void* cast to char* not lvalue?

Question

I'm trying to make a function that takes a void*, copies some memory to it, and then moves the pointer. Since it is a void pointer, I thought I'd cast it to char* and move that, like so:

PVOID SendAndMoveHead(PVOID dest, const Message& message, const size_t& size)
{
    PVOID ret = CopyMemory(dest, (PVOID)message.msg.c_str(), size);
    ((char*)dest) += size;
    return ret;
}

However, VS complains about ((char*)dest) saying

expression must me a modifiable lvalue

which I thought it was, since the following works:

PVOID SendAndMoveHead(PVOID dest, const Message& message, const size_t& size)
{
    PVOID ret = CopyMemory(dest, (PVOID)message.msg.c_str(), size);
    char* d = (char*)dest;
    d += size;
    return (PVOID)d;
}

If someone could shed some light on why the first version shouldnt work I'd really appreciate it.

Possible duplicate of [Why does an lvalue cast work?](https://stackoverflow.com/questions/15343276/why-does-an-lvalue-cast-work) — Adrian Shum, Sep 08 '17 at 08:43
@nwp - `const Message&` and `message.msg.c_str()` leaves no room for argument. — StoryTeller - Unslander Monica, Sep 08 '17 at 08:56

score 2 · Accepted Answer · answered Sep 08 '17 at 08:40

((char*)dest) gives you a new temporary char *. ((char*)dest) += size; would change the temporary and have no effect, which causes a compilation failure.

In the second example d is not a temporary and lives long enough to get returned.

Alternatively you could write return (char*)dest + size;.

void* cast to char* not lvalue?

1 Answers1