在使用它的类的构造函数中初始化std :: array的大小

Question

Is it possible to use the std::array<class T, std::size_t N> as a private attribute of a class but initialize its size in the constructor of the class?

class Router{
    std::array<Port,???> ports; //I dont know how much ports do will this have
public:
    Switch(int numberOfPortsOnRouter){
        ports=std::array<Port,numberOfPortsOnRouter> ports; //now I know it has "numberOfPortsOnRouter" ports, but howto tell the "ports" variable?
    }
}

I might use a pointer, but could this be done without it?

Answer 1

You have to make your class Router a template class

template<std::size_t N> 
class Router{
    std::array<Port,N> ports; 

...
}

in case you want to be able to specify the size of ports at Router level . By the way, N must be a constant known from compile time.

Otherwise you need std::vector .

Answer 2

No, the size must be known at compile time. Use std::vector instead.

class Router{
    std::vector<Port> ports;
public:
    Switch(int numberOfPortsOnRouter) : ports(numberOfPortsOnRouter) {
    }
};

Answer 3

The size of an std::array<T, N> is a compile-time constant which can't be changed at run-time. If you want an array with flexible bounds you can use a std::vector<T> . If the size of your array doesn't change and you somehow know the size from its context, you might consider using std::unique_ptr<T[]> . It is a bit more light-weight but also doesn't help with copying or resizing.

Answer 4

std::array is an array of fixed length. Therefore the length must be known at compile time. If you need an array with dynamic length, you want to use std::vector instead:

class Router{
    std::vector<Port> ports;
public:
    Switch(int numberOfPortsOnRouter):ports(numberOfPortsOnRouter){}
};