c++ (Type) property name

Question

void* SendMsg(void* msg) {
    char *result;
    char tmpBuf[1000]={0};
    message send_msg =*(message*)msg;//?

The problem is * and (*)type.

(message)"parameterName";

"message" is custom made struct. I know "parameterName*" is pointer, receives address. but "*(parameter)" phrase gets value from address. How I should call this?

Leave a space between `=` and `*` in memoriam for the pre-historic C operators =*, =+, etcetera. — Joop Eggen, Oct 28 '21 at 07:32
I would say `declare a new variable called send_msg of type message and initialize it with msg, cast to message star dereferenced`. — mch, Oct 28 '21 at 07:33
In data streams (file or net), data is often serialized as bytes (for which `char` was used in C). You then send the type-erased pointer (`void*`) and the length. To convert such a pointer back to the original type here, it's first cast to a pointer of that type `message*` and then dereferenced, using the `*`-operator. — JHBonarius, Oct 28 '21 at 07:42
`(message*)msg` is a cast from type `void*` to type `message*` (pointer to message). Then: `*` is access operator - it gives you the access to the value pointed by a pointer. Take a look also [here](https://stackoverflow.com/questions/10061332/c-asterisks-ampersand-and-star) — pptaszni, Oct 28 '21 at 07:54

stefaanv · Answer 1 · 2021-10-28T07:37:08.353

1

The code is in C-style whwere a void pointer is passed to the function. To use it it is cast in C-style to message* and immediately dereferenced (in this case assigned to a local variable named send_msg).

This code in C++ is technically known as a red herring for possible issues.

In C++, I would expect a base-class (interface) as parameter or a template parameter.

edited Oct 28 '21 at 07:37

answered Oct 28 '21 at 07:32

stefaanv

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so , this is same meaning to "*message Parameter". Am I right? – I am not englishman Oct 28 '21 at 10:03
Yes, if the parameter was passed as `message* param`, then `message send_msg = *param;` would be the same. Of course, with the code in the question things could go wrong if `void*` doesn't really point to a "message" struct. – stefaanv Oct 28 '21 at 12:16

score 1 · Answer 2 · answered Oct 28 '21 at 14:01

(message*)msg says "pretend that msg is a pointer to an object of type message. message send_msg = *(message*)msg; says "pretend that msg is a pointer to an object of type message and copy the message object that it points at into send_msg. That will work fine if msg in fact points at an object of type message.

c++ *(Type*) property name

2 Answers2

c++ (Type) property name