Count the number of words in the string. A word is one or more non-blank characters separated by one or more blanks. My suggesti
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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 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:
