C++双指针？ 我不知道怎么发音

Question

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

The problem is * and (*)type.

*(message*)"parameterName";

how I can search it? what is that?

"message" is custom made struct.

Answer 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.

Answer 2

(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 .