Answer:
Check the explanation
Explanation:
To solve the question above, we will be doing some arithmetic and bit wise operations on enum values. in addition to that, we are asserting the array with INTCOUNT Â and also printing the size of array i.e 12 and we are adding BIG_COUNT enum value to 5 and the result is 25.
#include<stdio.h>
enum EnumBits
{
ONE = 1,
TWO = 2,
FOUR = 4,
EIGHT = 8
};
enum Randoms
{
BIG_COUNT = 20,
INTCOUNT = 3
};
int main()
{
// Basic Mathimatical operations
printf("%d\n",(ONE + TWO)); //addition
printf("%d\n",(FOUR - TWO)); //subtraction
printf("%d\n",(TWO * EIGHT)); //multiplication
printf("%d\n",(EIGHT / TWO)); //division
// Some bitwise operations
printf("%d\n",(ONE | TWO)); //bitwise OR
printf("%d\n",(TWO & FOUR)); //bitwise AND
printf("%d\n",(TWO ^ EIGHT)); //bitwise XOR
printf("%d\n",(EIGHT << 1)); //left shift operator
printf("%d\n",(EIGHT >> 1)); //right shift operator
// Initialize an array based upon an enum value
int intArray[INTCOUNT]; // declaring array
// Have a value initialized be initialized to a static value plus
int someVal = 5 + BIG_COUNT; //adding constant and BIG_COUNT variable
printf("%d\n",someVal); //value will be 25
printf("%d",sizeof(intArray)); //value will be 12
return 0;
}
Kindly check the attached image below for the Code and Output Screenshots:
