c ++使用std :: enable_if有条件地将吸气剂添加到可变参数的变体模板中 [英] c++ Use std::enable_if to conditionally add getters to a variadic variant template
问题描述
在我的变体具有 int
, float
, bool
等中的任何一个作为模板参数的情况下,我尝试添加特殊化
I am trying to add specializations for the case where my variant has any of int
, float
, bool
, and others as template arguments.
到目前为止,我的尝试是:
#include <iostream>
#include <variant>
#include <string>
#include <type_traits>
template<typename... Types>
struct variant : std::variant<Types...> {
using std::variant<Types...>::variant;
template<typename T>
const T& get() const { return std::get<T>(*this); }
#define VARGET(X) typename std::enable_if<(std::is_same<Types, X>::value || ... ), X>::type get_##X() const { return get<X>(); }
VARGET(int)
VARGET(float)
VARGET(double)
VARGET(bool)
using std_string = std::string;
VARGET(std_string)
};
int main()
{
variant<int, float, bool> var = 3;
std::cout << var.get_int() << std::endl;
variant<int, float, bool> var2 = 0.1f;
std::cout << var2.get_float() << std::endl;
return 0;
}
但是在gcc 9上会出现错误:
But on gcc 9 this gives the error:
error: failed requirement 'std::is_same<int, double>::value || std::is_same<float, double>::value || std::is_same<bool, double>::value'; 'enable_if' cannot be used to disable this declaration
这为什么让我感到困惑:
根据我对SFINAE的理解,当参数之一是指定的类型之一时,宏模板的计算结果为:(以int为例)
From my understanding of SFINAE, when one of the arguments is one of the specified types, the macro template evaluates to: (For the example of int)
std :: enable_if<(std :: is_same< Types,int> :: value || ...),int> :: type
的计算结果为 int
因此表达式变为 int get_int()const {std :: get< int>(* this)}
如果它不是指定的类型之一:
And if it is not one of the specified types:
std :: enable_if<(std :: is_same< Types,size_t> :: value || ...),size_t> :: type
的计算结果为空
因此表达式变为 get_size_t()const {std :: get< size_t>(* this)}
这是无效的语法,因为该函数没有返回类型,但是由于SFINAE的缘故,由于X的其他替换确实会产生有效的语法,因此该函数仍应编译.
This is invalid syntax because the function does not have a return type, but because of SFINAE this should still compile because of other substitutions of X do produce valid syntax.
我的代码有什么问题,有没有办法得到我想要的结果?
What is wrong with my code, and is there a way to get the result I desire?
谢谢.
推荐答案
我在尝试编译代码时收到的错误与您的错误不同,请参见在...中没有名为type的类型",因为SFINAE在这种情况下不适用.
The error I am receiving when trying to compile your code differs from yours, see on Compiler Explorer; I receive the error "no type named type in ...
" which I would have expected, as SFINAE does not apply in this case.
SFINAE将应用于函数调用中.函数中没有任何东西可以替代,代码总是错误的/总是正确的,具体取决于结构的模板参数.
SFINAE would apply for instance in a function call if there is a substitution which whould cause a failure. Nothing were to be substituted in the function, the code is always wrong / always correct depending on the struct's template arguments.
在创建结构时,它要么总是格式错误,要么总是格式正确.您可以通过在函数中进行人工替换来解决此问题:
When creating the struct, it is either always malformed, or always well-formed. You might be able to work around this by imposing an artificial substitution in the function:
template <bool f=false>
std::enable_if_t<(std::is_same<Types, int>::value || ... ||f ), int> get_int() const { return get<int>(); }
然后它应该可以正常编译.
Then it should compile fine.
- 函数类型中使用的所有类型(包括返回类型和所有参数的类型)
- 模板参数声明中使用的所有类型
- 函数类型中使用的所有表达式
- 模板参数声明中使用的所有表达式(自C ++ 11起)
- 显式说明符中使用的所有表达式(自C ++ 20起)
事实上,代码被最佳压缩的原因是: std :: enable_if_t< false,int>func(){}
将始终无法编译.
The reason why the code is condensed best in the fact, that
std::enable_if_t<false, int> func() {}
will always fail to compile.
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