Question

此代码应输出0 0.25 0.5 0.75 1，而不是输出零。那是为什么？

定义函数u（x）= x;

void pde_advect_IC(double* x, double* u)
{
  int N = sizeof(x) / sizeof(x[0]); //size of vector u
  for (int i = 0; i <= N; i++)
    u[i] = x[i];
}

以下是实施：

int main()
{
  double a = 0.0;
  double b = 1.0;
  int   nx = 4;
  double dx = (b - a) / double(nx);
  double xx[nx + 1]; //array xx with intervals

  // allocate memory for vectors of solutions u0
  double* u0 = new double [nx + 1];

  //fill in array x
  for (int i = 0; i <= nx; i++)
    xx[i] = a + double(i) * dx;

  pde_advect_IC(xx, u0); // u0 = x (initial conditions)
  for (int i = 0; i <= nx; i++)
    cout<<u0[i]<<endl;

  // de-allocate memory of u0
  delete [] u0;
  delete [] u1;

  return 0;
}

Answer 1

你不能使用sizeof(x)因为它会返回指针的大小，而不是你认为传递给它的数组。您必须使用第三个参数指定大小，或使用更方便的内容，例如std::vector并使用size()。

Answer 2

这很有效。

void push_front( const Object& item )
{
    Node * new_node = new Node( item );
    ++size;

    if ( size == 1 )
    {
        new_node->next = nullptr;
        first = last = new_node;
        return;
    }

    new_node->next = first;
    first = new_node;
}

请注意，我向#include <iostream> #include <cstdlib> using namespace std; void pde_advect_IC(double* x, double* u, const int& N) { for (int i = 0; i < N; i++) u[i] = x[i]; } int main() { double a = 0.0; double b = 1.0; int nx = 4; double dx = (b - a) / double(nx); double xx[nx + 1]; //array xx with intervals // allocate memory for vectors of solutions u0 double* u0 = new double [nx + 1]; //fill in array x for (int i = 0; i <= nx; i++) xx[i] = a + double(i) * dx; pde_advect_IC(xx, u0, nx + 1); // u0 = x (initial conditions) for (int i = 0; i <= nx; i++) cout << u0[i] << endl; // de-allocate memory of u0 delete [] u0; return 0; }添加了const int& N，以便通过pde_advect_IC()将其传递给数组的大小，以确保它不会被错误地修改。请注意，使用const reference的诀窍不可以使用指针。

C ++：输出有什么问题？

2 个答案: