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C++ template argument as reference lvalue


C++ template argument as reference lvalue

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C++ template argument as reference lvalue
Tag : cpp , By : jehammon
Date : November 23 2020, 01:01 AM

will be helpful for those in need Floating point types cannot be non-type template parameters. Thus the author of the template took a reference to a double instead.
Your code has two problems. First of all, an argument to such a reference must have static storage duration and linkage, which your variable cannot have both as it is defined in block-scope. Secondly, your variable is const (as it is constexpr), and a non-const reference cannot be bound to a const object.

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template argument loses a lvalue reference, if not used directly


Tag : cpp , By : brij
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)

Does an lvalue argument prefer an lvalue reference parameter over a universal reference?


Tag : cpp , By : Reiner
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.

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" ) );

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; 
}

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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