C ++自动与自动& [英] C++ auto vs auto&

问题描述

如果我有一个功能:

Foo& Bar()
{
   return /// do something to create a non-temp Foo here and return a reference to it
}

这是为什么:

auto x = Bar(); /// probably calls copy ctor - haven't checked

与此不同吗?

auto &x = Bar(); /// actually get a reference here

(实际上，我希望第二个版本获得对引用的引用，这毫无意义.)

如果我明确地将x的类型指定为值或引用，那么我将得到期望的结果(当然).不过，我希望auto可以编译为Bar()的返回类型，在这种情况下，该返回类型是引用.

If I explicitly specified the type of x as a value or a reference, I'll get what I expect (of course). I would expect, though, that auto would compile to the return type of Bar(), which, in this case, is a reference.

在Foo和Foo&之间是否存在隐式强制转换?

Is there an implicit cast between Foo and Foo& that comes into play here?

(接受了规范参考，尽管我已经厌倦了阅读委员会演讲.)

(Spec references accepted, though I'm getting tired of reading committee-speak.)

(时间机器的第二次使用将默认使C ++通过引用传递.使用#pragma compatibility触发器来编译C代码.ARGH.)

(Second use of time machine will be making C++ pass by reference by default. With a #pragma compatibility trigger for compiling C code. ARGH.)

推荐答案

auto的类型推导与模板完全相同:

The type deduction for auto works exactly the same as for templates:

• 推论auto时，您将得到一个值类型.
• 推论auto&时，您将获得非常量引用类型
• 当您推断出const auto&时，您将获得一个const引用
• 推论auto&&时，您将得到
  • 如果您分配非常量引用，则为非常量引用
  • 如果您分配了const引用，则为const引用
  • 分配临时值时的值
  • when you deduce auto you will get a value type.
  • when you deduce auto& you wil get a non-const reference type
  • when you deduce const auto& you will get a const reference
  • when you deduce auto&& you will get
    • a non-const reference if you assign a non-const reference
    • a const reference if you assign a const reference
    • a value when you assign a temporary

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