++ means 2++ so i=3)

Step 6: printf("%d, %d, %d", i, j, m); It prints the value of the variables i, j, m

Hence the output of the program is 3, 2, 15