将auto关键字替换为推导类型（clang或VS2010） [英] Replace auto keyword with deduced type (clang or VS2010)

问题描述

有没有人编写一个脚本，插件或可执行文件，用编译器推导的类型替换每个'auto'实例？我需要端口一些C ++ 11代码，使用自动所有的地方。

Clang是我的第一个候选人。有没有人修改它做这样的事情？

另一种方法是从编译器解析错误，因为预期类型可能在错误输出中。我可以 -Dauto = int 并且可能回到无法将std :: vector< int> :: iterator转换为'int' code>

解决方案

不幸的是，这在一般情况下是不可能的。考虑：

 模板< typename T> void foo（T& t）
 {
 auto it = t.find（42）; 
 ... 
} 
 ... 
 std :: map< int，int> m; 
 std :: set< int> s; 
 ... 
 foo（m）; 
 foo（s）; 
  

当然是一个无意义的例子，但它表明没有办法知道什么替换auto依赖于模板参数。 std :: map 和 std :: set ，附带地包含相同名称的typedef > iterator ），因此 typename T :: iterator it 将在这里工作，但你可以实例化 foo

在标准库类中添加了很多typedef，以允许在 auto 被发明/重新使用之前编写这样的模板，你可以做同样的事情来处理一个没有 auto 的编译器。但它不是你可以自动化，至少不是没有努力相当于添加支持 auto 到编译器...

即使 auto 不依赖于模板类型，将其替换为对用户有意义并且可移植的一个难题。取：

  std :: map< int，int& m; 
 auto it = m.find（42）; 
  

auto 的合理替换为 std :: map< int，int> :: iterator ，但如果使用 -Dauto = int 错误消息，你可以用 std :: _ Rb_tree_iterator< std :: pair< const int，int> > 。这是标准库的实现细节，难以阅读，显然不可移植 - 您不想在代码中 。

我的编译器（GCC 4.4.6）说：

错误：无法转换 __ gnu_cxx :: __ normal_iterator ; int *，std :: vector





Has anyone written a script, plugin, or executable that replaces each instance of 'auto' with the compiler-deduced type? I need to port some C++11 code that uses auto all over the place.

Clang is my first candidate. Has anyone modified it to do something like this?

An alternative is to parse the errors from a compiler as the expected type might be in the error output. I could -Dauto=int and possibly get back "could not convert std::vector<int>::iterator to 'int'"

解决方案

Unfortunately, this is impossible in the general case. Consider:

template <typename T> void foo(T & t)
{
    auto it = t.find(42);
    ...
}
...
std::map<int, int> m;
std::set<int> s;
...
foo(m);
foo(s);

Admittedly a pointless example, but it shows that there's no way to know what to replace auto with, when dependent on a template argument. std::map and std::set, incidentally, contain typedefs of the same name (iterator) that represent the type of the respective iterator, so typename T::iterator it would work here, but you can instantiate foo for a T that does not have such a typedef.

The numerous typedefs in the standard library classes were added exactly to allow such templates to be written before auto was invented/re-purposed, and you can do the same thing to deal with a compiler that doesn't have auto. But it's not something you can automate, at least not without an effort comparable to adding support for auto to a compiler...

Even when auto is not dependent on a template type, it is a difficult problem to replace it with something that makes sense to the user and is portable. Take:

std::map<int, int> m;
auto it = m.find(42);

The reasonable replacement for auto is std::map<int, int>::iterator, but if you use -Dauto=int and look at the compiler error messages, you'd replace it with something like std::_Rb_tree_iterator<std::pair<const int, int> >. That's implementation detail of the standard library, hard to read and obviously not portable -- you don't want that in your code.

In your very example, my compiler (GCC 4.4.6) says:

error: cannot convert __gnu_cxx::__normal_iterator<int*, std::vector<int, std::allocator<int> > > to int in initialization

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