Question

char *recvmsg(){
    char buffer[1024];
    return buffer;
}

int main(){
    char *reply = recvmsg();
    .....
}

I get a warning:

warning C4172: returning address of local variable or temporary

Answer 1

You need to dynamically allocate your char array:

char *recvmsg(){
   char* buffer = new char[1024];
   return buffer;
}

for C++ and

char *recvmsg(){
   char* buffer = malloc(1024);
   return buffer;
}

for C.

What happens is, without dynamic allocation, your variable will reside on the function’s stack and will therefore be destroyed on exit. That’s why you get the warning. Allocating it on the heap prevents this, but you will have to be careful and free the memory once done with it via delete[].