c++ - 当参数是参数包时，右值引用不起作用

Question

当我从模板参数包中创建一个右值时，它不会编译，但如果它是一个“简单”模板参数，它编译得很好。

在此代码 for_each_in_tup1编译正常，但 for_each_in_tup3才不是。我不明白为什么这不会编译，但 GCC 9.2 和 VC v142 都同意这是错误的。

为什么这不是有效的语法？

这个例子:

#include <tuple>
using namespace std;

template< typename Tuple, size_t N = 0>
void for_each_in_tup1( Tuple&& tup ) {
    if constexpr(N < tuple_size<decay_t<Tuple>>::value)
        for_each_in_tup1<Tuple, N + 1>(forward<Tuple>(tup));
}

template< template<typename...> typename Tuple, typename... Ts>
void for_each_in_tup2( const Tuple<Ts...>& tup) {
}

template< template<typename...> typename Tuple, typename... Ts>
void for_each_in_tup3( Tuple<Ts...>&& tup) {
}

void test_lazy() {
    tuple<uint32_t, uint32_t> tup;
    for_each_in_tup1(tup);
    for_each_in_tup2(tup);
    for_each_in_tup3(tup);
}

失败:

<source>:20:22: error: cannot bind rvalue reference of type 'std::tuple<unsigned int, unsigned int>&&' to lvalue of type 'std::tuple<unsigned int, unsigned int>'
   20 |     for_each_in_tup3(tup);
      |                      ^~~

<source>:15:39: note:   initializing argument 1 of 'void for_each_in_tup3(Tuple<Ts ...>&&) [with Tuple = std::tuple; Ts = {unsigned int, unsigned int}]'
   15 | void for_each_in_tup3( Tuple<Ts...>&& tup) {
      |                        ~~~~~~~~~~~~~~~^~~

godbolt

Answer 1

在 for_each_in_tup1您正在使用转发引用:

template< typename Tuple, size_t N = 0>
void for_each_in_tup1( Tuple&& tup ) 

当你传递左值时， Tuple推导出为 Tuple& , 在引用折叠后你会得到 for_each_in_tup1(Tuple&) .左值可以绑定(bind)到左值引用，这就是这段代码有效的原因。如果您将右值传递给 for_each_in_tup1 , 参数为 Tuple&& ，并且它只能接受右值。

在 for_each_in_tup3有正常的右值引用，它只能绑定(bind)右值。
调用 for_each_in_tup3你必须投 tup右值，例如 std::move :

for_each_in_tup3(std::move(tup));