Answer:
#include <iostream>
using namespace std;
void CoordTransform(int *ptr1, int *ptr2);
int main()
{
int xVal;
int yVal;
cout<<"please enter two valid integers";
cin>>xVal;
cin>>yVal;
CoordTransform(&xVal , &yVal);
int xValNew=xVal;
int yValNew=yVal;
cout<<xValNew<<yValNew;
return 0;
}
void CoordTransform(int *ptr1, int *ptr2)
{
int a = *ptr1;
*ptr1=(*ptr1+1)*2;
*ptr2=(*ptr2+1)*2;
}
Explanation:
It will return new values in previously defined variables
coding language: c++
Answer:
#include <iostream>
using namespace std;
void CoordTransform(int *xVal, int *yVal) {
// since we have passed x and y by adress:
// any change in xVal or yVal will also change the value of x and y
*xVal = (*xVal + 1) * 2;
*yVal = (*yVal + 1) * 2;
}
int main() {
int x, y;
// geting x from user
cout << "Enter x: ";
cin >> x;
// getting y from user
cout << "Enter y: ";
cin >> y;
// passing x and y to function CoordTransform() by adress.
CoordTransform(&x, &y);
cout << "new x: " << x << endl;
cout << "new y: " << y << endl;
return 0;
}
Explanation:
a pointer points to a memory location of the veraible so, when we pass a veriable by adress to a function, any change at that memory location also effect the value of the veriable that is local to the calling function.
