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 x=12, y=7, z; here variable x, y and z are declared as an integer and variable x and y are initialized to 12, 7 respectively.

Step 2: z = x!=4 || y == 2;
becomes z = 12!=4 || 7 == 2;
then z = (condition true) || (condition false); Hence it returns 1. So the value of z=1.

Step 3: printf("z=%d\n", z); Hence the output of the program is "z=1".

Yes, the associativity of an operator is either Left to Right or Right to Left.

Yes, Each and every operator has an associativity.

The associativity (or fixity) of an operator is a property that determines how operators of the same precedence are grouped in the absence of parentheses. Operators may be left-associative, right-associative or non-associative.

An operation with only one operand is called unary operation.
Unary operators:
! Logical NOT operator.
~ bitwise NOT operator.
sizeof Size-of operator.

&& Logical AND is a logical operator.

Therefore, 1, 2, 3 are unary operators.

Because, comma operator used in the expression i (1, 2, 3, 4, 5). The comma operator has left-right associativity. The left operand is always evaluated first, and the result of evaluation is discarded before the right operand is evaluated. In this expression 5 is the right most operand, hence after evaluating expression (1, 2, 3, 4, 5) the result is 5, which on adding to i results into 7.

Step 1: int a=100, b=200, c;
Step 2: c = (a == 100 || b > 200);
becomes c = (100 == 100 || 200 > 200);
becomes c = (TRUE || FALSE);
becomes c = (TRUE);(ie. c = 1)
Step 3: printf("c=%d\n", c); It prints the value of variable i=1
Hence the output of the program is '1'(one).

Step 1: static int a[20]; here variable a is declared as an integer type and static. If a variable is declared as static and it will be automatically initialized to value '0'(zero).

Step 2: int i = 0; here vaiable i is declared as an integer type and initialized to '0'(zero).
Step 3: a[i] = i ; becomes a[0] = 0;
Step 4: printf("%d, %d, %d\n", a[0], a[1], i);
Here a[0] = 0, a[1] = 0(because all staic variables are initialized to '0') and i = 0.
Step 4: Hence the output is "0, 0, 0".