returning static pointer to local variable from function

Question

I've found this code on web as an example, but I think this is not correct. An address to automatic variable is returned and this is just coincidence that it might work sometimes:

returning a pointer to an destroyed local variable, which becomes invalid memory location, is undefined behavior.

My only little hesitancy is about the pointer being static, but I think this changes nothing as this is the variable that should be static not a pointer: local variable is going to be destroyed. Can you please confirm or deny?

double *& showNumber()
{
    double n = 1550.85;
    static double *v = &n;
    return v;
}

int main(int argc, char *argv[])
{
    double sn = *showNumber();
    sn = *showNumber();
    //...
}

static persists for the entire duration of the program as soon as you initialize them so this should be okay for v, not for n — Marco A.
– Marco A., Commented Mar 1, 2014 at 13:07

NPE · Accepted Answer · 2014-03-01 13:06:52Z

2

For this code to be well-defined, both n and v would need to be static.

Right now, the *showNumber() has undefined behaviour as it dereferences a dangling pointer.

answered Mar 1, 2014 at 13:06

NPE

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4 Comments

bolov Over a year ago

only n should be static

4pie0 Over a year ago

v should also be static as intended use of this function is also to return a pointer I think

NPE Over a year ago

@bolov: Why is that (bearing in mind that v is returned by reference)?

bolov Over a year ago

I am tired, didn't see the return by reference. You are right

Vlad from Moscow · Accepted Answer · 2014-03-01 14:00:50Z

2

Your code has still undefined behaviour because the value of the static pointer is invalid after exiting the function. The local variable refered to by the pointer will be destroyed.And any next time when the function will be called the address of this local variable can be different.

You could write your function the following way

double * showNumber()
{
    static double n = 1550.85;
    return &n;
}

In this case the returned pointer would contain the same valid value.

edited Mar 1, 2014 at 14:00

answered Mar 1, 2014 at 13:17

Vlad from Moscow

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1 Comment

gnasher729 Over a year ago

In C, the code static double* v = &n; would be illegal, because &n is not a compile-time constant. In C++, it is legal; the initialisation happens the first time the static variable is used, and only once. So even if you wrote *v = 1.0; inside that function, it would probably crash on the second call.

Marco A. · Accepted Answer · 2014-03-01 13:08:28Z

1

Static variables persist for the entire duration of the program as soon as you initialize them. You're all set for v, but not for n's address.

answered Mar 1, 2014 at 13:08

Marco A.

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Comments

Pranit Kothari · Accepted Answer · 2014-03-01 13:08:05Z

0

If pointer and it's variable both are static then only code will be OK. Otherwise local variable will anyway going to die.

answered Mar 1, 2014 at 13:08

Pranit Kothari

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Comments

rajendra shinde · Accepted Answer · 2016-10-20 05:05:29Z

0

Independent of local or global, static variable will live till the end of program. but it is recommenced that to use global when we return the value of that variable. because static local may get affected due to threads.

answered Oct 20, 2016 at 5:05

rajendra shinde

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