如何在 C++ 初始值设定项列表中将常量引用成员初始化为另一个成员（标准向量）

Question

I did the following as a cheap way to allow read-only access to a member container _numbers via numbers :

class Foo {
    Foo() : _numbers({}), numbers(_numbers) {
    // some code that populates `numbers` via `numbers.push_back(...)`
}

private:
    std::vector<int> _numbers;
public:
    const std::vector<int>& numbers;
}

However, doing so I see that numbers is empty, while in other cases it will contain the same elements as _numbers .

To be more precise, it seems to be undefined behavior. In my real example (of which this a simplified version) I have multiple reference-container pairs with this scheme, where the populated data is visible in the const-reference member for some pairs, and for some it is not.

Any idea whats wrong with this? Any help is deeply appreciated.

EDIT Here is a minimal working example:

#include <vector>

struct Foo2 {
public:
     const int max1;
     const int center1;

    Foo2(const int max1_);
private:
    std::vector<int> _numbers1, _numbers2;

public:
    const std::vector<int>& numbers1, numbers2;
};

Foo2::Foo2(const int max1_)
    : max1(max1_), center1(max1_/2),
      _numbers1({}), _numbers2({}),
      numbers1(_numbers1),
      numbers2(_numbers2)
{
    cout << max1 << endl;

    for (int i=0; i<center1; i++) {
        _numbers1.push_back(i);
        cout << "A: " << i << endl;
    }
    for (int i=center1; i<max1; i++) {
        _numbers2.push_back(i);
        cout << "B: " << i << endl;
    }

    for (int i: numbers1) {
        cout << "A: " << i << endl;
    }
    for (int i: numbers2) {
        cout << "B: " << i << endl;
    }
}

which gives the following Output when initializing Foo2 f(8) :

8
A: 0
A: 1
A: 2
A: 3
B: 4
B: 5
B: 6
B: 7
A: 0
A: 1
A: 2
A: 3

ie numbers2 does not see the contents of _numbers2 while for numbers1 it seems to work.

Answer 1

const vector<int>& numbers1, numbers2; — Here, only the first variable is a reference. You need & before the second variable to make it a reference as well. Then the code should work.

But I have to say that what you're doing is a really bad idea . You're paying for a convenient syntax with memory overhead, non-assignability, and possibly speed overhead.

Use getters instead: const vector<int>& numbers1() const {return _numbers1;} . Yes, you will have to type the extra () every time.

Answer 2

Hmm... The main cause of the problem was given by @HolyBlackCat: numbers2 was not a reference but an independant copy.

But IMHO there is a more fundamental problem. With:

public:
    const std::vector<int>& numbers;

You promise the compiler, that once initialized, the vector referenced by numbers will no longer change . Which is a lie, because the underlying vector changes...

Lying to the compiler is a sure path to UB: it will work sometimes, and will suddenly break because the compiler is free to change its actual code so that no changes to _member will be reflected in members .

Possible fixes:

• use std::vector<int>& numbers = _numbers; (no const ). This one will be fine, but you loose the ability to present a read only reference

• get a fully initialized reference when you need it through a getter (ie a method):

   const std::vector& numbers() { return _numbers; }

  Again it is fine, provided _number no longer changes after complete initialization of Foo

• use a dedicated object, implicitely convertible to a vector that will be fully initialized before being used:

   struct Bar { std::vector<int>_numbers; Bar(bool first) : Bar(0, first) {}; Bar(int max, bool first) { cout << max << endl; int center = max / 2; if (first) { for (int i = 0; i < center; i++) { _numbers.push_back(i); cout << "A: " << i << endl; } } else { for (int i = center; i < max; i++) { _numbers.push_back(i); cout << "B: " << i << endl; } } } operator const std::vector<int>& () { return _numbers; } }; struct Foo2 { public: const int max1; const int center1; Foo2(); Foo2(const int max1_); private: Bar _numbers1, _numbers2; public: const std::vector<int>& numbers1, &numbers2; };