reference-pointer sample

[digrev]

hi everybody i am new to progamming my question is how can i make this sample with using a pointer



#include <iostream>
#include<conio.h>
using namespace std;
#include <iostream>
#include<conio.h>
using namespace std;
double &f(int i); // return a reference

double vals[] = { 1.1, 2.2, 3.3, 4.4, 5.5 };

int main()
{
int i;

cout << "Here are the original values: ";
for(i=0; i < 5; i++)
cout << vals[i] << ' ';


f(1) = 8.23; // change 2nd element
f(3) = -8.8; // change 4th element

cout << "\nHere are the changed values: \n";
for(i=0; i < 5; i++)
cout << vals[i] << ' ';

return 0;
}

double &f(int i)
{
return vals[i]; // return a reference to the ith element
}

[ResidentBiscuit]

#include <iostream>
#include<conio.h>
using namespace std;
#include <iostream>
#include<conio.h>
using namespace std;
double &f(int i); // return a reference 

double vals[] = { 1.1, 2.2, 3.3, 4.4, 5.5 }; 

int main() 
{ 
int i; 
int *ptr = &i;

cout << "Here are the original values: "; 
for(i=0; i < 5; i++) 
cout << vals[i] << ' '; 


//random pointer stuff
*ptr = 12; //Might break something, I don't know.

f(1) = 8.23; // change 2nd element 
f(3) = -8.8; // change 4th element 

cout << "\nHere are the changed values: \n"; 
for(i=0; i < 5; i++) 
cout << vals[i] << ' '; 

return 0; 
} 

double &f(int i) 
{ 
return vals[i]; // return a reference to the ith element 
}

This what you looking for?

Be generic, you get what you get

[vlad from moscow]

double * f(int i); // return a pointer


double * f(int i) 
{ 
   return &vals[i]; // return a pointer to the ith element 
}

[digrev]

hi again i am trying this way but i get erros



#include <iostream>
#include<conio.h>
using namespace std;
#include <iostream>
#include<conio.h>
using namespace std;
double *f(int i); // return a reference

double vals[] = { 1.1, 2.2, 3.3, 4.4, 5.5 };

int main()
{
int i;

cout << "Here are the original values: ";
for(i=0; i < 5; i++)
cout << vals[i] << ' ';


f(1) = 8.23; // change 2nd element
f(3) = -8.8; // change 4th element

cout << "\nHere are the changed values: \n";
for(i=0; i < 5; i++)
cout << vals[i] << ' ';

return 0;
}

double *f(int i)
{
return &vals[i]; // return a reference to the ith element
}

[vlad from moscow]

You should change these statements

f(1) = 8.23; // change 2nd element
f(3) = -8.8; // change 4th element

to

*f(1) = 8.23; // change 2nd element
*f(3) = -8.8; // change 4th element