C ++的向量如何分配内存

Question

Consider my code, I have vector P which type is Particle. Then inside the Particle there's also has vector x, v and xBest.

So, It's vectors inside of vector.

struct Particle
{
    vector<double> x;
    vector<double> v;
    vector<double> xBest;
    double fitness;
    double pBest;
};

class Swarm
{
  public:
    vector<Particle> P;
    Swarm();
};

Since the compiler won't let me reserve memory for vector when declaring a class or struct. like this:

class Swarm
{
  public:
    vector<Particle> P(30);
    Swarm();
};

So, I do it in the constructor like this:

Swarm::Swarm()
{
    P.reserve(30);
    for(size_t i = 0; i < 30; i++)
    {
        P[i].x.reserve(10);
        P[i].v.reserve(10); 
        P[i].xBest.reserve(10);         
    }
}

And it's work.

I'm very curious about this. Since the size of the vectors in struct Particle haven't been initialize yet so the size of Particle is unknown. But I can reserve the memory for 30 Particles even before reserving the memory for 3 vector in struct Particle !!

How is that possible?

Answer 1

This is undefined behavior. When you reserve a vector, you don't create the objects, so the loop:

for(size_t i = 0; i < 30; i++)
{
    P[i].x.reserve(10);
    P[i].v.reserve(10); 
    P[i].xBest.reserve(10);         
}

Is calling reserve on vectors that DO NOT exist.

You can't reserve capacity on vectors that don't exist. You need to create your Particle s first.

Answer 2

In C++ the storage needed by every object is known statically, ie at compile time. There is no such thing as VLAs in C++. std::vector doesn't store the objects directly, it stores a pointer to an array on the heap:

template <typename T>
class vector {
    // For illustration purposes only
    T *array;
    std::size_t size;
};

As you can see, the size of the vector is always constant, no matter how many elements it points to which is why your example (minus the UB you have as mentioned in the comments) works.