[英]How std::initializer_list Works?
I am currently studying c++11 i did not understand constructor of std::initializer_list it looks like this我目前正在研究 c++11 我不明白 std::initializer_list 的构造函数它看起来像这样
constexpr initializer_list() noexcept : _First(nullptr), _Last(nullptr) {}
constexpr initializer_list(const _Elem* _First_arg, const _Elem* _Last_arg) noexcept
: _First(_First_arg), _Last(_Last_arg) {}
But How it works with但它是如何工作的
std::initializer_list<int> v{1,2,3,4,5,6,7,8,9,0};
and i tried this我试过这个
constexpr init(const _Elem* _First_arg, const _Elem* _Last_arg) noexcept
: _First(_First_arg), _Last(_Last_arg) {}
but this shows error但这显示错误
init<int> ob{1,2,3,4,5,6,7,8,9,0}; //this shows error
note: candidate: 'constexpr init<_Elem>::init(const _Elem*, const _Elem*) [with _Elem = int]'
constexpr init(const _Elem* _First_arg, const _Elem* _Last_arg) noexcept
^~~~
note: candidate expects 2 arguments, 10 provided
and i changed {} to () like我将 {} 更改为 () 喜欢
std::initializer_list<int> v(1,2,3,4,5,6,7,8,9,0);
This shows error.这显示错误。
1)How std::initializer_list works? 1)std::initializer_list 是如何工作的?
2)What is behind {}? 2) {} 后面是什么?
Thanks.谢谢。
std::initializer_list
is special. std::initializer_list
是特殊的。 It is impossible to write a class that could be used as a constructor argument in the same way.不可能以相同的方式编写可以用作构造函数参数的 class。 The language rules specify how std::initializer_list
works - or rather, how constructors that accept std::initializer_list
ie initializer-list constructors work.语言规则指定std::initializer_list
如何工作 - 或者更确切地说,接受std::initializer_list
的构造函数,即初始化列表构造函数如何工作。 And the language implementation makes it work as specified.语言实现使其按规定工作。
PS Identifiers such as _Elem
are reserved to the language implementation. PS 标识符(例如_Elem
)保留给语言实现。 Since your class init
is not part of the language implementation, using reserved identifiers results in undefined behaviour.由于您的 class init
不是语言实现的一部分,因此使用保留标识符会导致未定义的行为。 Don't use reserved identifiers.不要使用保留的标识符。
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