2015-03-19 62 views
11

是否有可能?用initializer_list初始化一个数组?

#include <array> 
#include <initializer_list> 

struct A 
{ 
    A (std::initializer_list<int> l) 
     : m_a (l) 
    { 
    } 

    std::array<int,2> m_a; 
}; 

int main() 
{ 
    A a{ 1,2 }; 
} 

但这会导致这个错误:

t.cpp: In constructor ‘A::A(std::initializer_list<int>)’: 
t.cpp:7:19: error: no matching function for call to ‘std::array<int, 2ul>::array(std::initializer_list<int>&)’ 
     : m_a (l) 
       ^
t.cpp:7:19: note: candidates are: 
In file included from t.cpp:1:0: 
/usr/lib/gcc/x86_64-pc-linux-gnu/4.8.2/include/g++-v4/array:81:12: note: std::array<int, 2ul>::array() 
    struct array 
      ^
/usr/lib/gcc/x86_64-pc-linux-gnu/4.8.2/include/g++-v4/array:81:12: note: candidate expects 0 arguments, 1 provided 
/usr/lib/gcc/x86_64-pc-linux-gnu/4.8.2/include/g++-v4/array:81:12: note: constexpr std::array<int, 2ul>::array(const std::array<int, 2ul>&) 
/usr/lib/gcc/x86_64-pc-linux-gnu/4.8.2/include/g++-v4/array:81:12: note: no known conversion for argument 1 from ‘std::initializer_list<int>’ to ‘const std::array<int, 2ul>&’ 
/usr/lib/gcc/x86_64-pc-linux-gnu/4.8.2/include/g++-v4/array:81:12: note: constexpr std::array<int, 2ul>::array(std::array<int, 2ul>&&) 
/usr/lib/gcc/x86_64-pc-linux-gnu/4.8.2/include/g++-v4/array:81:12: note: no known conversion for argument 1 from ‘std::initializer_list<int>’ to ‘std::array<int, 2ul>&&’ 
+0

这是不可能的,作为一种解决方法,手动复制'initializer_list'的元素:'A(std :: initializer_list l)/ *不要初始化m_a */{std :: copy(l。 begin(),l.end(),m_a.begin());}' – Mankarse 2015-03-19 11:38:50

+0

请注意,这不是特殊的构造函数; 'std :: initializer_list x = {1,2}; std :: array m {?????? };'。您只能将这些东西放在聚合初始化的一部分,即每个元素的初始化符一个接一个。 – 2015-03-19 11:49:34

+0

请参阅http://stackoverflow.com/q/5549524/1436796,它也提供了一种可能的解决方案,即使成员数组应该被限定为'const'(这会使得它不为空,然后再复制列表内容)。 – 2015-03-19 12:37:56

回答

15

不在此例。您可以初始化array列表初始化

std::array<int, 2> a{1,2}; 

,但你不能用initializer_list初始化array,因为array仅仅是一个聚合类型上只有默认和拷贝构造函数。

您可以将该数组留空,然后将initializer_list的内容复制到其中。