/* Code for exercise 22.8. |  | "C++ Solutions--Companion to The C++ Programming Language, Third Edition" | by David Vandevoorde; Addison-Wesley 1998; ISBN 0-201-30965-3. | | Permission to use, copy, modify, distribute and sell this software | and its documentation for any purpose is hereby granted without fee. | The author makes no representations about the suitability of this | software for any purpose.  It is provided "as is" without express or | implied warranty. `----------------------------------------------------------------------*/#include <assert.h>#include <stddef.h>#include <string.h>struct SimpleArray {   SimpleArray(): s_(0) {}   SimpleArray(ptrdiff_t s): a_(new double[s]), s_(s) {}   SimpleArray(SimpleArray const &a)      : a_(new double[a.s_]), s_(a.s_) {      copy(a);   }   ~SimpleArray() { if (s_!=0) delete[] a_; }   SimpleArray& operator=(SimpleArray const&a) {      if (&a!=this)         copy(a);      return *this;   }   ptrdiff_t size() const { return s_; }   void size(ptrdiff_t s) {      assert(s_==0 and s>0);      s_ = s;      a_ = new double[s];   }   double const& operator[](ptrdiff_t k) const      { return a_[k]; }   double& operator[](ptrdiff_t k) { return a_[k]; }   void copy(SimpleArray const &a) {      memcpy(a_, a.a_, s_*sizeof(double));   }private:   double *a_;   ptrdiff_t s_;};SimpleArray operator+(SimpleArray const &a,                      SimpleArray const &b) {   SimpleArray result(a.size());   for (ptrdiff_t k = 0; k!=a.size(); ++k)      result[k] = a[k]+b[k];   return result;}SimpleArray operator*(SimpleArray const &a,                      SimpleArray const &b) {   SimpleArray result(a.size());   for (ptrdiff_t k = 0; k!=a.size(); ++k)      result[k] = a[k]*b[k];   return result;}SimpleArray operator*(double a, SimpleArray const &b) {   SimpleArray result(b.size());   for (ptrdiff_t k = 0; k!=b.size(); ++k)      result[k] = a*b[k];   return result;}int main() {   SimpleArray x(10000), y(10000), z(10000);   for (int i = 0; i<10000; ++i) {      x[i] = y[i] = z[i] = 1.0/i;   }   for (int i = 0; i<10000; ++i) {      x = 0.5*(x+y)+z;   }   return 0;}