What is printed by the following ANSI C program?#include<stdio.h> int main(int argc, char *argv[]){ char a = 'P'; char b = 'x'; char c = (a & b) + '*'; char d = (a | b) - '-'; char e = (a ^ b) + '+'; printf("%c %c %c ", c, d, e); return 0; }ASCII encoding for relevant characters is given below

First, find the ASCII integer values of 'P' and 'x':

• 'P' is the $16^{\text{th}}$ uppercase letter: $\text{ASCII}('P^{\prime}) = 65 + 15 = 80 = 01010000_2$

• 'x' is the $24^{\text{th}}$ lowercase letter: $\text{ASCII}('x^{\prime}) = 97 + 23 = 120 = 01111000_2$

Now, perform the bitwise operations and arithmetic:

1. Bitwise AND (&):

   $$a \ \&\ b = 01010000_2 \ \&\ 01111000_2 = 01010000_2 = 80$$

   $$c = (a \ \&\ b) + '*' = 80 + 42 = 122$$

   The character with ASCII value $122$ is 'z' ($97 + 25 = 122$).

2. Bitwise OR (|):

   $$a \mid b = 01010000_2 \mid 01111000_2 = 01111000_2 = 120$$

   $$d = (a \mid b) - '-' = 120 - 45 = 75$$

   The character with ASCII value $75$ is 'K' ($65 + 10 = 75$).

3. Bitwise XOR (^):

   $$a \oplus b = 01010000_2 \oplus 01111000_2 = 00101000_2 = 32 + 8 = 40$$

   $$e = (a \oplus b) + '+' = 40 + 43 = 83$$

   The character with ASCII value $83$ is 'S' ($65 + 18 = 83$).

Since printf uses the %c specifier, it outputs the characters 'z', 'K', and 'S' separated by spaces: z K S.

Hence, (A) is the correct option.