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basic_string
assignment vs. basic_string::assign
Section: 23.4.3.7.3 [string.assign] Status: C++17 Submitter: Marshall Clow Opened: 2016-01-05 Last modified: 2017-07-30
Priority: 0
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Discussion:
In issue 2063(i), we changed the Effects of basic_string::assign(basic_string&&)
to match the
behavior of basic_string::operator=(basic_string&&)
, making them consistent.
basic_string::assign(const basic_string&)
, and its Effects differ from
operator=(const basic_string&)
.
Given the following definition:
typedef std::basic_string<char, std::char_traits<char>, MyAllocator<char>> MyString; MyAllocator<char> alloc1, alloc2; MyString string1("Alloc1", alloc1); MyString string2(alloc2);
the following bits of code are not equivalent:
string2 = string1; // (a) calls operator=(const MyString&) string2.assign(string1); // (b) calls MyString::assign(const MyString&)
What is the allocator for string2
after each of these calls?
If MyAllocator<char>::propagate_on_container_copy_assignment
is true, then it should be alloc2
,
otherwise it should be alloc1
.
alloc2
23.4.3.7.3 [string.assign]/1 says that (b) is equivalent to assign(string1, 0, npos)
, which eventually calls
assign(str.data() + pos, rlen)
. No allocator transfer there.
[2016-02, Issues Telecon]
P0; move to Tentatively Ready.
Proposed resolution:
This wording is relative to N4567.
Modify 23.4.3.7.3 [string.assign] p.1 as indicated:
basic_string& assign(const basic_string& str);-1- Effects: Equivalent to
-2- Returns:*this = str
.assign(str, 0, npos)*this
.