Using C. Please fill out the code and do not make modifications to the functions

Question

Using C. Please fill out the code and do not make modifications to the functions or headers :

#include
#include

int foo(int* a, int* b, int c){
/*Swap the addresses stored in the pointer variables a and b*/
  
/*Decrement the value of integer variable c*/
  
/*Return c*/
}

int main(){
/*Declare three integers x,y and z and initialize them randomly to values in [0,10] */
  
/*Print the values of x, y and z*/
  
/*Call foo() appropriately, passing x,y,z as parameters*/
  
/*Print the values of x, y and z*/
  
/*Print the value returned by foo*/

  
return 0;
}

1) The function int foo(int *a, int *b, int c) should perform the following computations -Swap the addresses of the integers pointed to by a and b (not the values of the integers pointed to by a and b). - Decrement the value of c - Return the value of c. ) In the main function, declare three integers x, y, and z, and assign them random integer values in the interval [O, 10] 3) Print the values of the integers x, y, and z. Call foo() appropriately passing x, y, and z as arguments. Print out the values of integers x, y, and z after calling the function foo(.). Also, print the value returned by foo(). 4) Answer the following questions in a comment at the bottom of the file: a) Is the return value different than the value of integer z? Why or why not? b) Are the values of integers x and y different before and after calling the function foo(.)? Why or why not?

Explanation / Answer

#include <stdio.h>

int foo(int* a, int* b, int c){

printf(" Inside foo, Address of a = %d Address of b = %d c = %d",a,b,c);
/*Swap the addresses stored in the pointer variables a and b*/
int *temp;
temp = a;
a = b;
b = temp;

printf(" Inside foo After Swapping addresses , Address of a = %d Address of b = %d c = %d",a,b,c);
/*Decrement the value of integer variable c*/
c--;

/*Return c*/
return c;

}
int main(){
/*Declare three integers x,y and z and initialize them randomly to values in [0,10] */
int x,y,z;

x = 5;
y = 7;
z = 4;

/*Print the values of x, y and z*/
printf("x = %d y = %d z = %d",x,y,z);

/*Call foo() appropriately, passing x,y,z as parameters*/
int result = foo(&x,&y,z);

/*Print the values of x, y and z*/
printf(" x = %d y = %d z = %d",x,y,z);


/*Print the value returned by foo*/
printf(" Value returned from foo : %d",result);

return 0;
}

Output:

x = 5 y = 7 z = 4
Inside foo, Address of a = 309520424 Address of b = 309520428 c = 4
Inside foo After Swapping addresses , Address of a = 309520428 Address of b = 309520424 c = 4
x = 5 y = 7 z = 4
Value returned from foo : 3

4 a) yes , the return value is different than the value of integer z. It will be the decremented value of z by 1.

4 b) No, the value of x and y will be same after the call to foo() because the scope of a and b is local inside the function foo() . Inside the function , the addresses are swapped but will not reflect outside the function .

Do ask if any doubt. Please upvote.

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.