考虑这个代码示例：
#include <initializer_list>
#include <iostream>
int main()
{
for(auto e: []()->std::initializer_list<int>{return{1,2,3};}())
std::cout<<e<<std::endl;
return 0;
}
我试图用g(gcc版本4.9.2(Debian 4.9.2-10))编译它)
输出正确.
在clang(Debian clang版本3.5.0-9(标签/ RELEASE_350 / final)(基于LLVM 3.5.0))输出例如：
0 2125673120 32546
其中第一行始终为0,最后两行为“随机”.
这是cl ang还是别的东西？我认为这个代码示例是正确的.
更新：
解决方法
5 An object of type
std::initializer_list<E>is constructed from an initializer list as if the implementation allocated an array ofNelements of typeE,whereNis the number of elements in the initializer list. Each element of that array is copy-initialized with the corresponding element of the initializer list,and thestd::initializer_list<E>object is constructed to refer to that array. If a narrowing conversion is required to initialize any of the elements,the program is ill-formed.6 The lifetime of the array is the same as that of the
initializer_listobject.
您的lambda的return语句初始化一个临时std :: initializer_list< int>并返回其副本.这一切都很好,除了其引用的数组的生命期在完整表达式的结尾结束.通过lambda以外的initializer_list访问死亡数组会导致未定义的行为.
initializer_list不是容器,它是对临时容器的引用.如果你试图像一个容器一样使用它,你将会有一个糟糕的时间.
在C 14(引用N4140)第6段中澄清：
6 The array has the same lifetime as any other temporary object (12.2),except that initializing an
initializer_listobject from the array extends the lifetime of the array exactly like binding a reference to a temporary.
通过CWG issue 1290的解决方案.这个澄清使得不可能使用initializer_list作为例如C 11的意图的成员变量.然而,即使在C14中,您的程序也有未定义的行为.
