pointers and reference variables

[khulme]

i am mixing up my reference and pointer variables. need a bit of guidance. i am trying to return a value by reference from a function.

//

#include "stdafx.h"
#include <iostream>
#include <iomanip>

using namespace std;

// FUNCTION PROTOTYPE
double convert(double *measurement);

int main()
{
	double measurement;

	cout << "Enter a length in inches and I will convert it to centimeters:\n";
	cin >> measurement;
	
	convert(&measurement);

	cout << setprecision(4);
	cout << fixed << showpoint;
	cout << "Value in centimeters: " << measurement << endl;

	system("pause");
	return 0;
}

/***********
convert
***********/

double convert(double *measurement) 
{
return *measurement * 2.54;
}

[ResidentBiscuit]

If you want to return by reference, you need to change your function header to show that.

int& foo(int* ptr);

There's an example that takes a pointer to an int as a parameter, and returns a reference to an int as a return value.

[Texan40]

In convert you are passing in a pointer and returning a value. If you want to work with a reference then perhaps:

void convert(double &measurement)
{
  measurement *= 2.54;
}

then you would just pass by convert(measurement) instead of passing a pointer

[khulme]

ok. those are both helpful. thank you ResidentBiscuit and Texan40. let me see what I can do.

[khulme]

ok. i got it i think. this one compiles and runs.

//

#include "stdafx.h"
#include <iostream>
#include <iomanip>

using namespace std;

// FUNCTION PROTOTYPE
void convert(double *);

int main()
{
	double measurement;

	cout << "Enter a length in inches and I will convert it to centimeters:\n";
	cin >> measurement;
	
	convert(&measurement);

	cout << setprecision(4);
	cout << fixed << showpoint;
	cout << "Value in centimeters: " << measurement << endl;

	system("pause");
	return 0;
}

/***********
convert
***********/

void convert(double *measurement) 
{
*measurement *= 2.54;
}