Lambda表達式鍛鍊

Question

我一直在試圖更多地瞭解lambda表達式最近，和有趣的練習的想法...

是有辦法簡化C++集成功能是這樣的：

// 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; 
} 

通過使用c＃和lambda表達式來使用

？

Answer 1

這個怎麼樣：

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; 
} 

測試：

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

Answer 2

真正的力量來自，如前所述，調用它時。例如，在C＃

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; 
    } 

然後：

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... 

Answer 3

LAMBDA POWA！不知道這是否是正確的（沒有C＃程序員！只是喜歡它的拉姆達的東西）

(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; 
} 

好了，我想，而代碼是C++，爲什麼不把它的C++和獲得拉姆達？這裏是它如何尋找C++ 0x，希望很快作爲標準發佈：

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; }); 
}