c++ - 没有空构造函数编译失败

Question

我是 C++ 的新手，并试图准确理解我的代码发生了什么。

这些类都在它们自己的头文件中定义。代码如下。

队列:

template<class T> class Queue
{
  public:
    Queue(unsigned int size)
    {
        _buffer = new T[size]; //need to make sure size is a power of 2
        _write = 0;
        _read = 0;
        _capacity = size;
    }
    
    /* other members ... */
   
   private:
    unsigned int _capacity;
    unsigned int _read;
    unsigned int _write;
    T *_buffer;
};

连续剧:

template<class T> class Queue;
template<class T> class Serial
{
  public:
    Serial(unsigned int buffer_size)
    {
        _queue = Queue<T>(buffer_size); //<---here is the problem
    }
  private:
    Queue<T> _queue;
};

当我尝试像这样创建一个 Serial 实例时:

Serial<unsigned char> s = Serial<unsigned char>(123);

编译器提示没有参数为零的 Queue 构造函数，至少我认为这是错误的意思:

In instantiation of 'Serial<T>::Serial(unsigned int) [with T = unsigned char]':

no matching function for call to 'Queue<unsigned char>::Queue()'

ambiguous overload for 'operator=' in '((Serial<unsigned char>*)this)->Serial<unsigned char>::_queue = (operator new(16u), (((Queue<unsigned char>*)<anonymous>)->Queue<T>::Queue<unsigned char>(buffer_size), ((Queue<unsigned char>*)<anonymous>)))' (operand types are 'Queue<unsigned char>' and 'Queue<unsigned char>*')

invalid user-defined conversion from 'Queue<unsigned char>*' to 'const Queue<unsigned char>&' [-fpermissive]

invalid user-defined conversion from 'Queue<unsigned char>*' to 'Queue<unsigned char>&&' [-fpermissive]

conversion to non-const reference type 'class Queue<unsigned char>&&' from rvalue of type 'Queue<unsigned char>' [-fpermissive]

当我向 Queue 添加一个空的构造函数时，它编译没有问题。当我单步调试调试器时，我看到它进入带有参数的构造函数，而不是空参数。

为什么会这样？

Answer 1

Serial(unsigned int buffer_size)缺乏member init list ，因此 _queue必须首先使用默认构造函数创建，然后为其分配一个值 ( Queue<T>(buffer_size) )。

这个:

Serial(unsigned int buffer_size) : _queue(buffer_size) {}

将使用 Queue(unsigned int)构造函数代替，并且在没有默认构造函数的情况下也能工作。