Implement a function that returns an integer that is one gre
Solution
Exercise 1:
#include<stdio.h>
 #include <assert.h>
 int main()
 {
 int addOne(int k);
 //If expression evaluates to TRUE, assert() does nothing.
 //If expression evaluates to FALSE, assert() displays an error message on stderr
 //and aborts program execution.
 assert(addOne(-3) == -2);
 printf(\"\  Result: %d\", addOne(-3));
 assert(addOne(0) == 1);
 printf(\"\  Result: %d\", addOne(0));
 assert(addOne(1) == 2);
 printf(\"\  Result: %d\", addOne(1));
 }
 int addOne(int k)
 {
 return (++k);
 }
Output:
Result: -2
 Result: 1
 Result: 2
Exercise 2:
#include<stdio.h>
 #include <assert.h>
 void addOne(int *k)
 {
 (++*(k));
 }
int main()
 {
 //If expression evaluates to TRUE, assert() does nothing.
 //If expression evaluates to FALSE, assert() displays an error message on stderr
 //and aborts program execution.
 int x = 10;
 //Address of X is passed
 addOne(&x);
 assert((x) == 11);
 printf(\"\  Result: %d\", x);
 x = 20;
 addOne(&x);
 assert((x) == 21);
 printf(\"\  Result: %d\", x);
 }
Output:
Result: 11
 Result: 21
Exercise 3:
#include<stdio.h>
 #include <assert.h>
 #include<stdbool.h>
 bool isEven(int k)
 {
 if(k%2==0)
 return (true);
 else
 return (false);
 }
 int main()
 {
 //If expression evaluates to TRUE, assert() does nothing.
 //If expression evaluates to FALSE, assert() displays an error message on stderr
 //and aborts program execution.
 assert(isEven(12) == true);
 printf(\"\  Result: %d\", isEven(12));
 assert(isEven(6) == true);
 printf(\"\  Result: %d\", isEven(6));
 assert(isEven(5) == false);
 printf(\"\  Result: %d\", isEven(5));
 }
Output:
Result: 1
 Result: 1
 Result: 0


