CODEWITHSUNDEEP
c++arrays
Ben Hymers
Ben Hymers

Reputation: 26536

Cast pointer to fixed-size array in return statement

The simplest way to ask this question is with some code:

struct Point
{
    int x;
    int y;
    int z;

    int* as_pointer() { return &x; }        // works
    int (&as_array_ref())[3] { return &x; } // does not work   
};

as_pointer compiles, as_array_ref does not. A cast seems to be in order but I can't figure out the appropriate syntax. Any ideas?

Upvotes: 11

Views: 7428

Answers (3)

Antoine
Antoine

Reputation: 14064

I find that array types are easier to deal with with a typedef:

typedef int ints[3];

Then your as_array_ref must be written so that &as_array_ref() == &x.

The following syntaxes are possible:

  1. plain C-style cast from int* to ints*:

    ints& as_array_ref() { return *( (ints*)(&x) ); }

  2. C++ style reinterpret_cast (suggested by @Mike Seymour - see also his answer). It is often considered a better practice in C++:

    ints& as_array_ref() { return *reinterpret_cast<ints*>(&x); }

  3. Cast from int& to ints& which is slightly shorter but (for me) less intuitive:

    ints& as_array_ref() { return reinterpret_cast<ints&>(x); }

Upvotes: 12

Mike Seymour
Mike Seymour

Reputation: 254451

The cast you need to reinterpret a reference to a variable as a reference to an array is:

reinterpret_cast<int(&)[3]>(x);

Be aware that using this gives undefined behaviour; it will probably work on any sensible implementation, but there's no guarantee that there won't be padding between the members of the class, while arrays are not padded.

Upvotes: 9

Nico Brailovsky
Nico Brailovsky

Reputation: 318

I think that what you're trying to do would be easier (and clearer/cleaner) with an union.

Upvotes: 2

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