CODEWITHSUNDEEP
c++arraystemplatesreferencesize
Jonathan Mee
Jonathan Mee

Reputation: 38919

Using the Array Template Parameter for a Return

(I can't seem to find a name for this paradigm, but you'll get a bonus upvote if you can!) This template paradigm is used to prevent an array from decaying into a pointer:

template <size_t N>
void foo(const char (&bar) [N]);

Live Example

We are able to use N in the body of foo to get the size off bar. Given const char array[] = "lorem ipsum" we use this call foo(array) to define bar as a const char (&) [12], effectively passing 2 parameters for the price of one.

I want to use this same idea but in a return, for example:

template <size_t N>
const char (&) [N] foo(const char (&bar) [N]);

Such that I could pass the array through, and have another reference to it, such as: auto result = foo(bar)

The code I have here yields the error:

unrecognizable template declaration/definition

Is this possible?

EDIT:

For clarification, I want to be able to have a function return that will result in the equivalent of:

auto& result = bar;

Such that result will be a const char (&) [*] where * is the size of bar.

Upvotes: 0

Views: 59

Answers (1)

Kerrek SB
Kerrek SB

Reputation: 477110

C++ syntax is wonderful:

template <std::size_t N>
const char (&f(const char (&arr)[N]))[N]    // take that, Lisp
{
    return arr;
}

For the less masochistic:

template <std::size_t N> using CarrN = const char[N];

template <std::size_t N>
CarrN<N> & f(CarrN<N> & arr)
{
    return arr;
}

As for terminology: We usually say that "arr is passed by reference", though the more discerning user might say "passed as an lvalue" (and also returned as such).

Upvotes: 4

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