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c++ - 为什么重载成员函数访问元组和递归累积结果失败?

转载 作者:塔克拉玛干 更新时间:2023-11-03 07:21:38 27 4
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我想访问元组成员并不断累积结果。但不起作用,它看起来像访问越界(不完整类型的无效使用 struct std::tuple_element<0u, std::tuple<> > )

#include <iostream>
#include <tuple>

template<typename...ARGS>
struct NextTest
{
std::tuple<ARGS...> data;
template<std::size_t I,bool=I<sizeof...(ARGS)>
struct Dispatch{};
template<std::size_t I,typename T>
T next3(T t,Dispatch<I,false>)const
{
return t;
}
template<std::size_t I,typename T>
auto next3(T t,Dispatch<I,true>)const->decltype(this->next3<I+1>(std::get<I>(data),Dispatch<I+1>{}))
{

return this->next3<I+1>(std::get<I>(data),Dispatch<I+1>{});
}
template<std::size_t I,typename T>
auto next3(T t)const->decltype(this->next3<I>(t,Dispatch<I>{}))
{
return this->next3<I>(t,Dispatch<I>{});
}
template<std::size_t I,typename T,typename std::enable_if<(I>=sizeof...(ARGS))>::type* =nullptr>
T next(T t)const
{
return t;
}
template<std::size_t I,typename T,typename std::enable_if<(I<sizeof...(ARGS))>::type* =nullptr>
auto next(T t)const->decltype(this->next<I+1>(std::get<I>(data)))
{

return this->next<I+1>(std::get<I>(data));
}
template<std::size_t I,typename std::enable_if<(I>=sizeof...(ARGS))>::type* =nullptr>
void next2()const
{
std::cout<<"end!";
}
template<std::size_t I,typename std::enable_if<(I<sizeof...(ARGS))>::type* =nullptr>
void next2()const
{
std::cout<<"in seq ";
next2<I+1>();
}

};
void testexpr1()
{
NextTest<int> nt;
nt.next<0>( 1);//fail
nt.next3<0>( 1);//fail
nt.next2<0>();//pass!
}

mingw gcc 4.8.1 为什么报错?我该怎么办?

**update1:**这符合:

...\include\c++\tuple||In instantiation of 'struct std::tuple_element<1u, std::tuple<int> >':|
...\include\c++\tuple|771| required by substitution of 'template<unsigned int __i, class ... _Elements> constexpr typename std::__add_r_ref<typename std::tuple_element<__i, std::tuple<_Elements ...> >::type>::type std::get(std::tuple<_Elements ...>&&) [with unsigned int __i = 1u; _Elements = {int}]'|
...\test.cpp|83| required by substitution of 'template<unsigned int I, class T> decltype (this->.next1<(I + 1)>(get<I>(this->.data))) NextTest<ARGS>::next1(T&, typename std::enable_if<(I < sizeof (ARGS ...))>::type*) [with unsigned int I = I; T = T; ARGS = {int}] [with unsigned int I = 1u; T = <missing>]'|
...\test.cpp|83| required by substitution of 'template<unsigned int I, class T> decltype (this->.next1<(I + 1)>(get<I>(this->.data))) NextTest<ARGS>::next1(T&, typename std::enable_if<(I < sizeof (ARGS ...))>::type*) [with unsigned int I = I; T = T; ARGS = {int}] [with unsigned int I = 0u; T = int]'|
...\test.cpp|104|required from here|
...\include\c++\tuple|680|error: invalid use of incomplete type 'struct std::tuple_element<0u, std::tuple<> >'|
...\include\c++\utility|84|error: declaration of 'struct std::tuple_element<0u, std::tuple<> >'|

最佳答案

代码中似乎有两个错误:

  • 正如@Nawaz 提到的,this-> 在 decltype 部分中是必需的,以访问成员函数和成员变量(至少在 gcc 4.7.2 中)。
  • 第二个是 off-by-one error在访问元组元素时(请参阅 enable_if 元函数调用中“I”更改为“I+1”的位置以及 Dispatch 结构中的默认参数)。

解决这两个问题后,代码可以在 gcc 4.7.2 上正常编译。

固定代码如下:

#include <iostream>
#include <tuple>

template<typename...ARGS>
struct NextTest
{
std::tuple<ARGS...> data;

template<std::size_t I,bool=(I+1)<sizeof...(ARGS)>
struct Dispatch{};

template<std::size_t I,typename T>
T next3(T t,Dispatch<I,false>)const
{
return t;
}
template<std::size_t I,typename T>
auto next3(T t,Dispatch<I,true>)const->decltype(this->next3<I+1>(std::get<I>(this->data),Dispatch<I+1>{}))
{
return this->next3<I+1>(std::get<I>(data),Dispatch<I+1>{});
}
template<std::size_t I,typename T>
auto next3(T t)const->decltype(this->next3<I>(t,Dispatch<I>{}))
{
return this->next3<I>(t,Dispatch<I>{});
}

template<std::size_t I,typename T,typename std::enable_if<((I+1)>=sizeof... (ARGS))>::type* =nullptr>
T next(T t)const
{
return t;
}
template<std::size_t I,typename T,typename std::enable_if<((I+1)<sizeof...(ARGS))>::type* =nullptr>
auto next(T t)const->decltype(this->next<I+1>(std::get<I>(this->data)))
{
return this->next<I+1>(std::get<I>(data));
}

template<std::size_t I,typename std::enable_if<((I+1)>=sizeof...(ARGS))>::type* =nullptr>
void next2()const
{
std::cout<<"end!";
}
template<std::size_t I,typename std::enable_if<((I+1)<sizeof...(ARGS))>::type* =nullptr>
void next2()const
{
std::cout<<"in seq ";
next2<I+1>();
}
};

void testexpr1()
{
NextTest<int> nt;
nt.next<0>( 1);//pass
nt.next3<0>( 1);//pass
nt.next2<0>();//pass!
}

关于c++ - 为什么重载成员函数访问元组和递归累积结果失败?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/25301302/

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