Is it possible to pass an “unnamed” variable to a function?

Question

Let's say there is the following function:

void SetTheSize(const SIZE *size) { ... }

Is there any way to call that function without specifying the SIZE variable? eg,

SetTheSize((const SIZE*)&{10, 10});

edit: I should had mentioned that the SIZE is a struct, without SIZE(int, int) constructor.

Answer 1

No, only thing nearest in C++ is to do like this:

void SetTheSize(const SIZE& size); //Change the pointer to const reference

And call it using an unnamed temporary variable: SetTheSize(SIZE(10,10));

Answer 2

SetTheSize(&SIZE{10, 10}); would "work" with the improved initialization syntax in C++0x.

It "only" deserves a warning: taking address of a temporary.

Legal things to do would be to bind the temporary to references:

void SetTheSize(SIZE&&);

or

void SetTheSize(const Size&);

with usage

SetTheSize(Size{10, 10});

Answer 3

You can't do it via a pointer, although you can pass a temporary via a const reference. And you can't create arrays like that (if that's what your syntax is trying to do). So, no.

Answer 4

None of the answers so far have been pedantic, and somebody's got to do it. It's actually really quite easy:

void SetTheSize(const SIZE *size);
SIZE const newSize(10, 10);
SetTheSize(&newSize);

The variable being passed to SetTheSize is indeed an unnamed temporary. It is the result of the expression &newSize and has type SIZE const* .

If you wanted the SIZE object itself to be an unnamed temporary, instead of the pointer passed to SetTheSize , that's possible too:

void SetTheSize(const SIZE *size);
SIZE const& newSize = SIZE(10, 10);
SetTheSize(&newSize);

Now newSize is a reference to an unnamed temporary variable.