Question

我最近一直试图更多地了解lambda表达，并想到了一个有趣的练习......

有没有办法简化这样的c ++集成函数：

// Integral Function
double integrate(double a, double b, double (*f)(double))
{
    double sum = 0.0;

    // Evaluate integral{a,b} f(x) dx
    for(int n = 0 ; n <= 100; ++n)
    {
        double x = a + n*(b-a)/100.0;
        sum += (*f)(x) * (b-a)/101.0;
    }
    return sum;
}

使用c＃和lambda表达式？

Answer 1

这个怎么样：

public double Integrate(double a,double b, Func<double, double> f)
{
    double sum = 0.0;

    for (int n = 0; n <= 100; ++n)
    {
        double x = a + n * (b - a) / 100.0;
        sum += f(x) * (b - a) / 101.0;
    }
    return sum;
}

测试：

    Func<double, double> fun = x => Math.Pow(x,2);        
    double result = Integrate(0, 10, fun);

Answer 2

Lambda Powa！不确定这是否正确（没有C＃程序员！只是喜欢它的lambda东西）

(a, b, c) => {
    double sum = 0.0;
    Func<double, double> dox = (x) => a + x*(b-a)/100.0;

    // Evaluate integral{a,b} f(x) dx
    for(int n = 0 ; n <= 100; ++n)
        sum += c(dox(n)) * (b-a)/101.0;

    return sum;
}

好的，所以我认为虽然代码是C ++，为什么不保留它C ++并获得lambda？这就是它如何查找c ++ 0x，希望很快就会以标准形式发布：

static double Integrate(double a, double b, function<double(double)> f)
{
    double sum = 0.0;

    // Evaluate integral{a,b} f(x) dx
    for(int n = 0; n < 100; ++n) {
        double x = a + n * (b - a) / 100.0;
        sum += f(x) * (b - a) / 101.0;
    }
    return sum;
}  

int main() {
    Integrate(0, 1, [](double a) { return a * a; });
}

Answer 3

如上所述，真正的力量来了。例如，在C＃

中

    static double Integrate(double a, double b, Func<double, double> func)
    {
        double sum = 0.0;

        // Evaluate integral{a,b} f(x) dx
        for(int n = 0 ; n <= 100; ++n)
        {
            double x = a + n*(b-a)/100.0;
            sum += func(x) * (b - a) / 101.0;
        }
        return sum;
    }

然后：

    double value = Integrate(1,2,x=>x*x); // yields 2.335
    // expect C+(x^3)/3, i.e. 8/3-1/3=7/3=2.33...

Lambda表达练习

3 个答案: