> go generic full speed What does this mean? As a simple example, suppose I use malloc to allocate memory (because I don't want to allocate with new which wou

> go generic full speed What does this mean? As a simple example, suppose I use malloc to allocate memory (because I don't want to allocate with new which would oblige me to handle exceptions and prevent me from handing ownership over to a C client or using an allocator provided by a C client (without extra work)). I have to cast the return value as in struct Foo *x = (struct Foo*)malloc(n*sizeof(*x)); instead of struct Foo *x = malloc(n*sizeof(*x)); This looks purely cosmetic, but in C, I can write the macro #define MallocA(n,p) (!(*(p) = malloc((n)*sizeof(**(p)))) which then works with a standard error checking convention struct Foo *arr; err = MallocA(n,&arr);CHK(err); This doesn't work in C++ without non-portable typeof or evil and less safe (not conforming for function pointers) *(void**)(p) = malloc((n)*sizeof(**(p))) Similarly, if a client registers a callback with a context, I store their context in a void* and pass it back to them int UserCallback(void *ctx,...) { struct User *user = (struct User*)ctx; instead of int UserCallback(void *ctx,...) { struct User *user = ctx; I understand that this is just cosmetic. I don't see how "going generic full speed" helps with this. Also note that aggregate returns are slower for all but trivially small structures, and downright bad for big structures.