The initializing constructor looks like an assignment, but it isn’t

Raymond

Some time ago, I warned about the perils of the accidental C++ conversion constructor: A single-parameter constructor is considered by default to be a conversion constructor; you can opt out of this by marking the constructor explicit.

I gave as an example this class:

class Buffer
{
public:
  Buffer(size_t capacity);
  Buffer(std::initializer_list<int> values);
};

The size_t constructor is not marked as explicit, so it is a conversion constructor. And that permits weird things like this:

Buffer b = 1; // um...

What exactly is happening here?

A common misconception is that what’s happening is that a temporary Buffer is created (with the capacity 1), and then that temporary buffer is assigned to the destination buffer b.

That’s not what’s happening. You can prove this by deleting the assignment operators.

class Buffer
{
public:
  Buffer(size_t capacity);
  Buffer(std::initializer_list<int> values);
  Buffer& operator=(Buffer const&) = delete;
  Buffer& operator=(Buffer&&) = delete;
};

Buffer b = 1; // still compiles

(Deleting the move assignment operator is redundant because declaring the copy assignment operator automatically suppresses the implicit move assignment operator. But I deleted it explicitly for emphasis.)

Even though there is an equal-sign in the statement, there is no actual assignment.

There can’t be an assignment, if you think about it, because the assignment operator assumes that this refers to an already-constructed object. But we don’t have a constructed object yet.

According to the language rules,

Buffer b = 1;

is a copy-initialization, and the copy initialization is performed by taking the thing on the right-hand side and, if the types don’t match,¹ it looks for a conversion constructor.

The equals sign doesn’t mean assignment here. It’s just a quirk of the syntax.

¹ If the types do match, then “the initializer expression is used to initialize the destination object.” At this point copy elision kicks in:

extern Buffer get_buffer();

Buffer b = get_buffer();

The Buffer returned by get_buffer() is placed directly into the memory occupied by b.

Copy elision also means that

Buffer b = Buffer(1);

does not create a temporary Buffer of capacity 1, and then construct b from that temporary buffer. Instead, the Buffer of capacity 1 is constructed directly into b. The result is the same as Buffer b(1);.

Since the copy elision rule can be repeated,

Buffer b = Buffer(Buffer(Buffer(1)));

is also the same as Buffer b(1);, because each repetition of the rule strips away one of the calls to Buffer(...).

2 comments

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  • Martin Ba

    “… copy elision rule …” – is copy elison now required by The Language in this case, or is this still at the discretion of the compiler?

    • 紅樓鍮

      Copy elision in C++17 and later is an intuitive way to say “initialization by prvalue”. You can read tartanllama’s article “Guaranteed Copy Elision Does Not Elide Copies” if you want to dive deeper.

      If you’re not striving to be a language lawyer, then yes, copy elision from C++17 onwards is mandatory regardless of optimization.