我正在尝试元组,并遇到一个创建元组的问题.
代码示例如下.
代码示例如下.
//a.cpp
#include <tuple>
using namespace std;
int main() {
auto te = make_tuple(); //this line is ok
auto tte = make_tuple(te); //this line gives an error.
return 0;
}
我用g 4.5(g -std = c 0x a.cpp)和MS VC 2010编译它.
两个编译器都在main()中的第二行给我一个错误.
我的问题是这样的：
由于’te’是一个明确定义的变量,为什么不能用te作为内容来创建另一个元组.这个语义是否正确？
我想这是一种边界的情况,但是如果算术是正确的,应该允许零.
来自gcc的错误消息是：
$gcc -std=c++0x a.cpp
In file included from a.cpp:1:0:
c:\mingw\bin\../lib/gcc/mingw32/4.5.2/include/c++/tuple: In constructor
'std::tuple<_Elements>::tuple(std::tuple<_UElements ...>&) [with _UElements = {},_Elements = {std::tuple<>}]':
c:\mingw\bin\../lib/gcc/mingw32/4.5.2/include/c++/tuple:551:62: instantiated from
'std::tuple<typename std::__decay_and_strip<_Elements>::__type ...>
std::make_tuple(_Elements&& ...) [with _Elements = {std::tuple<>&},typename
std::__decay_and_strip<_Elements>::__type = <type error>]'
a.cpp:6:27: instantiated from here
c:\mingw\bin\../lib/gcc/mingw32/4.5.2/include/c++/tuple:259:70: error: invalid
static_cast from type 'std::tuple<>' to type 'const std::_Tuple_impl<0u>&'
解决方法
这看起来像编译器已经匹配了你的std :: tuple<>反对std :: tuple< std :: tuple>的以下构造函数(见N3242中的20.4.2p15-17)：
template <class... UTypes> tuple(const tuple<UTypes...>& u);@H_502_24@Requires:
sizeof...(Types) == sizeof...(UTypes)@H_502_24@.
is_constructible<Ti,const Ui &>::value@H_502_24@ is true for alli@H_502_24@.Effects:
Constructs each element of*this@H_502_24@ with
the corresponding element of u.Remark: This constructor shall not
participate in overload resolution
unlessconst Ui &@H_502_24@ is implicitly
convertible toTi@H_502_24@ for alli@H_502_24@.
我认为这是编译器执行std :: tuple的一个bug; “注释”意味着这个构造函数不应该被考虑,因为它不会被编译.
