Address of return value of method

Question

I have a get-method that returns a value:

int Foo::getValue() const
{
    return value_;
}

I'm using the following code to write this value to a binary file:

void Bar::write(const Foo& foo)
{
    /* some code... */

    file_.write(reinterpret_cast<char*>(&foo.getValue()), sizeof(int));
}

And it seems that everything works well. But I am not sure about that &foo.getValue() thing. Will it always return an address of return value of getValue() or are there any pitfalls?

EDIT1: I dont want to create a temporary var for this.

EDIT2: Yes, it compiles and works fine.

EDIT3: write is a ostream::write

EDIT4: With /Wall and /Za it will not compile: error C2102: '&' requires l-value

`reinterpret_cast(&foo.getValue())` This seems dangerous to me. A `std::string` is not just a null terminated `char *` ... cant you just use `foo.getValue().c_str()`? — Fantastic Mr Fox, Dec 21 '15 at 18:16
That shouldn't even compile. If a function returns by value then it returns an rvalue. You can't take the address of an rvalue. — Brian Bi, Dec 21 '15 at 18:17
Opps, i made a mistake. It's not std::string but simple int. And yes, its compiles and works! — NekoUnderscore, Dec 21 '15 at 18:21
@aschepler agreed, that will be the proper way to do this. But im still wondering about doing it like i did. It works, but make me feel unsure... — NekoUnderscore, Dec 21 '15 at 18:29
It cannot compile, as Brian says you cannot take the address of an rvalue. What compiler are you using? — edmz, Dec 21 '15 at 18:34
@black im using VS2015 and just compiled my code again. Build: 1 succeeded, 0 failed, 0 up-to-date, 0 skipped — NekoUnderscore, Dec 21 '15 at 18:36
@NekoUnderscore See a conforming compiler failing [here](http://coliru.stacked-crooked.com/a/92c30404741264d0) — edmz, Dec 21 '15 at 18:38
@black im not trying to trick you or something. It can write the value to binary file and even can read it back. I tested it. — NekoUnderscore, Dec 21 '15 at 18:42
legal or not, I don't see what is the problem catching the temporary with const reference then write it? better be smart and safe then correct. — David Haim, Dec 21 '15 at 19:05

score 4 · Accepted Answer · edited May 23 '17 at 12:25

The standard forbids you to take the address of an rvalue: https://stackoverflow.com/a/28459180/2642059 A compiler that is not standard compliant may allow this behavior, but assuredly that is not behavior that will be cross platform, nor is it guaranteed to succeed even between updates to the compiler that may more closely align it with the C++ standard.

Since you are not programming for a specific compiler, but for C++, finding something that works in a specific compiler should not be your objective. Finding a way to accomplish this within the standard would be preferable.

A couple easy options for doing that would be:

Turn it into an lvalue by assigning it to a temporary value. They're free! The compiler will optimize them right out
Change the return of getValue to const int& this will allow you to work with an lvalue. Since ints are typically copied by value this probably will not have repercussions in your code base, but it could
Create another method which could directly return (const char*)&value_, you could get in trouble abusing this one though so be careful

score 3 · Answer 2 · answered Dec 21 '15 at 19:05

3

Microsoft Compiler for whatever reason has a `"feature" - it implicitly converts temporary to lvalue. You can check it by this code:

int &ref = foo.getValue();

this code as well as your should not compile. Proper code would be:

void Bar::write(const Foo& foo)
{
    /* some code... */
    int temp = foo.getValue();
    file_.write(static_cast<const char*>(&temp), sizeof(int));
}

answered Dec 21 '15 at 19:05

Slava

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The rvalueness can be worked around using `std::addressof()`. The question remains valid. – Ben Voigt Dec 21 '15 at 19:09
1

Why copy? Just `int&& ref = foo.getValue() => &ref`. – edmz Dec 21 '15 at 19:27
@black probably const reference then, but it is just matter of taste, I doubt there would be any difference in generated code. – Slava Dec 21 '15 at 19:34
@BenVoigt Can it? When I use code similar to `std::addressof(foo.getValue())` from this example, I get an `expression must be lvalue` error. Or is that not the usage of `addressof()` you had in mind? – spaaarky21 Mar 05 '18 at 07:40
@spaaarky21: It was possible in C++14, but a new deleted overload was added in C++17. – Ben Voigt Mar 05 '18 at 15:52

Address of return value of method

2 Answers2