C++ basic array assignment is not working - c++

Question

So I have a function
f(int D[],int A[],int len){
for(int k = 0; k < len; k++)
D[k] = A[k];
and if I output D the numbers are all wrong. The function f is called with D initialised as int* D = new int[100000]; in the main function and A is all good because I output it in the function and it looks ok. So... can't understand where the problem is... I also tried memcpy(D+k,A+k,sizeof(int)); and it doesn't work.

Answer 1

Your loop works perfectly. The problem must be somewhere else in your code.
Here is an example program which copies the data in three different ways: a for loop, memcpy, and std::copy:
#include <algorithm>
#include <cstring>
#include <iostream>
#include <iterator>
void copy1(int D[], int A[], int len) {
for(int k = 0; k < len; k++)
D[k] = A[k];
}
void copy2(int D[], int A[], int len) {
std::memcpy(D, A, len*sizeof(int));
}
void copy3(int D[], int A[], int len) {
std::copy(A, A+len, D);
}
int main () {
int a[10] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
int *d = new int[10];
std::ostream_iterator<int> out(std::cout, ",");
// First, print the initial values
std::copy(d, d+10, out);
std::cout << "\n";
// Next do the copies and print the values again
copy1(d, a, 10);
std::copy(d, d+10, out);
std::cout << "\n";
copy2(d, a, 10);
std::copy(d, d+10, out);
std::cout << "\n";
copy3(d, a, 10);
std::copy(d, d+10, out);
std::cout << "\n";
}
The output I get is:
0,0,0,0,0,0,0,0,0,0,
1,2,3,4,5,6,7,8,9,10,
1,2,3,4,5,6,7,8,9,10,
1,2,3,4,5,6,7,8,9,10,

Answer 2

Run your code in debugger and see if you do not have garbage in A[] in the first place (it could go wrong after function call). Also I suggest you pass reference (like int & D[] and const int & A[] — const to prevent altering of A).

Answer 3

Start with a minimal working example such as the following, then integrate your other code into it until something breaks. That’ll be the source of your error. Without more information, this is pretty much all I can tell you; when you encounter wrong values in a C++ program, you’re likely reading from an uninitialised variable. I suggest you try stepping through your program in a debugger such as GDB.
#include <algorithm>
#include <iostream>
#include <iterator>
void f(int D[], const int A[], int len){
for(int k = 0; k < len; k++)
D[k] = A[k];
}
int main(int argc, char** argv) {
int A[] = { 1, 2, 3, 4, 5 };
int D[5];
f(D, A, 5);
std::copy(A, A + 5, std::ostream_iterator<int>(std::cout, " "));
std::cout << '\n';
std::copy(D, D + 5, std::ostream_iterator<int>(std::cout, " "));
return 0;
}
Unless you have a very good reason to do so, prefer std::vector and std::array over raw arrays. Learning how arrays and pointers work is a good reason to use them, though.

Answer 4

I believe the problem is that you are passing in a pointer to an array
int* D = new int[100000];
however, your function takes two integer arrays, which is not the same as a pointer to an array.
Here's a SSCCE of a snippet that behaves like you want it:
#include <iostream>
using namespace std;
void f(int * d, int * a, int length){
for(int k = 0; k < length; k++){
d[k] = a[k];
}
}
int main() {
int* a = new int[9];
for(int i = 0; i < 9; i++){a[i] = i;}
int* d = new int[9];
f(d, a, 9);
for(int i = 0; i < 9; i++){
cout << d[i] << " ";
}
}