自定义数组类:列表初始化的构造方法 [英] Custom Array Class: Constructor for Initialization of List
问题描述
我正在使用C ++进行自定义数组类的研究(作为一个自我指导的练习),我不确定如何创建一个允许我按照以下方式做一些事情的构造函数:
I'm working on a custom array class in C++ (as a self-led exercise), and I'm not sure how to create a constructor that allows me to do something along the lines of:
#include "array.h"
#include <iostream>
int main()
{
array<int> test = {1, 2, 3, 4};
std::cout << test(1) << std::endl;
return 0;
}
编译器(VS Express 2013)给出的错误是没有构造函数array :: array [with T = int]的实例"与参数列表匹配.参数类型为(int,int,int,int)."
The error that the compiler (VS Express 2013) gives me is "no instance of constructor array::array [with T = int]" matches the argument list. argument types are (int, int, int, int)."
我不确定调用一组元素枚举的构造函数是什么.我知道我已经正确地重载了 operator()(const int&)
.我也知道这(由于我不清楚的原因)有效:
I'm not sure what the constructor that takes the enumeration of a set of elements is called. I know I've properly overloaded operator()(const int&)
. I also know that this (for a reason that's not clear to me) works:
#include "array.h"
#include <iostream>
int main()
{
array<char> test = "abcd";
std::cout << test(1) << std:: endl; // prints 'a', as expected.
std::cout << test(4) << std::endl; // prints 'd', as expected.
return 0;
}
这是通过 array(const T [])
构造函数实现的:将为 array< int>解决方案吗?test = {1,2,3,...,n}
的情况是否相似?
This is achieved with a array(const T[])
constructor: will the solution for the array<int> test = {1, 2, 3, ..., n}
case be similar?
在此先感谢您的指导.
编辑:如果有帮助,请添加以下代码.
Including the code below, in case it's helpful.
template<typename T>
class array
{
public:
typedef T* iterator;
typedef const T* const_iterator;
private:
iterator head;
unsigned long elems;
public:
array()
: head(nullptr)
, elems(0) {}
array(const unsigned long &size)
: head(size > 0 ? new T[size] : nullptr)
, elems(size) {}
array(const T[]);
array(const array&);
~array() { delete[] head; }
iterator begin() const { return head; }
iterator end() const { return head != nullptr ? &head[elems] : nullptr; }
unsigned long size() const { return elems; }
array& operator=(const array&);
T& operator()(const unsigned long&);
};
template<typename T>
array<T>::array(const T rhs[])
{
unsigned long size = sizeof(rhs) / sizeof(T);
head = new T[size];
iterator pointer = begin();
for (const_iterator i = &rhs[0]; i != &rhs[0] + size; i++)
*pointer++ = *i;
}
template<typename T>
array<T>::array(const array<T> &rhs)
{
head = new T[rhs.size()];
iterator pointer = begin();
for (const_iterator i = rhs.begin(); i != rhs.end(); i++)
*pointer++ = *i;
}
template<typename T>
array<T>& array<T>::operator=(const array<T> &rhs)
{
if (this != &rhs)
{
delete[] head;
head = new T[rhs.size()];
iterator pointer = begin();
for (const_iterator i = rhs.begin(); i != rhs.end(); i++)
*pointer++ = *i;
}
return *this;
}
template<typename T>
T& array<T>::operator()(const unsigned long &index)
{
if (index < 1 || index > size())
{
// Add some error-handling here.
}
return head[index - 1];
}
推荐答案
#include <initializer_list>
// ...
template <typename T>
class array
{
// ...
array(std::initializer_list<T> il);
// ...
template <typename T>
array<T>::array(std::initializer_list<T> il)
{
unsigned long size = il.size();
head = new T[size];
iterator pointer = begin();
for (const T& i : il)
*pointer++ = i;
}
// ...
array<int> test = {1, 2, 3, 4};
建议的改进:
-
array(const T rhs []);
等效于array(const T * rhs);
,即一个指针,这意味着sizeof(rhs)/sizeof(T)
表达式不会为您提供项目数.如果要为const char *
使用特殊的构造函数,则考虑整个array< char>
专业化,或者如果T,则至少从重载分辨率中禁用此构造函数
不是char
array(const T rhs[]);
is an equivalent ofarray(const T* rhs);
, that is, a pointer, which means thatsizeof(rhs) / sizeof(T)
expression will not give you the number of items. If you want a special constructor forconst char*
, then consider either an entirearray<char>
specialization or at least disabling this constructor from the overload resolution ifT
is notchar
head = new T [size];
默认初始化所有元素(为每个 T
类型的实例调用默认构造函数).然后调用一个赋值操作: * pointer ++ = * i;
.这可以通过使用 placement-new 来改进,例如 :: new((void *)ptr)T(* i);
其中 ptr
是指向未初始化的原始内存缓冲区的指针,例如 new char [sizeof(T)* size]
或从
head = new T[size];
default-initializes all elements (calls default constructor for each instance of type T
). Then you call an assignment operation: *pointer++ = *i;
. This can be improved by utilizing a placement-new, like ::new ((void*)ptr) T(*i);
where ptr
is a pointer to a raw, uninitialized memory buffer, like new char[sizeof(T)*size]
or returned from get_temporary_buffer
.
如果您想知道为什么下面的方法有效 array< char>test = {"abcd"};
与您当前的实现,这是一个解释:
And if you are wondering why the following works array<char> test = { "abcd" };
with your current implementation then here is an explanation:
-
array< T>
类具有一个构造函数,该构造函数采用const T []
,对于T = char
,其实例化为array< char> :: array(const char *)
.
array<T>
class has a constructor takingconst T[]
, which forT=char
is instantiated asarray<char>::array(const char*)
.
列表初始化可用于调用对象的构造函数.
List-initialization can be used to call object's constructor.
您的 const T []
构造函数不是显式的,这意味着您可以使用 copy-initialization 语法,如下所示:
Your const T[]
constructor is not explicit, which means that you can use copy-initialization syntax like below:
array<char> test = { "test" };
对于 T = char
,表达式 sizeof(rhs)/sizeof(T)
尽管如上所述无效,但其计算结果为 sizeof(char *)/sizeof(char)
,(最有可能是 4/1 = 4
.
The expression sizeof(rhs) / sizeof(T)
, although not valid as explained above, for T=char
evaluates to sizeof(char*) / sizeof(char)
, which is (most-probably) 4 / 1 = 4
.
用于初始化的"test"
恰好有4个字母,这很幸运.
Your "test"
used for initialization has exactly 4 letters, your luck.
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