std :: vector< type>的类型要求
我仍然困惑在C ++ 11中使用 std :: vector
的类型的要求,但这可能是由错误的编译器(gcc 4.7.0)。此代码:
I am still confused about the requirements for a type to be used with a std::vector
in C++11, but this may be caused by a buggy compiler (gcc 4.7.0). This code:
struct A {
A() : X(0) { std::cerr<<" A::A(); this="<<this<<'\n'; }
int X;
};
int main()
{
std::vector<A> a;
a.resize(4);
}
工作正常,并产生预期的输出,表示默认的ctor )被调用(而不是隐式复制ctor)。但是,如果我向类添加一个删除的副本ctor,
works fine and produces the expected output, indicating that the default ctor (explicitly given) is called (and not an implicit copy ctor). However, if I add a deleted copy ctor to the class, viz
struct A {
A() : X(0) { std::cerr<<" A::A(); this="<<this<<'\n'; }
A(A const&) = delete;
int X;
};
gcc 4.7.0不编译,但尝试使用删除的ctor。这是正确的行为还是一个错误?如果是前者,如何让代码工作?
gcc 4.7.0 does not compile, but tries to use the deleted ctor. Is that correct behaviour or a bug? If the former, how to get the code working?
C ++ 11标准确实需要 CopyInsertable
,因为其他人已经指出。但是这是C ++ 11标准中的一个错误。此后在 N3376 中已更正 MoveInsertable
和 DefaultInsertable
。
The C++11 standard does indeed require CopyInsertable
as others have pointed out. However this is a bug in the C++11 standard. This has since been corrected in N3376 to MoveInsertable
and DefaultInsertable
.
向量< T,A> :: resize(n)
成员函数需要 MoveInsertable
和 DefaultInsertable
。当分配器 A
时,这些大致转换为 DefaultConstructible
和 MoveConstructible
使用默认的构造
定义。
The vector<T, A>::resize(n)
member function requires MoveInsertable
and DefaultInsertable
. These roughly translate to DefaultConstructible
and MoveConstructible
when the allocator A
uses the default construct
definitions.
以下程序使用clang / libc ++编译:
The following program compiles using clang/libc++:
#include <vector>
#include <iostream>
struct A {
A() : X(0) { std::cerr<<" A::A(); this="<<this<<'\n'; }
A(A&&) = default;
int X;
};
int main()
{
std::vector<A> a;
a.resize(4);
}
并为我打印:
A::A(); this=0x7fcd634000e0
A::A(); this=0x7fcd634000e4
A::A(); this=0x7fcd634000e8
A::A(); this=0x7fcd634000ec
$ b $ p
如果您删除上面的move构造函数并将其替换为删除的拷贝构造函数, c $ c> A 不再 MoveInsertable
/ MoveConstructible
使用删除的拷贝构造函数,如OP的问题中正确显示的那样。
If you remove the move constructor above and replace it with a deleted copy constructor, A
is no longer MoveInsertable
/MoveConstructible
as move construction then attempts to use the deleted copy constructor, as correctly demonstrated in the OP's question.