isSet（）或operator void *（）或显式操作符bool（）或别的？ [英] isSet() or operator void*() or explicit opertor bool() or something else?

问题描述

什么是最先进的功能，检查值是否已设置？

例如，下面的迭代器解析单元格。
一些单元格包含值，其他单元格为空。

For example, the below iterator parses cells. Some cells contain a value, other cells are empty.

最方便的方法是什么？
$ b

What is the most convenient way?

struct iterator 
{                                  //usage:
  bool isset() const               // if (it.isset()) 
  bool isSet() const               // if (it.isSet()) 
  bool empty() const               // if (it.empty()) 

  bool is_set()   const            // if (it.is_set()) 
  bool is_valid() const            // if (it.is_valid()) 

  operator void*() const;          // if (it) 

  explicit operator bool() const;  // if ((bool)it) or if(it) //thanks @stijn
  operator          bool() const;  // if (it) //why not implicit conversion?

  bool operator!() const;          // if (!!it)

  //throwing exception as pointed out by @MatthieuM
  Type get() { if (isSet()) return value_; else throw; }
  //usage:
  //     try {                    // if (it.isSet()) {
  //        Type x = it.get();    //    Type x = it.get();
  //     }                        // }
  //     catch (...) {            // else {
  //        //empty               //    //empty
  //     }                        // }

  //please feel free to propose something different
  ...
};

反映：

1. 我的老板不明白 isset（） =>已重命名为 isSet（）


2. empty（）更多关于容器收集，而不只是一个单一的单元格：（


3. operator void * 似乎是合乎逻辑的方式，但已弃用在C ++ 11流


4. 显式运算符是尚未支持（我的代码必须符合旧的编译器）

1. my boss does not understand isset() => renamed to isSet()
2. empty() is more about container collection, not just one single cell :(
3. operator void* seems to be the logical way but deprecated in C++11 streams
4. explicit operator is not yet supported (my code have to be compliant with old compilers)

$ b b

我正在阅读：

I am reading:

• http://en.wikibooks.org/wiki/More_C%2B%2B_Idioms/Safe_bool


• （ http://www.artima.com/cppsource/safebool.html ）


• C ++ 11中的安全bool成语已过时？


• 安全bool习语和显式运算符bool之间的不兼容性


• 安全bool成语在boost？


• 提升源代码...

• http://en.wikibooks.org/wiki/More_C%2B%2B_Idioms/Safe_bool
• ( http://www.artima.com/cppsource/safebool.html )
• Is the safe-bool idiom obsolete in C++11?
• Incompatibilities between safe bool idiom and explicit operator bool
• Safe bool idiom in boost?
• boost source code...

推荐答案

我印象深刻的是 explicit_cast< T> =http://rads.stackoverflow.com/amzn/click/0321228774 =nofollow>不完全的C ++：实用解决方案[...] 尚未提及。这个概念很简单 - 你设置一个伪关键字，它实际上是一个模板类，实现你想要的转换。我一直在使用此自己的C ++后端库，没有任何重要问题。 p>

I'm impressed that explicit_cast<T> from Imperfect C++: Practical Solutions [...] hasn't been mentioned. The concept is very simple - you set up a pseudo-keyword that actually is a templated class implementing the conversion that you want. I've been using this in my own C++ backports library without any important issue.

class MyClass {
  ...implementation
  operator explicit_cast<bool> () const { 
      (return a bool somehow)
  }
};

您使用伪关键字，就像您预期的那样：

You use the pseudo-keyword just as you would expect it to work:

MyClass value;
...
if ( explicit_cast<bool>(myobject) )  {
  do something
} else {
  do something else
}
...

最好的部分是，这个解决方案可以完美映射到原生显式运算符在C ++ 11中，导致基本上零开销和本地语法。因此，它也更通用，比试图找出是否调用isset，is_set，is_Set，isSet等等...

The best part of all, this solution can be mapped perfectly to native explicit operator in C++11, resulting in essentially zero overhead and native syntax. As a result, it's also more generic than trying to figure out if to call "isset", "is_set", "is_Set", "isSet", etc...

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