Question

I’m learning to use pointer to copy char array.

I have the following code in C++. What I’m trying to do is to transfer and array (set1) using pointer to another pointer array (temp).

But when I try to print out (temp), it is not the same as (set1).

Transfer an array via pointer to another temp array pointer.

#include <iostream>
#include <cstdlib>

using namespace std;

int main()
{
    char set1[] = "ABC";

    char* p = &set1[0];

    int tempSize = 0;
    char* temp = new char[256];

    for (int i = 0; i < 3; i++)
    {
        *temp = *p;
        cout << *temp;   // ABC
        ++temp;
        ++tempSize;

        ++p;
    }

    cout << "\n";

    for (int i = 0; i < tempSize; i++)
    {
        cout << temp[i]; // Why ABC is not printed?
    }

    delete [] temp;

    return 0;
}

Answer 1

// Why ABC is not printed?

Because your pointer is travelling in undefined behavior region:

char* temp = new char[256];
...
++temp;  // gone !!

On top of that,

1. you are not terminating the string with \0 in the end (may not be needed in your code)
2. delete[]ing this corrupt pointer in the end.

Since you are writing for learning purpose, I would suggest simple fix to your code:

char* const temp = new char[256];
      ^^^^^ ensures `temp` is not modifiable

Now use temp[i] for traversing purpose.