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Why does std::is_rvalue_reference not do what it is advertised to do?


Why does std::is_rvalue_reference not do what it is advertised to do?

By : Nemtendo
Date : November 20 2020, 04:01 AM
this will help Because for forwarding reference, T will never be deduced as an rvalue reference. Suppose passing an object of type int to OwnershipReceiver, if the object is an lvalue, T will be deduced as an lvalue-reference, i.e. int&; if the object is an rvalue, T will be deduced as an non-reference, i.e. int. That's why std::is_rvalue_reference::value won't work because it's always false.
Note that the purpose of the code is to make sure the parameter type of OwnershipReceiver is an rvalue-reference, it doesn't mean the type of T is an rvalue-reference too.
code :
template <typename T>
auto receive_ownership(T&& t) -> typename std::enable_if<std::is_rvalue_reference<decltype(t)>::value>::type      
{
   // taking file descriptor of t, and clear t
}


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std::is_rvalue_reference inside template function with rvalue reference parameter

std::is_rvalue_reference inside template function with rvalue reference parameter


By : Kevin Meier
Date : March 29 2020, 07:55 AM
this will help V is not the rvalue-reference - decltype(element) is (if element is an rvalue). V is simply the general type of element. Specifically:
code :
typename std::remove_reference<decltype(element)>::type; // V
If I accept a parameter via universal reference, is exactly one of is_rvalue_reference and is_lvalue_reference true?

If I accept a parameter via universal reference, is exactly one of is_rvalue_reference and is_lvalue_reference true?


By : VuNguyen
Date : March 29 2020, 07:55 AM
it helps some times Is it ever possible for this code to print "neither"? ,
Is it ever possible for this code to print "neither"?
code :
foo(42);  // "neither" is printed because T is deduced as int
int i=0;
// "neither" is printed because T is explicitly specified as int:
foo<int>(std::move(i));
SFINAE with std::is_rvalue_reference for variadic template argument list fails

SFINAE with std::is_rvalue_reference for variadic template argument list fails


By : user3024415
Date : March 29 2020, 07:55 AM
I hope this helps . I try to implement a function taking a variable list of rvalue references. It looks like this: , Args is not an rvalue reference, Args&& is.
code :
#include <type_traits>
#include <utility>

template <
    typename... Args,
    typename = std::enable_if_t<std::conjunction<
        std::is_rvalue_reference<Args&&>...>::value>
>
void myfunc(Args&&... args) { }

struct probe {};

int main() {
    probe x;

    // this compiles
    myfunc(std::move(x));

    // this does not
    myfunc(x);

    return 0;
}

Advertised shortcuts vs. non-advertised shortcuts in windows setup project

Advertised shortcuts vs. non-advertised shortcuts in windows setup project


By : Jason
Date : March 29 2020, 07:55 AM
seems to work fine Advertised shortcuts are used by advertised features:
when a feature is installed as advertised, it is only registered with Windows Installer; it's content is not actually copied on the machine when launching an advertised shortcut which targets something in this advertised feature, the feature content is finally copied on the machine
Advertised Shortcuts

Advertised Shortcuts


By : Benjamin
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further Not really. The target of an advertised shortcut is the keyfile of an installed product's component. This is so it can do a health check when you click it.
All of this is impossible to have that without having an MSI.
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