我有一个高度可配置的类,有许多模板参数,如下所示：
template<bool OptionA = false,bool OptionB = false,bool OptionC = false,class T = Class1B>
class MyClass
{
}
现在,如果我想创建类类型并且我只想将OptionB设置为true,我必须执行以下操作：
MyClass< false,true>
特别是对于许多模板参数,这变得很麻烦.
不,我的问题是,是否有任何可用于使用构建器模式创建基于模板的类类型的示例？
我正在寻找这样的东西：
class Builder
{
uSEOptionA();
uSEOptionB();
uSEOptionC();
useClass2B(); //instead of Class1B
create();
}
最后,调用Builder.uSEOptionB().uSEOptionC().useClass2B.create()应该返回MyClass< false,true,Class2B>.这可能吗？
编辑：向模板参数列表添加了类.
解决方法
正如其他人所说,做你想做的最简单的方法是使用enum而不是Builder.但是,如果你想要一个Builder,你可以尝试这样的事情：
template<bool OptionA = false,typename T = Class1B>
struct Builder_t
{
Builder_t() = default;
// ~Builder_t() { std::cout << "Builder dtor." << std::endl; }
auto uSEOptionA() -> Builder_t<true,OptionB,OptionC,T> { return {}; }
auto uSEOptionB() -> Builder_t<OptionA,T> { return {}; }
auto uSEOptionC() -> Builder_t<OptionA,T> { return {}; }
auto useClass2B() -> Builder_t<OptionA,Class2B> { return {}; }
MyClass<OptionA,T> create() { return {}; }
};
using Builder = Builder_t<>;
// ...
// Build MyClass<true,false,Class2B>:
auto ma2 = Builder{}.uSEOptionA().useClass2B().create();
这会导致每个函数返回一个不同的Builder,其模板将被下一个函数使用;最终模板用作MyClass的模板.每个函数都修改其指定的模板参数,允许编译时版本的Builder模式.但是,它确实有成本,如果取消注释用户定义的析构函数,这一点就变得很明显了.
考虑这个简单的测试程序：
#include <iostream>
#include <typeinfo>
class Class1B {};
class Class2B {};
template<bool OptionA = false,typename T = Class1B>
class MyClass
{
public:
MyClass() {
std::cout << "MyClass<"
<< OptionA << ","
<< OptionB << ","
<< OptionC << ","
<< "type " << typeid(T).name() << ">"
<< std::endl;
}
};
template<bool OptionA = false,T> create() { return {}; }
};
using Builder = Builder_t<>;
int main()
{
std::cout << std::boolalpha;
std::cout << "Default:\n";
std::cout << "Direct: ";
MyClass<> m;
std::cout << "Builder: ";
auto mdefault = Builder{}.create();
std::cout << std::endl;
std::cout << "Builder pattern:\n";
std::cout << "A: ";
auto ma = Builder{}.uSEOptionA().create();
std::cout << "C: ";
auto mc = Builder{}.uSEOptionC().create();
std::cout << "---\n";
std::cout << "AB: ";
auto mab = Builder{}.uSEOptionA().uSEOptionB().create();
std::cout << "B2: ";
auto mb2 = Builder{}.uSEOptionB().useClass2B().create();
std::cout << "---\n";
std::cout << "ABC: ";
auto mabc = Builder{}.uSEOptionA().uSEOptionB().uSEOptionC().create();
std::cout << "AC2: ";
auto mac2 = Builder{}.uSEOptionA().uSEOptionC().useClass2B().create();
std::cout << "---\n";
std::cout << "ABC2: ";
auto mabc2 = Builder{}.uSEOptionA().uSEOptionB().uSEOptionC().useClass2B().create();
}
通常,输出如下(使用GCC)：
Default: Direct: MyClass<false,type 7Class1B> Builder: MyClass<false,type 7Class1B> Builder pattern: A: MyClass<true,type 7Class1B> C: MyClass<false,type 7Class1B> --- AB: MyClass<true,type 7Class1B> B2: MyClass<false,type 7Class2B> --- ABC: MyClass<true,type 7Class1B> AC2: MyClass<true,type 7Class2B> --- ABC2: MyClass<true,type 7Class2B>
但是,如果我们取消注释析构函数……
Default: Direct: MyClass<false,type 7Class1B> Builder dtor. Builder pattern: A: MyClass<true,type 7Class1B> Builder dtor. Builder dtor. C: MyClass<false,type 7Class1B> Builder dtor. Builder dtor. --- AB: MyClass<true,type 7Class1B> Builder dtor. Builder dtor. Builder dtor. B2: MyClass<false,type 7Class2B> Builder dtor. Builder dtor. Builder dtor. --- ABC: MyClass<true,type 7Class1B> Builder dtor. Builder dtor. Builder dtor. Builder dtor. AC2: MyClass<true,type 7Class2B> Builder dtor. Builder dtor. Builder dtor. Builder dtor. --- ABC2: MyClass<true,type 7Class2B> Builder dtor. Builder dtor. Builder dtor. Builder dtor. Builder dtor.
Builder_t :: create()之前的每个调用都会创建一个不同的Builder_t,所有这些都会在创建实例后被销毁.这可以通过使Builder_t成为constexpr类来减轻,但如果有大量参数要处理,则可能会减慢编译速度：
template<bool OptionA = false,typename T = Class1B>
struct Builder_t
{
// Uncomment if you want to guarantee that your compiler treats Builder_t as constexpr.
// size_t CompTimeTest;
constexpr Builder_t()
// Uncomment if you want to guarantee that your compiler treats Builder_t as constexpr.
// : CompTimeTest((OptionA ? 1 : 0) +
// (OptionB ? 2 : 0) +
// (OptionC ? 4 : 0) +
// (std::is_same<T,Class2B>{} ? 8 : 0))
{}
constexpr auto uSEOptionA() -> Builder_t<true,T> { return {}; }
constexpr auto uSEOptionB() -> Builder_t<OptionA,T> { return {}; }
constexpr auto uSEOptionC() -> Builder_t<OptionA,T> { return {}; }
constexpr auto useClass2B() -> Builder_t<OptionA,Class2B> { return {}; }
constexpr MyClass<OptionA,T> create() { return {}; }
};
using Builder = Builder_t<>;
// ....
// Uncomment if you want to guarantee that your compiler treats Builder_t as constexpr.
// char arr[Builder{}/*.uSEOptionA()/*.uSEOptionB()/*.uSEOptionC()/*.useClass2B()/**/.CompTimeTest];
// std::cout << sizeof(arr) << '\n';
