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Conversion operator to pointer-to-array

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  Conversion operator to pointer-to-array Juha Nieminen <nospam@thanks.invalid> - 2019-09-18 07:37 +0000
    Re: Conversion operator to pointer-to-array Tim Rentsch <tr.17687@z991.linuxsc.com> - 2019-09-18 08:58 -0700
    Re: Conversion operator to pointer-to-array Öö Tiib <ootiib@hot.ee> - 2019-09-18 09:32 -0700
    Re: Conversion operator to pointer-to-array Frederick Gotham <cauldwell.thomas@gmail.com> - 2019-09-26 00:12 -0700
      Re: Conversion operator to pointer-to-array Juha Nieminen <nospam@thanks.invalid> - 2019-09-26 08:39 +0000
      Re: Conversion operator to pointer-to-array Manfred <noname@add.invalid> - 2019-09-26 16:34 +0200

#68489 — Conversion operator to pointer-to-array

Is there a way to express this without using the type alias?
This is one instance where my C++-fu isn't strong enough.

struct S
{
    using DArray4 = double(*)[4];
    operator DArray4() { return nullptr; }
};

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#68508

Juha Nieminen <nospam@thanks.invalid> writes:

> Is there a way to express this without using the type alias?
>
> struct S
> {
>     using DArray4 = double(*)[4];
>     operator DArray4() { return nullptr; }
> };

There is.

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#68509

On Wednesday, 18 September 2019 10:37:23 UTC+3, Juha Nieminen  wrote:
> Is there a way to express this without using the type alias?
> This is one instance where my C++-fu isn't strong enough.
> 
> struct S
> {
>     using DArray4 = double(*)[4];
>     operator DArray4() { return nullptr; }
> };

You may not use () or [] in type identifier of conversion operator.
It is still possible without type alias but then the solutions
that I can produce look far worse than what you had. Example:

    #include <type_traits>

    struct S
    {
        operator std::common_type_t<double(*)[4]>() { return nullptr; }
    }; 

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#68687

On Wednesday, September 18, 2019 at 8:37:23 AM UTC+1, Juha Nieminen wrote:
> Is there a way to express this without using the type alias?
> This is one instance where my C++-fu isn't strong enough.
> 
> struct S
> {
>     using DArray4 = double(*)[4];
>     operator DArray4() { return nullptr; }
> };


I'm still stuck in C++03, I haven't learned the C++11 stuff yet...

But I must ask...

Why the use of "using" above when we can simply do. . .

typedef double (*DArray4)[4];

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#68688

Frederick Gotham <cauldwell.thomas@gmail.com> wrote:
> On Wednesday, September 18, 2019 at 8:37:23 AM UTC+1, Juha Nieminen wrote:
>> Is there a way to express this without using the type alias?
>> This is one instance where my C++-fu isn't strong enough.
>> 
>> struct S
>> {
>>     using DArray4 = double(*)[4];
>>     operator DArray4() { return nullptr; }
>> };
> 
> 
> I'm still stuck in C++03, I haven't learned the C++11 stuff yet...
> 
> But I must ask...
> 
> Why the use of "using" above when we can simply do. . .
> 
> typedef double (*DArray4)[4];

I think you answered your own question...

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#68699

On 9/26/2019 9:12 AM, Frederick Gotham wrote:
> On Wednesday, September 18, 2019 at 8:37:23 AM UTC+1, Juha Nieminen wrote:
>> Is there a way to express this without using the type alias?
>> This is one instance where my C++-fu isn't strong enough.
>>
>> struct S
>> {
>>      using DArray4 = double(*)[4];
>>      operator DArray4() { return nullptr; }
>> };
> 
> 
> I'm still stuck in C++03, I haven't learned the C++11 stuff yet...
> 
> But I must ask...
> 
> Why the use of "using" above when we can simply do. . .
> 
> typedef double (*DArray4)[4];
> 

In C++11 the 'using' construct can do more than this, however in this 
case 'using' and typedef' are equivalent.

'Simply' is a matter of taste. In the context of C++ and later, the 
'using' form is a more generic type alias declaration, which includes 
template aliases.

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