初始化数组的std ::的大小在使用它的类的构造函数

问题描述

是否可以使用的std ::阵列＆LT; T类，的std ::为size_t N'GT; 作为类的私有属性，但是初始化它的大小在类的构造函数

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?
    }
}

我可能会使用一个指针，但会这样没有它怎么办？

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

推荐答案

没有，尺寸必须在编译时是已知的。使用的std ::矢量代替。

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) {
    }
};

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