std::vector<type> 的类型要求 [英] type requirements for std::vector&lt;type&gt;

问题描述

我仍然对在 C++11 中与 std::vector 一起使用的类型的要求感到困惑，但这可能是由错误的编译器 (gcc 4.7.0) 引起的.这段代码:

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<<'
'; }
  int X;
};

int main()
{
  std::vector<A> a;
  a.resize(4);
}

工作正常并产生预期的输出，表明调用了默认的构造函数(显式给出)(而不是隐式复制构造函数).但是，如果我向类中添加一个已删除的复制构造函数，即

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<<'
'; }
  A(A const&) = delete;
  int X;
};

gcc 4.7.0 无法编译，但会尝试使用已删除的 ctor.这是正确的行为还是错误?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

正如其他人指出的那样，C++11 标准确实需要 CopyInsertable.然而，这是 C++11 标准中的一个错误.这已在 N3376 到 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.

vector::resize(n) 成员函数需要 MoveInsertable 和 DefaultInsertable.当分配器 A 使用默认的 construct 定义时，这些大致转换为 DefaultConstructible 和 MoveConstructible.

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<<'
'; }
  A(A&&) = default;
  int X;
};

int main()
{
  std::vector<A> a;
  a.resize(4);
}

对我来说打印出来:

 A::A(); this=0x7fcd634000e0
 A::A(); this=0x7fcd634000e4
 A::A(); this=0x7fcd634000e8
 A::A(); this=0x7fcd634000ec

如果删除上面的移动构造函数并用删除的复制构造函数替换它，A 不再是 MoveInsertable/MoveConstructible 作为移动构造然后尝试使用已删除的复制构造函数，如 OP 问题中正确演示的那样.

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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