C++: What is wrong with the output? - c++

Question

This code should output 0 0.25 0.5 0.75 1, instead it outputs zeros. Why is that?
Define a function 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];
}
Here is the implementation:
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

You can't use sizeof(x) because that will return the size of the pointer, not the array you thought you passed to it. You have to specify the size with a third parameter or use something more convenient like an std::vector and use size().

Answer 2

This works.
#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;
}
Note that I added const int& N to pde_advect_IC() in order to pass it the size of the array, by const reference, to be sure it does not get modified by mistake.
Note that your trick with sizeof() does not work with pointers.