Lambda expression exercise

I have been trying to learn more about lambda expressions lately, and thought of a interesting exercise...

is there a way to simplify a c++ integration function like this:

// 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;
}

by using c# and lambda expressions?

From stackoverflow

• What about this:

  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;
  }
  

  Test:

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

  Marc Gravell : hehe - within seconds of each other ;-p

• The real power comes, as stated, when calling it. For example, in 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;
      }
  

  Then:

      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 Powa! Not sure whether this is right (No C# programmer! Just liking its lambda stuff)

  (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;
  }
  

  Ok, so i think while the code is C++, why not keep it C++ and get lambda in? Here it is how it looks for c++0x, being hopefully released as a Standard very soon :

  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; });
  }