代码之家 › 专栏 › 技术社区 › manuel

现代C中空指针的替代方案++

type-safety polymorphism pointers c++

0

manuel · 技术社区 · 7 月前

我想从一个源中检索数据,通过一组不同的函数通过一个函数提供。我来自c,所以我有这个例子:

#include <iostream>

using namespace std;

void getInt(void *v, const double &src) {*static_cast<int*>(v) = src; }
void getDouble(void *v, const double &src) {*static_cast<double*>(v) = src; }

struct twoInt{int a; int b;};
void getTwoInt(void *v, const double &src)
{
    static_cast<twoInt*>(v)->a = src;
    static_cast<twoInt*>(v)->b = static_cast<int>(src*10) % 10;
}

template <typename T>
void getData(T f, void* val)
{
    double dataSource = 2.1;
    f(val, dataSource);
}


int main()
{
    int valI;
    double valD;
    twoInt valTwoInt;

    getData(getInt, static_cast<void*>(&valI));
    cout << "Int: " << valI << endl;

    getData(getDouble, static_cast<void*>(&valD));
    cout << "Double: " << valD << endl;

    getData(getTwoInt, static_cast<void*>(&valTwoInt));
    cout << "Int1: " << valTwoInt.a << ", Int2: " << valTwoInt.b <<endl;
}

但我想用一种现代的C++方式来做,没有空指针,以获得更多的类型安全性。

1 回复 | 直到 7 月前

1

Jarod42 7 月前

你不需要在这里删除类型,所以正确的类型和模板可以完成这项工作:

void getInt(int& out, double src) { out = src; }
void getDouble(double& out, double src) { out = src; }

struct twoInt{int a; int b;};
void getTwoInt(twoInt& v, double src)
{
    v.a = src;
    v.b = static_cast<int>(src * 10) % 10;
}

template <typename F, typename T>
void getData(F f, T& val)
{
    double dataSource = 2.1;
    f(val, dataSource);
}

int main()
{
    int valI;
    getData(getInt, valI);
    std::cout << "Int: " << valI << std::endl;

    double valD;
    getData(getDouble, valD);
    std::cout << "Double: " << valD << std::endl;

    twoInt valTwoInt;
    getData(getTwoInt, valTwoInt);
    std::cout << "Int1: " << valTwoInt.a << ", Int2: " << valTwoInt.b <<std::endl;
}

Demo