return reference to local variable

Question

Possible Duplicate:
Can a local variable's memory be accessed outside its scope?!

Here is a code:

    #include <iostream>
using namespace std;

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

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

the nonRef is printed OK as 46.5 the ref is not OK.

is the first output behavior is correct or just got lucky?

Thanks

This has been asked before (more than once). [Accepted answer](http://stackoverflow.com/questions/6441218/can-a-local-variables-memory-be-accessed-outside-its-scope/6445794#6445794) received 1642 upvotes (to date)... — Eran, Jul 14 '11 at 06:19
Eric Lippert makes perfect analogies. Coincidentally I didn't read the first comment on his answer until just now :) — Marlon, Jul 14 '11 at 06:46

Alok Save · Accepted Answer · 2011-09-17T05:54:25.880

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).

score 0 · Answer 2 · answered Jul 14 '11 at 06:06

0

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;
}

answered Jul 14 '11 at 06:06

Devin Garner

1,361
9
20

1

You're missing the point. Whether you use references or pointers is irrelevant in this case. – Fiktik Jul 14 '11 at 06:09
1

Returning address to local variable is Undefined Behavior. Undefined Behavior means anything can happen and a behavior cannot be defined. – Jaffa Jul 14 '11 at 06:10

score 0 · Answer 3 · answered Jul 14 '11 at 06:19

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.

return reference to local variable

3 Answers3

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