检测 C++ 中是否存在类型 [英] Detect if a type exists in C++

问题描述

我需要一个可以这样调用的模板:

I'd need a template which can be called like this:

int x = type_exists< std::vector<int> >::value;

如果 #include <vector> 在源代码中较早出现(显式或可传递)，则应将 x 设置为 1，否则应将 x 设置为 0.

This should set x to 1 if #include <vector> was present (either explicitly or transitively) earlier in the source, otherwise it should set x to 0.

是否可以用 C++ 来实现?我正在使用 GCC，所以 GCC 扩展也很好.

Is it possible to do it in C++? I'm using GCC, so GCC extensions are also fine.

稍微改变调用语法也可以.

It's also OK to change the call syntax a bit.

运行 C++ 编译器两次是不行的:首先只是为了确定我们是否遇到编译错误.

It's not OK to run the C++ compiler twice: first just to figure out if we get a compile error.

推荐答案

这不是您要找的，但它尽可能接近 type_exists 特性:

This is not what you are looking for, but it's as close as you can get to a type_exists trait:

template<class T> struct Void { typedef void type; };

template<class T, class U = void>
struct type_exists { enum { value = 0 }; };

template<class T>
struct type_exists<T, typename Void<T>::type> { enum { value = 1 }; };

显然，它有效:

static_assert(type_exists<int>::value, "int is not defined");
static_assert(type_exists<SomeNonexistingType>::value, "expected compile-time error");

这正是它应该做的.使用 GCC 5.4.0 测试.

This does exactly what it is supposed to do. Tested with GCC 5.4.0.

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