我有一个课程模板
template <class T> class Collection
{
private:
int size;
int type;
T* Arr;
int Case;
public:
void ArrayGenerating() {
switch(type) {
case 1:
Arr = new T[size];
for (int i = 0; i < size; i++) {
srand((unsigned)time(NULL));
Arr[i] = static_cast <T> (rand()) % size;
}
case 2:
Arr = new T[size];
for (int i = 0; i < size; i++) {
srand((unsigned)time(NULL));
Arr[i] = static_cast <T> (rand()) / (static_cast <T> (RAND_MAX/size));
}
case 3:
Arr = new T[size];
for (int i = 0; i < size; i++) {
srand((unsigned)time(NULL));
Arr[i].setNumerator(static_cast <int> (rand()) % size);
srand((unsigned)time(NULL));
Arr[i].setDenominator(static_cast <int> (rand()) % size);
}
}
}
};
我想创建一个通用数据类型的随机数组
类型1,即整数数组。类型2,浮点数组。类型3,我有一个自定义数据类型&#34;分数&#34;。但是当我编译程序时,有错误:
Error 1 error C2228: left of '.setNumerator' must have class/struct/union
Error 2 error C2228: left of '.setDenominator' must have class/struct/union
那么如果有这种并发症的解决方案吗?
答案 0 :(得分:2)
我猜,type
是一个常数,取决于T.否则当T是浮点数时,将T *指向int是没有意义的。如果这是真的,则根本没有必要。
我认为,您正在寻找的是模板专业化(未经测试的代码):
// this is common to all cases.
class CollectionBase {
protected:
int size;
};
// the general template is not defined
// the compiler will complain whenever T is neither int, nor float, nor fraction.
template<class T> class Collection;
// here come the specializations
template<> class Collection<int>: private CollectionBase
{
private:
int* Arr;
public:
void ArrayGenerating() {
Arr = new int[size];
for (int i = 0; i < size; i++) {
srand((unsigned)time(NULL));
Arr[i] = static_cast<int>(rand()) % size;
}
}
};
template<> class Collection<float>: private CollectionBase
{
private:
float* Arr;
public:
void ArrayGenerating() {
Arr = new float[size];
for (int i = 0; i < size; i++) {
srand((unsigned)time(NULL));
Arr[i] = static_cast<float>(rand()) / (static_cast<float>(RAND_MAX/size));
}
}
};
template<> class Collection<fraction>: private CollectionBase
{
private:
fraction* Arr;
public:
void ArrayGenerating() {
Arr = new fraction[size];
for (int i = 0; i < size; i++) {
srand((unsigned)time(NULL));
Arr[i].setNumerator(static_cast <int> (rand()) % size);
srand((unsigned)time(NULL));
Arr[i].setDenominator(static_cast <int> (rand()) % size);
}
}
};
请注意,这种代码很危险。考虑std::vector<>
而不是自己管理动态分配的数组。
另外请注意,根据经验,只要构造函数完成,您的类的所有方法都应该可以安全地调用。在您的代码中,访问Arr
的任何函数都会在ArrayGenerating()
运行之前使用指向某个内存的随机指针。每当你因某种原因打电话给ArrayGenerating()
两次时,你的代码就会泄漏内存,因为你在创建一个新数组之前从不费心delete[]
。
C ++为内存管理提供的最佳工具是构造函数和析构函数。当您封装每个资源时,您最好在一段时间内在处理程序对象中释放。在这种情况下,std::vector
已经满足您的需求。
所以这是一个完整的(但未经测试的)最通用的解决方案。我从一个自由函数开始创建随机数:
template<typename T> struct dist{
using uniform = std::uniuniform_int_distribution<T>;
};
template<> struct dist<float> {
using uniform = std::uniuniform_real_distribution<float>;
};
template<typename T>
std::vector<T> createRandomNumbers(size_t s) {
auto e1 = std::default_random_engine{std::random_device{}()};
auto u = dist<T>::uniform{0, static_cast<T>(s)};
auto r = std::vector<T>(s, 0);
for( auto& i: r ) i = u(e1);
return r;
}
// fraction need a specialization
template<>
std::vector<fraction> createRandomNumbers<fraction>(size_t s) {
auto e1 = std::default_random_engine{std::random_device{}()};
auto u = dist<int>::uniform{0, static_cast<int>(s)};
auto r = std::vector<fraction>(s, 0);
for( auto& i: r ) {
i.setNumerator(u(e1));
i.setDenominator(u(e1));
}
return r;
}
现在我们实现了一个类似你的Collection
类模板,如果我们真的需要它:
template <typename T> Collection {
private:
// this will handle all your memory management needs
std::vector<T> randoms;
public:
Collection(size_t s) :
randoms{createRandomNumbers<T>(s)}
{};
createNewRandoms(size_t s) {
std::swap(randoms, createRandomNumbers<T>(s));
};
// whatever else is necessary
};
答案 1 :(得分:0)
你为什么要这样做,让你的生活变得更加艰难? 它可以这么简单:
#include <iostream>
#include <chrono>
#include <random>
template<class type_t, std::size_t size>
class Array
{
private:
type_t arr[size];
public:
Array()
{
for (std::size_t i = 0; i < size; ++i)
{
//nice C++ random number generation
auto seed = static_cast<unsigned>(std::chrono::system_clock::now().time_since_epoch().count());
std::minstd_rand0 randm(seed);
arr[i] = randm();
}
}
//test function
void print()
{
for (int i = 0; i < size; ++i)
std::cout << arr[i] << " ";
}
};
int main()
{
Array<int, 4> arr;
arr.print();
std::cin.get();
}
尝试摆脱C风格的C ++。加入黑暗面。
答案 2 :(得分:0)
注意:我不会评论您对C函数的使用或其他问题。其他人已经告诉过你如何避免它们。
类模板中的type
成员以某种方式违背了泛型编程的目的,不是吗?您应该删除type
并将switch
替换为模板专业化。
这是一个简单的例子,可以帮助您入门:
// Collection for all T except of `fraction`
template <class T> class Collection
{
private:
int size;
T* Arr;
int Case;
public:
void ArrayGenerating() {
Arr = new T[size];
for (int i = 0; i < size; i++) {
srand((unsigned)time(NULL));
Arr[i] = static_cast <T> (rand()) % size;
}
};
};
// Collection for `fraction`
template <> class Collection<fraction>
{
private:
int size;
fraction* Arr;
int Case;
public:
void ArrayGenerating() {
Arr = new fraction[size];
for (int i = 0; i < size; i++) {
srand((unsigned)time(NULL));
Arr[i].setNumerator(static_cast <int> (rand()) % size);
srand((unsigned)time(NULL));
Arr[i].setDenominator(static_cast <int> (rand()) % size);
}
}
};
这是最简单的模板专业化,但可能导致大量代码重复。不过,你也可以解决这个问题。例如,您可以将所有公共部分提取到公共私有基类中,如下所示:
namespace detail
{
template <class T> class CollectionBase
{
protected:
int size;
T* Arr;
int Case;
};
}
template <class T> class Collection : detail::CollectionBase<T>
{
public:
void ArrayGenerating() {
Base::Arr = new T[Base::size];
for (int i = 0; i < Base::size; i++) {
srand((unsigned)time(NULL));
Base::Arr[i] = static_cast <T> (rand()) % Base::size;
}
};
private:
using Base = detail::CollectionBase<T>;
};
template<> class Collection<fraction> : detail::CollectionBase<fraction>
{
public:
void ArrayGenerating() {
Base::Arr = new fraction[Base::size];
for (int i = 0; i < size; i++) {
srand((unsigned)time(NULL));
Arr[i].setNumerator(static_cast <int> (rand()) % size);
srand((unsigned)time(NULL));
Arr[i].setDenominator(static_cast <int> (rand()) % size);
}
}
private:
using Base = detail::CollectionBase<fraction>;
};
通常,阅读有关模板专业化的更多信息,您一定会找到正确的解决方案: