如何区分函数参数中的右值和右值引用 [英] How to distiguish between an rvalue and rvalue reference in a function parameter
问题描述
传递参数时,我想在函数参数中区分这两种情况,如下所示:
int rvalue();
int&&& rvalue_ref();
f(rvalue());
f(rvalue_ref());但是,当我尝试转发这样的引用:
$ b $ b int rvalue()
{
return 1;
}
int&& rvalue_ref(int i)
{
return std :: move(i);
}
template< class T>
void f(T& x)
{
if(std :: is_rvalue_reference< T>())
{
std :: cout< ; Rvalue reference<< std :: endl;
}
else if(std :: is_lvalue_reference< T>())
{
std :: cout< Lvalue reference<< std :: endl;
}
else
{
std :: cout< 不是参考<< std :: endl;
}
}
int main()
{
f(rvalue()); // should printNot a reference
f(rvalue_ref(1)); //应打印Rvalue reference
}
它打印出对于这两种情况。有没有办法在C ++中区分两种情况?
解决方案我不知道如何单独使用函数参数。 xvalue和prvalue之间的区别可能会在函数调用中丢失。
但是你可以通过调用 decltype
对参数,在调用函数之前。这里有一个例子,调用你的函数,把相关信息作为第二个参数。我借用了此主题的代码。
#include< iostream>
int rvalue()
{
return 1;
}
int&& rvalue_ref(int&& i)//修改签名以避免返回引用局部变量(ty。user657267)
{
return std :: move(i);
}
template< typename T>
struct value_category {
//或者可以是一个积分或枚举值
static constexpr auto value =prvalue;
};
template< typename T>
struct value_category< T&> {
static constexpr auto value =lvalue;
};
template< typename T>
struct value_category< T&&> {
static constexpr auto value =xvalue;
};
//用于确保检查表达式的双括号,
//不是实体
#define VALUE_CATEGORY(expr)value_category< decltype((expr))> ::值
#define f(X)f_(X,VALUE_CATEGORY(X))
template< class T>
void f_(T&& x,char const * s)
{
std :: cout< s<< '\\\
';
}
int main()
{
f(rvalue()); // should printNot a reference
f(rvalue_ref(1)); //应该打印Rvalue reference
int j; f(j);
}
输出:
prvalue
xvalue
lvalue
当然,你可以修改字符串以适应,或者用枚举等替换它们。
When passed a parameter, I would like to distinguish between these two cases in a function parameter, like this:
int rvalue();
int&& rvalue_ref();
f(rvalue());
f(rvalue_ref());
However, when I try with forwarding references like this:
int rvalue()
{
return 1;
}
int&& rvalue_ref(int i)
{
return std::move(i);
}
template<class T>
void f(T&& x)
{
if (std::is_rvalue_reference<T>())
{
std::cout << "Rvalue reference" << std::endl;
}
else if (std::is_lvalue_reference<T>())
{
std::cout << "Lvalue reference" << std::endl;
}
else
{
std::cout << "Not a reference" << std::endl;
}
}
int main()
{
f(rvalue()); // Should print "Not a reference"
f(rvalue_ref(1)); // Should print "Rvalue reference"
}
It prints out "Not a reference" for both cases. Is there a way to distinguish both cases in C++?
解决方案 I don't see how to do this solely using function parameter. The distinction between xvalue and prvalue may be lost in the function call.
But you can do it with a macro that calls decltype
on the argument, before calling the function. Here is an example that calls your function with the relevant information as a second parameter. I borrowed code from this thread.
#include <iostream>
int rvalue()
{
return 1;
}
int&& rvalue_ref(int &&i) // Modified signature to avoid return reference to local variable (ty. user657267)
{
return std::move(i);
}
template<typename T>
struct value_category {
// Or can be an integral or enum value
static constexpr auto value = "prvalue";
};
template<typename T>
struct value_category<T&> {
static constexpr auto value = "lvalue";
};
template<typename T>
struct value_category<T&&> {
static constexpr auto value = "xvalue";
};
// Double parens for ensuring we inspect an expression,
// not an entity
#define VALUE_CATEGORY(expr) value_category<decltype((expr))>::value
#define f(X) f_(X, VALUE_CATEGORY(X))
template<class T>
void f_(T&& x, char const *s)
{
std::cout << s << '\n';
}
int main()
{
f(rvalue()); // Should print "Not a reference"
f(rvalue_ref(1)); // Should print "Rvalue reference"
int j; f(j);
}
Output:
prvalue
xvalue
lvalue
Of course you can trivially modify the strings to suit, or replace them with enums etc.
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