在C ++中，当我想在初始化类对象数组时通过构造函数初始化类成员时，我该怎么办？

Question

I thought I could do the following:

#include <iostream>

using namespace std;

class cl
{
  public:
    double* Arr;
    cl(int);
};

cl::cl(int i)
{
    Arr=new double[i];
    Arr[0]=11;
};

int main()
{
    cl* BB;
    BB=new cl(3)[3];      /* I want to initialize a class member Arr for each element of the class cl array BB when I initialize this array by means of constructor. What should I do? */
    cout << cl[1].Arr[0] << endl;                                                                                                   

    return 0;
}

but obviously something is wrong with the line where I noted. The compiler wouldn't compile.

Answer 1

In C++11 you can use brace initializer:

cl* BB = new cl[3] {1, 2, 3};

or more verbose, so it's clear that the numbers are passed as arguments to constructors of different objects:

cl* BB = new cl[3] {{1}, {2}, {3}}; 

Though as you are allocating memory dynamically anyway, it's better to use std::vector , also it's more convenient if you want to initialize large number of objects with the same parameters:

std::vector<cl> BB(300, 3);

However, std::vector initialization won't compile if cl doesn't have a copy constructor. In that case you can use emplace_back() :

vector<cl> BB;
BB.reserve(300);
for(int i = 0; i < 300; ++i)
    BB.emplace_back(3);

This in turn can be wrapped into a template back_emplacer_iterator to use with std::fill_n

Answer 2

There are several approaches:

• C++11 initializer list

   cl* BB = new cl[3] {42, 42, 42}; 

• STL-vector (recommended)

   std::vector<cl> BB( 3, cl( 42 ) ); 

As pre-C++11 there are also other more involved solutions relying on placement new operator although I'd not recommend this.

Answer 3

The format of a new-expression is a type, followed by an initializer. If you want to allocate an array of 3 cl objects then the type is cl[3] . You can't say cl(3)[n] because cl(3) is not a type.

In C++03 the only valid initializer for a dynamically-allocated array is () which value-initializes each element, but you can't do that as your type doesn't have a default constructor.

In C++11 you can use an initializer-list to pass arguments to each array element:

cl* ptr = new cl[3]{ 3, 3, 3};

Answer 4

Each element in the array will be initialized by the default constructor (no argument constructor). After that you may call any function that does the desired task. Corrected program example:

class cl
{
  public:
    double* Arr;
    cl(int i = 0);
    ~cl();
    void allocArr(int i=0);
};

cl::cl(int i)
{
    allocArr(i);
}

void cl::allocArr(int i)
{
    if (i <= 0) {
        Arr = (double *) NULL;
    }
    else {
        Arr=new double[i];
        Arr[0]=11;
    }
};

cl::~cl() {
    if (Arr)
      delete [] Arr;
}

int main()
{
    cl* BB;
    BB=new cl[3]; // default constructor 
    for (int i = 0; i < 3; i++) {
        BB[i].allocArr(3);
    }

    cout << BB[1].Arr[0] << endl;                                                                                                   

    return 0;
}