std :: vector< type>的类型要求 [英] type requirements for std::vector<type>

问题描述

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?

推荐答案

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.

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