#include <iostream> #include <string> using namespace std; void swap(int * a, in

Question

#include <iostream>
#include <string>

using namespace std;

void swap(int * a, int * b)
{
   int temp = *a;
   cout<<"the contents of the memories pointed to by a and b have the following values at the start of swap"<<endl;
   cout<<"*a = "<<*a<<" and *b = "<<*b<<endl;
   *a=*b;
   *b=temp;
   cout<<"the contents of the memories pointed to by a and b have the following values at the end of swap"<<endl;
   cout<<"*a = "<<*a<<" and *b = "<<*b<<endl;

}
int main()
{
   int a=111, b=222;

   cout<<"a and b have the following values before swap is called"<<endl;
   cout<<"a = "<<a<<" and b = "<<b<<endl;
   swap(&a,&b);
   cout<<"a and b have the following values after swap is called"<<endl;
   cout<<"a = "<<a<<" and b = "<<b<<endl;
   return 0;
}

Question 5:   What is the purpose of the

Explanation / Answer

5) & means you are sending address location of variables

* means pointing to particular memory location

6)Are the variables a and b used in the function main the same a and b variables that are used in the function swap in the program in Step 2?

variables a and b are passed by reference so those will be same in function swap. so function swap will have same variables as in function main.
so both are same variables.

Question 7:   Were the values stored in the variables a and b in the function main swapped after the execution of the function swap in the program in Step 2?
Yes.. because those variables a and b are passed by reference so those will be same in function swap. so variables in function swap are same variables a and b of
function main. since we changed thier values in function swap, they will be swapped after function call in main.

8) pass by referrence

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

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