从指针typedef常量引用 [英] Typedef const reference from a pointer

问题描述

可能重复：

为什么编译：

class Bar {};

int main() {
  Bar i;
  Bar *b = &i;
  typedef const Bar& type;
  type t = type(b);
}

G ++（ 4.5 ，4.7个快照）， Comeau 和MSVC所有人都很高兴但警告未使用的变量。

G++ (4.5, 4.7 snapshots), Comeau and MSVC all are happy but warn about unused variables.

编译器认为它意味着什么？是UB吗？为什么不是一个错误？

What does the compiler think it means? Is it UB? Why isn't it an error?

我认为这应该是一个错误，因为我不小心做了一个 Bar * 进入 const Bar& ，无需取消引用或疯狂投放。

I think it should be an error because I've accidentally made a Bar* into const Bar& without dereferencing or crazy casts. I thought every part of this was completely safe.

推荐答案

C风格的转换会依次尝试不同的C ++类型： p>

A C-style cast tries different C++ cast types in turn:

[C ++ 11：5.4 / 5]：

• a const_cast （5.2.11） li>
  
• a static_cast （5.2.9），


• a static_cast 后跟 const_cast ，


• a reinterpret_cast （5.2.10）或


• a reinterpret_cast 可以使用显式类型转换的转换符号执行const_cast ，

• a const_cast (5.2.11),
• a static_cast (5.2.9),
• a static_cast followed by a const_cast,
• a reinterpret_cast (5.2.10), or
• a reinterpret_cast followed by a `const_cast,

>相同的语义限制和行为适用，除了在以下情况下执行static_cast时，即使基类不可访问，转换也是有效的：

can be performed using the cast notation of explicit type conversion. The same semantic restrictions and behaviors apply, with the exception that in performing a static_cast in the following situations the conversion is valid even if the base class is inaccessible:

• [..]

然后遵循各种复杂规则，

And then follows various complex rules that I can't be bothered to parse in detail.

你得到必要的警告，这是一个愚蠢的施法，但因为它是你要求的是它的尝试。

You get the requisite warnings that it's a stupid cast, but since it's what you asked for it's what's attempted.

比较：

class Bar {};

int main() {
  Bar *b = 0;
  typedef const Bar& type;
  const type t = static_cast<type>(b);
}

// In function 'int main()':
// Line 6: error: invalid static_cast from type 'Bar*' to type 'const Bar&'
// compilation terminated due to -Wfatal-errors.

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