Clarify that std::launder is not needed when using the result of std::memcpy
Jan Schultke

Created on 2024-04-05.00:00:00 last changed 2 months ago


Date: 2024-04-07.17:52:38

Proposed resolution:

This wording is relative to N4971.

  1. Modify [cstring.syn] as indicated:

    -3- The functions memcpy and memmove are signal-safe ([support.signal]). Both functions implicitly create objects ([intro.object]) in the destination region of storage immediately prior to copying the sequence of characters to the destination. Both functions return a pointer to a suitable created object.

Date: 2024-04-11.17:55:57
int x = 0;
alignas(int) std::byte y[sizeof(int)];
int z = *static_cast<int*>(std::memcpy(y, &x, sizeof(int)));

This example should be well-defined, even without the use of std::launder. std::memcpy implicitly creates an int inside y, and https://www.iso-9899.info/n3047.html# states that

The memcpy function returns the value of [the destination operand].

In conjunction with [cstring.syn] p3, this presumably means that std::memcpy returns a pointer to the (first) implicitly-created object, and no use of std::launder is necessary.

The wording should be clarified to clearly support this interpretation or reject it.

Date User Action Args
2024-04-05 12:16:18adminsetmessages: + msg14053
2024-04-05 00:00:00admincreate