how to pass char variable thorough a function with pointers(example already sent)?

Question

I am practicing c++ and this code is supposed to copy an array to another array.output types are character and which is passed by pointer through the function 'f'. my output type is pointer which is pointing to a char array.here is my code:

using namespace std;
void f(char*, char*);
int main()
    {
    char *a1;
    char a2[] = "1234";

    f(a1, a2);
    cout << a1<< endl;
    return 0;

    }

void f(char *a, char *b)
    {
    int len;
    len = sizeof(b) / sizeof(b[0]);
    a = new char[len];
    for (; (*a=*b )!= '\0'; a++, b++)
    cout<<*a<<endl;

    for (int i=0;i<4;i++)
    cout<<a[i]<<endl;
    }

1. in the function each time for loop runs 'a' elements changes to the unreadable value. at first, it shows right value and then it changes. for example, in the first loop inside the function, it prints right value of 'a' but in second loop it changes it for no reason!
2. this function doesn't return any value. as I expect by calling by pointer it has to change the variable but nothing happens. deep explanation is appreciated. I am completely new in c++ programming.

Answer 1

Without changing the basic style of your code, here is a fixed version, with comments:

#include <iostream>

using namespace std;
void f(char**, char*); // You need to pass a pointer to the pointer!
void f2(char*&, char*); // Alternative passing the pointer by reference
int main()
{
    char* a1;
    char a2[] = "1234";

//  f(&a1, a2);     // Pass the address of the 'a1' pointer
    f2(a1, a2);     // a1 pointer is now passed by reference
    cout << a1 << endl;
    delete a1[]; // Here, we free the memory allocated by the new call!
    return 0;
}

void f(char** a, char* b)
{
    int len;
//  len = sizeof(b) / sizeof(b[0]); // You can't get the array length like this!
    len = strlen(b); // Assuming your string is null-terminate!
    *a = new char[len + 1]; // Make room for the null terminator!
    // Make a copy of the pointer (atemp) - otherwise you are changing the address of the new character
    // array you just created - and this will then fail in the 2nd loop ...
    for (char* atemp = *a; (*atemp = *b) != '\0'; atemp++, b++)
        cout << *atemp << endl;

    for (int i = 0; i < len; i++) // Length won't always be 4!
//  for (int i = 0; i < 4; i++)
        cout << (*a)[i] << endl;
}

void f2(char*& a, char* b)
{
    size_t len = strlen(b); // Assuming your string is null-terminate!
    a = new char[len + 1];
    for (char* atemp = a; (*atemp = *b) != '\0'; atemp++, b++)
        cout << *atemp << endl;

    for (int i = 0; i < len; i++) // Length won't always be 4!
        cout << a[i] << endl;
}

Feel free to ask for more details about why the changes were made!

EDIT: Added alternative using argument as a reference.

Answer 2

A lot of mistakes in your program:

void f(char *a, char *b)
{
    len = sizeof(b) / sizeof(b[0]);

b is a pointer to char. As this it has the pointer size of the machine type, typically it is 32 or 64 bit. So you never get the size of the array!

If you really want to pass a raw pointer to char to a function and want to handle the data stored at the position where the pointer points to, you have either to pass also the size ( number of elements ) or you have to know an end mark. In case of a c-sytle string you have a '\0' at the end of the string which makes it possible to use it.

So you have first to count the number of elements until '\0' is reached, allocate a new array, copy the elements and than return the pointer.

Next failure: If you allocate memory with new, you have to use delete later in your program.

But: What you did is bad sytle C code! It has really nothing todo with C++ nor with good C. There is no need to write a own string copy function as there is already strcpy and strlen to get the size.

In C++ your code can be something like that:

void f( const std::string& src, std::string& target )
{
    target = src;
}

int main()
{
    std::string a2{"abcd"};
    std::string a1;
    f( a2, a1 );
    std::cout << a1 << std::endl;
}