c++ - 自动变量初始化和复制/移动构造函数

Question

我有一个片段:

struct MyCass2 {
  MyCass2() {}

  MyCass2(MyCass2 const&) = delete;

  MyCass2(MyCass2&&) = delete;
};

int
main() {
  auto a = MyCass2();
}

这导致

main.cpp:43:8: error: call to deleted constructor of 'MyCass2'
  auto a = MyCass2();
       ^   ~~~~~~~~~
main.cpp:38:3: note: 'MyCass2' has been explicitly marked deleted here
  MyCass2(MyCass2&&) = delete;
  ^
1 error generated.

为什么我认为毕竟会有模板类型推导和直接初始化？有人可以解释一下在这种情况下自动变量初始化是如何工作的吗？

Answer 1

Why I thought there will be a template type deduction

auto 使用模板参数推导规则推导变量的类型。在这种情况下，类型将被推断为 MyCass2。

and a direct initialization after all?

a 不是 direct-initialized ，因为你使用了 copy-initialization - 请参阅标记为 (1) 的语法。

how the automatic variable initialization work in this case?

a 是从 = 右侧的临时变量复制初始化的。但是，由于该类型既不可复制也不可移动，因此不允许进行复制初始化。

But I [defined] the move/copy constructors and neither of them was called, how is that?

默认构造函数用于初始化临时对象。复制初始化中对移动构造函数的调用允许为elided .

I was sure that this auto var=initializer is kind of a syntax overloading. Like with copy initialization T var=initializer, where instead of operator= a copy constructor is called.

嗯，事实并非如此。这里，auto 用于且仅用于推断类型。一旦推断出类型，表达式就完全等同于

MyCass2 a = MyCass2();