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double riemann(double (*pf)(double t), double a, double b, int n) { double s = 0, h = (b-a)/n, x; int i; for (x = a, i = 0; i < n; x += h, i++) s += (*pf)(x); return s*h; } double cube(double t) { return t*t*t; } 3.24 3.9204 3.992 3.9992 In this test run, we are integrating the function y = x3 over the interval [0, 2]. By elementary calculus, the value of this integral is 4.0. The call riemann(cube,0,2,10) approximates this integral using 10 subintervals, obtaining 3.24. The call riemann(cube,0,2,100) approximates the integral using 100 subintervals, obtaining 3.9204. These sums get closer to their limit 4.0 as n increases. With 10,000 subintervals, the Riemann sum is 3.9992. Note that the only significant difference between this riemann() function and the sum() function in Example 7.18 is that the sum is multiplied by the subinterval width h before being returned. This derivative() function is similar to the sum() function in Example 7.18, except that it implements the formula for the numerical derivative instead. It has three arguments: a pointer to the function f, the x value, and the tolerance h. In this test run, we pass it (pointers to) the cube() function and the sqrt() function. #include <iostream> #include <cmath> using namespace std; double derivative(double (*)(double), double, double); double cube(double); int main() { cout << derivative(cube, cout << derivative(cube, cout << derivative(cube, cout << derivative(sqrt, cout << derivative(sqrt, cout << derivative(sqrt, }
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// Returns an approximation to the derivative f'(x): double derivative(double (*pf)(double t), double x, double h) { return ((*pf)(x+h) - (*pf)(x-h))/(2*h); } double cube(double t) { return t*t*t; }
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3.01 3.0001 3 0.500628 0.500006 0.5 The derivative of the cube() function x3 is 3x2, and its value at x = 1 is 3, so the numerical derivative should be close to 3.0 for small h. Similarly, the derivative of the sqrt() function x is 1 ( 2 x ) , and its value at x = 1 is 1/2, so its numerical derivative should be close to 0.5 for small h. The pointer pmax is used to locate the maximum float. It is initialized to have the same value as p[0] which points to the first float. Then inside the for loop, the float to which p[i] points is compared to the float to which pmax points, and pmax is updated to point to the larger float when it is detected. So when the loop terminates, pmax points to the largest float: float* max(float* p[], int n) { float* pmax = p[0]; for (int i = 1; i < n; i++) if (*p[i] > *pmax) pmax = p[i]; return pmax; } void print(float [], int); void print(float* [], int); int main() { float a[8] = {44.4, 77.7, 22.2, 88.8, 66.6, 33.3, 99.9, 55.5}; print(a, 8); float* p[8]; for (int i = 0; i < 8; i++) p[i] = &a[i]; // p[i] points to a[i] print(p, 8); float* m = max(p, 8); cout << m << ", " << *m << endl; } 44.4, 77.7, 22.2, 88.8, 66.6, 33.3, 99.9, 55.5 44.4, 77.7, 22.2, 88.8, 66.6, 33.3, 99.9, 55.5 0x3fffcd4, 99.9 Here we have two (overloaded) print() functions: one to print the array of pointers, and one to print the floats to which they point. After initializing and printing the array a, we define the array p and initialize its elements to point to the elements of a. The call print(p, 8) verifies that p provides indirect access to a. Finally, the pointer m is declared and initialized with the address returned by the max() function. The last output verifies that m does indeed point to the largest float among those accessed by p.
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