Date: Thu, 10 Jun 2021 15:19:27 +0300

So do I understand correctly that if there are the following overloaded functions

void f( const int( & )[] ) // #1

{

std::cout << "void f( int & )[] )\n";

}

void f( const int( & )[2] ) // #2

{

std::cout << "void f( int & )[2] )\n";

}

then

f( { 1, 2 } ); // calls #2

f( { 1 } ); // calls #1

With best regards,

Vlad from Moscow

You can meet me at http://cpp.forum24.ru/ or www.stackoverflow.com or http://ru.stackoverflow.com

>Четверг, 10 июня 2021, 14:41 +03:00 от Edward Catmur via Std-Discussion <std-discussion_at_[hidden]>:

>

>On Thu, 10 Jun 2021 at 12:18, Vladimir Grigoriev via Std-Discussion < std-discussion_at_[hidden] > wrote:

>>In the paragraph # 3 of the section 12.4.3.2 Ranking implicit conversion sequences of the C++ 20 Standard there is written

>>

>>(3.1) — List-initialization sequence L1 is a better conversion sequence than list-initialization sequence L2 if

>>

>>and

>>

>>(3.1.2) — L1 and L2 convert to arrays of the same element type, and either the number of elements n1 initialized by L1 is less than the number of elements n2 initialized by L2, or n1 = n2 and L2 converts to an array of unknown bound and L1 does not,

>>

>>This description is entirely unclear. For starters what is «the number of initialized elements»? Is it the size of the initialized array or is it the number of initializers in the initializer ;list?

>

>It is the size of the initialized array, which is the extent of the array if the array is an array of known bound, and the number of initializers in the initializer list if the array is an array of unknown bound.

>

>>Secondly, how can be n1 == n2 if the initialized arrays of unknown bounds?

>

>If both arrays are arrays of unknown bound, then both n1 and n2 will be the number of initializers in the initializer list, and thus n1 == n2 will hold.

>

>If one array is an array of unknown bound and the other is an array of known bound, then, supposing WLOG L1 converts to an array of size m1 and L2 converts to an array of unknown bound, n1 will equal m1 and n2 will equal the number of initializers in the initializer list, so n1 == n2 will hold precisely when the number of initializers in the initializer list is m1.

>

>>

>>Thirdly, from the first statement (3.1) it follows that L1 is better than L2 when «L2 converts to an array of unknown bound and L1 does not» However in this example in the C++ Standard

>>

>>void f(int (&&)[] ); // #1

>>void f(double (&&)[] ); // #2

>>void f(int (&&)[2]); // #3

>>f( {1} ); // Calls #1: Better than #2 due to conversion, better than #3 due to bounds

>>

>>the function #1 is selected instead of the function #3. Is it a typo?

>

>In that example, denoting n1 n2 n3 as the number of elements initialized in the conversion for #1 #2 #3 respectively, n1 == 1 whereas n3 == 2. So the first conjunct of the second disjunct of the second conjunct of [over.ics.rank]/3.1.2 does not hold and #3 is not better than #1.

>

>There is implementation variance; Clang erroneously selects #3.

>

>>With best regards,

>>Vlad from Moscow

>>

>>

>>

>>You can meet me at http://cpp.forum24.ru/ or www.stackoverflow.com or http://ru.stackoverflow.com --

>>Std-Discussion mailing list

>>Std-Discussion_at_[hidden]

>>https://lists.isocpp.org/mailman/listinfo.cgi/std-discussion

>--

>Std-Discussion mailing list

>Std-Discussion_at_[hidden]

>https://lists.isocpp.org/mailman/listinfo.cgi/std-discussion

void f( const int( & )[] ) // #1

{

std::cout << "void f( int & )[] )\n";

}

void f( const int( & )[2] ) // #2

{

std::cout << "void f( int & )[2] )\n";

}

then

f( { 1, 2 } ); // calls #2

f( { 1 } ); // calls #1

With best regards,

Vlad from Moscow

You can meet me at http://cpp.forum24.ru/ or www.stackoverflow.com or http://ru.stackoverflow.com

>Четверг, 10 июня 2021, 14:41 +03:00 от Edward Catmur via Std-Discussion <std-discussion_at_[hidden]>:

>

>On Thu, 10 Jun 2021 at 12:18, Vladimir Grigoriev via Std-Discussion < std-discussion_at_[hidden] > wrote:

>>In the paragraph # 3 of the section 12.4.3.2 Ranking implicit conversion sequences of the C++ 20 Standard there is written

>>

>>(3.1) — List-initialization sequence L1 is a better conversion sequence than list-initialization sequence L2 if

>>

>>and

>>

>>(3.1.2) — L1 and L2 convert to arrays of the same element type, and either the number of elements n1 initialized by L1 is less than the number of elements n2 initialized by L2, or n1 = n2 and L2 converts to an array of unknown bound and L1 does not,

>>

>>This description is entirely unclear. For starters what is «the number of initialized elements»? Is it the size of the initialized array or is it the number of initializers in the initializer ;list?

>

>It is the size of the initialized array, which is the extent of the array if the array is an array of known bound, and the number of initializers in the initializer list if the array is an array of unknown bound.

>

>>Secondly, how can be n1 == n2 if the initialized arrays of unknown bounds?

>

>If both arrays are arrays of unknown bound, then both n1 and n2 will be the number of initializers in the initializer list, and thus n1 == n2 will hold.

>

>If one array is an array of unknown bound and the other is an array of known bound, then, supposing WLOG L1 converts to an array of size m1 and L2 converts to an array of unknown bound, n1 will equal m1 and n2 will equal the number of initializers in the initializer list, so n1 == n2 will hold precisely when the number of initializers in the initializer list is m1.

>

>>

>>Thirdly, from the first statement (3.1) it follows that L1 is better than L2 when «L2 converts to an array of unknown bound and L1 does not» However in this example in the C++ Standard

>>

>>void f(int (&&)[] ); // #1

>>void f(double (&&)[] ); // #2

>>void f(int (&&)[2]); // #3

>>f( {1} ); // Calls #1: Better than #2 due to conversion, better than #3 due to bounds

>>

>>the function #1 is selected instead of the function #3. Is it a typo?

>

>In that example, denoting n1 n2 n3 as the number of elements initialized in the conversion for #1 #2 #3 respectively, n1 == 1 whereas n3 == 2. So the first conjunct of the second disjunct of the second conjunct of [over.ics.rank]/3.1.2 does not hold and #3 is not better than #1.

>

>There is implementation variance; Clang erroneously selects #3.

>

>>With best regards,

>>Vlad from Moscow

>>

>>

>>

>>You can meet me at http://cpp.forum24.ru/ or www.stackoverflow.com or http://ru.stackoverflow.com --

>>Std-Discussion mailing list

>>Std-Discussion_at_[hidden]

>>https://lists.isocpp.org/mailman/listinfo.cgi/std-discussion

>--

>Std-Discussion mailing list

>Std-Discussion_at_[hidden]

>https://lists.isocpp.org/mailman/listinfo.cgi/std-discussion

Received on 2021-06-10 07:19:35