在初始化数组构造函数不使用默认的构造函数或赋值 [英] Initialize array in constructor without using default constructor or assignment
问题描述
考虑：
struct A {
A (int);
A (const A &);
};
struct B {
A foo [2];
B (const A & x, const A & y)
: foo {x, y} /* HERE IS THE PROBLEM */
{}
};
我期待这一点，因为我用的C ++ 0x在GCC4.3，据称支持初始化器列表的支持工作。没有喜悦。
I was expecting this to work since I'm using C++0x support in GCC4.3, which allegedly supports initialiser lists. No joy.
我有没有默认构造函数A类。这是没有商量余地。分配默认后是不是一种选择。
I have a class A which has no default constructor. This is not negotiable. Assignment post-default is not an option.
我想创建B上使用A. B :: foo的可能不标准::向量。
I am trying to create B which uses A. B::foo may not be std::vector.
我怎么能初始化 B :: foo的在 B（...），构建它的元素恰好一次？
How can I initialise B::foo in B(...), constructing its elements exactly once?
目前，我condidering计算机来代替乙
At the moment, I am condidering replacing B with
struct B {
A foo_a;
B foo_b;
A * foo () {
assert ((&foo_b) - *&foo_a) == 1);
return &foo_a;
}
B (const A & x, const A & y) : foo_a(x), foo_b(y) {}
};
甚至使用字符富[2 * sizeof的（A）] 与安置新 - ！YUK
Or even using char foo [2*sizeof(A)] with placement new -- YUK!
肯定有这样做一个适当的方式？
Surely there's a proper way to do this?
推荐答案
不幸的是，实在是没有合适的，干净的方式做到这一点。考虑它的东西的语言的限制，从C ++的构造函数和C风格的数组一个尴尬的混合效果。在C ++ 11标准的解决这个问题，但在那之前你必须解决一个解决方法。
Unfortunately, there really is no proper, clean way to do this. Consider it something of a language limitation that results from an awkward mixing of C++ constructors and C style arrays. The C++11 standard addresses this issue, but until then you'll have to settle for a workaround.
由于 A 没有默认的构造函数，一个可能的解决方法是让 A * 指针数组，然后循环阵列上并初始化每一个与新。 （显然，别忘了删除 B中的析构函数数组中的每个项目，或者只是使用智能指针。）
Since A has no default constructor, one possible work-around is to have an array of A* pointers, and then loop over the array and initialize each one with new. (Obviously, don't forget to delete each item in the array in B's destructor, or just use smart pointers.)
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