char* passed to a function

PHP Code:

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BOOL pmRetrieveComboText (HWND hWndCombo, char* dest)
{
    UINT idx;        //current index
    UINT    len;    //length of index text
    char* ret = '\0';        //the value to return

    idx = SendMessage (hWndCombo, CB_GETCURSEL, NULL, NULL);
    len = SendMessage (hWndCombo, CB_GETLBTEXTLEN, idx, NULL);

    if ((idx == -1) || (len == -1))        return 0;

    ret = new char[len + 1];    //allocate the necessary memory

    //fill the allocated array with the item text
    SendMessage (hWndCombo, CB_GETLBTEXT, idx, (LPARAM) ret);
    dest = ret;

    delete[] ret; 
    return TRUE;
} 


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why wont this function change dest? i basically wanna to pass a char* by reference, but it doesm't change anything...can anyone help?
thanks

dest as a pointer is passed by value, and anything passed (in fact all function parameters are always passed by value and will be free'd from the stack one the function exits, the only changes the function can do to external is data is by passing pointers to the data to be manipulated) to return a pointer as data you would need a pointer to a pointer to a char.

What you are intending to do is allocating a data within a function without guarrantee that it will be deallocated, this is symptoms of earlier experience with java or COM objects in VB.

What you should do is use a string instead or do the operations directly on char and let the calling function be responsible for having the memory allocated, or alternatively pass the length of the text gained in a separate function call

Code:

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    char *buf = *dest;
    char *s =    *ret;
    while(s!=0x0) *buf++=*s++;
    s=0x0;

    delete[] ret;

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*sigh*

Someone has been using C way too much :p