Are objects bound to local references automatically destructed?

Question

Consider the following code.

struct MyImage
{
    MyImage(const int handle);
    MyImage(const CString& filePath);
    virtual ~MyImage();

    void Process();
    void SaveAs(const CString& filePath);

    // No copying!
    MyImage(const MyImage& other) = delete;
    MyImage& operator=(const MyImage& other) = delete;
}

void ProcessImageFile(const CString& inFilePath, const CString& outFilePath)
{
    MyImage& image = MyImage(-1); // initialized with invalid handle

    if (DecryptionRequired())
    {
        const CString tempFilePath = ::GetTempFileName();
        Decrypt(inFilePath, tempFilePath);
        image = MyImage(tempFilePath);
        _tremove(tempFilePath);
    }
    else
    {
        image = MyImage(inFilePath);
    }

    image.Process();
    image.SaveAs(outFilePath);
}

Will the object referenced by image be destructed when ProcessImageFile() returns?

Note that this code is invalid in standard C++, so this is a Visual-C++ specific question. — Brian Bi, Oct 08 '15 at 19:12
@Brian yes, I was supspecting that and in the meantime Nathan wrote an answer. Looks kinda fishy, but I wasnt sure. — 463035818_is_not_an_ai, Oct 08 '15 at 19:18
The title in the question is wrong. References are not destructed. Objects are destructed. — Christian Hackl, Oct 08 '15 at 19:44

score 6 · Accepted Answer · edited May 23 '17 at 11:44

6

MyImage& image = MyImage(-1); // initialized with invalid handle

Should not compile as you cannot take a non const reference to a temporary variable. If you had

const MyImage& image = MyImage(-1); // initialized with invalid handle

Then the liftime would be extended untill the reference's lifetime ends. Since the reference variable is an automatic object the lifetime will end when it goes out of scope. From [basic.stc.auto]

Block-scope variables explicitly declared register or not explicitly declared static or extern have automatic storage duration. The storage for these entities lasts until the block in which they are created exits.

As for why Visual studio is allowing this see Non-const reference bound to temporary, Visual Studio bug?

edited May 23 '17 at 11:44

Community

1
1

answered Oct 08 '15 at 19:13

NathanOliver

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when would we rather write `const MyImage& image = MyImage(-1); ` instead of `const auto image = MyImage(-1);` ? – Johan Lundberg Oct 08 '15 at 19:21
@JohanLundberg When writing C++03 or below compatible code. – dascandy Oct 08 '15 at 19:23
right, but what I meant was; why use a reference to hold a return value? – Johan Lundberg Oct 08 '15 at 19:27
@JohanLundberg this is a constructor call not a function call. – NathanOliver Oct 08 '15 at 19:30
IDK why the OP is doing it this way instead of `MyImage image(-1);` – NathanOliver Oct 08 '15 at 19:31
I have since refactored the code to avoid using references, but I was still curious enough to ask the question. Thank you for this informative answer. – piedar Oct 08 '15 at 19:40
@JohanLundberg: If you're using a factory that returns some subclass of the specified type... maybe one where the true type depends in some complex way on template parameters. – Ben Voigt Oct 08 '15 at 19:50
@BenVoigt; thank you. right, such as `D f(){ return D(); } ; const B& b = f(); ` where B is a base class of D. And its template generalizations. – Johan Lundberg Oct 09 '15 at 17:08

Are objects bound to local references automatically destructed?

1 Answers1