template argument loses a lvalue reference, if not used directly
Date : March 29 2020, 07:55 AM
I hope this helps you . The problem here is that type deduction takes priority over defaulted function template parameters. Therefore you get the T parameter deduced and T never deduces to a reference. You can prevent this by making the type not deducible. A generic identity type trait can do this. template <typename T>
struct identity { using type = T; };
template <typename T>
using NotDeducible = typename identity<T>::type;
template<typename T_orig, typename T=typename target<T_orig>::T>
void g(NotDeducible<T> a) { // blah
template<typename T_orig> void g(typename target<T_orig>::T a)
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Does an lvalue argument prefer an lvalue reference parameter over a universal reference?
Date : March 29 2020, 07:55 AM
This might help you The "universal reference" deduces the parameter to foo&. The first template also deduces the parameter to foo&.
C++ has a partial ordering rule for function templates that makes T& be more specialized than T&&. Hence the first template must be chosen in your example code.
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difference between rvalue reference and lvalue reference as argument
Tag : cpp , By : AJacques
Date : November 04 2020, 08:17 AM
Does that help Only constant lvalue references may be bound to temporary objects. So this function void printReference (const string& str)
{
cout << str;
}
const std::string s1( "constnat lvalue" );
printReference( s1 );
std::string s2( "non-constant lvalue" );
printReference( s2 );
printReference( "A temporary object of type std::string" );
printReference( static_cast<const std::string>( "A temporary object of type std::string" ) );
void printReference (string&& str)
{
cout << str;
}
printReference( "A temporary object of type std::string" );
printReference( static_cast<const std::string>( "A temporary object of type std::string" ) );
void printReference (const string&& str)
^^^^^
{
cout << str;
}
printReference( static_cast<const std::string>( "A temporary object of type std::string" ) );
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Converting possible lvalue reference template to lvalue
Tag : cpp , By : Piotr Balas
Date : March 29 2020, 07:55 AM
hope this fix your issue What you're finding is std::remove_reference: template< class T > struct remove_reference {typedef T type;};
template< class T > struct remove_reference<T&> {typedef T type;};
template< class T > struct remove_reference<T&&> {typedef T type;};
template <typename T>
typename remove_reference<T>::type Magic(T t) {
return t;
}
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Why a template argument of a rvalue reference type can be bound to a lvalue type?
Tag : cpp , By : kuba53280
Date : March 29 2020, 07:55 AM
I hope this helps you . You can read this article Universal References in C++11 to understand. Here some part of it: Widget&& var1 = someWidget; // here, “&&” means rvalue reference
auto&& var2 = var1; // here, “&&” does not mean rvalue reference
template<typename T>
void f(std::vector<T>&& param); // here, “&&” means rvalue reference
template<typename T>
void f(T&& param); // here, “&&”does not mean rvalue reference
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