升压变暧昧建设 [英] Boost variant ambiguous construction

问题描述

升压变的文件接受任意类型的构造以下称

 模板＆LT; typename的T＆GT;变种（T＆安培;操作数）;
 

• 要求： T必须明确地转换为有界类型之一（即T1，T2等）

同样是构造函数接受真正的常量T＆安培; 和 T＆放大器;＆安培; 。因此，我希望下面的code将不能编译：

 的boost ::变体LT;的std ::字符串，布尔＆GT; V =文本;
 

不过，code编译和 v 成为一个布尔值，这是我绝对不想。当然，解决办法是换行字符串在的std ::字符串构造函数。我的问题是：

1. 为什么这个code编译？


2. 它是如何选择的类型（如为const char * 可转换为两个的std ::字符串和布尔）？

解决方案

一般情况下，用户定义的转换失去重载解析过程中的标准转换。

有从为const char 指针布尔内置的转换是preferred在无内置转换从为const char * 到的std ::字符串（如见的隐式转换的）

。

的std ::字符串，而标准库的一部分，不是一个内置的类型，因此它的转换构造函数只考虑了转换后的内置类型。

一些参考文献：

• 为什么编译器选择通过字符串L＆QUOT的隐含类型转换布尔;＆QUOT;？


• 为什么我重载C ++的构造函数不叫？

The Boost Variant documentation says the following of the constructor that accepts arbitrary type:

template<typename T> variant(T & operand);

• Requires: T must be unambiguously convertible to one of the bounded types (i.e., T1, T2, etc.).

The same is true of the constructors accepting const T& and T&&. So I expect that the following code won't compile:

boost::variant<std::string, bool> v = "text";

But the code compile, and v becomes a bool, which is something I definitely did not want. Of course the solution is to wrap the string literal in a std::string constructor. My question is:

1. Why does this code compile?
2. How does it choose the type (as const char* is convertible to both std::string and bool)?

解决方案

Generally, user-defined conversions lose overload resolution process to standard conversions.

There is a built-in conversion from const char pointers to bool which is preferred over the non-built-in conversion from const char * to std::string (e.g. see Implicit conversions).

std::string, while part of the standard library, is not a built-in type so its conversion constructors are considered only after conversions to built-in types.

Some references:

• Why does the compiler choose bool over string for implicit typecast of L""?
• Why is my overloaded C++ constructor not called?

这篇关于升压变暧昧建设的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！