return reference to local variable [duplicate]

Answer 1

Normally you use * to declare a pointer variable and also to dereference the pointer back to the original variable afterwords. & returns the address from a variable

I haven't tested this, but does the following work any better?

#include <iostream>
using namespace std;

double* GetSomeData()
{
  double h = 46.50;
  return &h;
}

int main()
{
  double nonRef = GetSomeData();
  double* ref = GetSomeData();
  cout << "nonRef: " << nonRef << endl;
  cout << "ref: " << *ref << endl;
  return 0;
}

Answer 2

If you are merely testing the difference between these two lines:

double nonRef = GetSomeData();
double &ref = GetSomeData();

Try classes.

class test
{
public:
    int x;
    int& getx()
    {
        return x;
    }
};

//Some code
test c;
int nonRef = c.getx();
int &ref = c.getx();

ref = 8;
nonref = c.getx();

After this, c.getx() returns 8.

Answer 3

Yes you got lucky.

Returning reference to local variable is Undefined Behavior. Undefined Behavior means anything can happen and a behavior cannot be defined.

Regarding Undefined Behavior,

C++ Standard section 1.3.24 states:

Permissible undefined behavior ranges from ignoring the situation completely with unpredictable results, to behaving during translation or program execution in a documented manner characteristic of the environment (with or without the issuance of a diagnostic message), to terminating a translation or execution (with the issuance of a diagnostic message).