Returning address of a local variable.

[robezy]

I was reading Object oriented programming using c++ by Ira Pohl, when I came across following piece of code.

my_string& operator+(const my_string& a, const my_string& b)
{
   my_string* temp = new my_string(a.len + b.len);
   
   strcpy(temp->s, a.s);
   strcpy(temp->s, b.s);
   return *temp;
}

I remember reading somewhere that, it is an error to return the address of a local variable. Can anyone tell me how above code wroks? Isn't it wrong to return the address of temp?

[kevinchkin]

They are not returning the address of temp. *temp means, get the value stored at the address pointed by temp. They have actually de-referenced the pointer.


EDIT: And the memory pointed by temp is on heap, so it does not follow the scope rules, you will have to explicitly de-allocate it using delete.

[robezy]

I'm still confused. What is the return type here? Is it "my_string" ? What does "my_string&" return type means?

Here what is returned? value of temp or a reference to temp?

Thanks.

[Bazzy]

my_string& means a 'reference of type my_string'

[jsmith]

The local variable in the function is a pointer to a my_string.

What is being returned from the function is a reference to the my_string pointed to by the local variable.

That is quite a bizarre piece of code. Screams memory leak to me.

[Duthomhas]

Yes, it has a memory leak. It uses new to allocate something on the heap, then returns it like something referenced. At no time does the caller know it is supposed to delete it.

[guestgulkan]

For the record - I thought I would check out how such a peice of code slipped past the Author/and editor of that book - And I found this on the Author's homepage at:
http://users.soe.ucsc.edu/~pohl/

Notice about halfway down the page in the Acknowledgment section.

[robezy]

Thanks guys for clarifying this piece of code.

@guestgulkan

Thanks for linking author's homepage. I can use this to check for other corrections.