Struct hack equivalent in C++

Question

The struct hack where you have an array of length 0 as the last member of a struct from C90 and C99 is well known, and with the introduction of flexible array members in C99

Answer 1

C++ does not have the concept of "flexible arrays". The only way to have a flexible array in C++ is to use a dynamic array - which leads you to use int* things. You will need a size parameter if you are attempting to read this data from a file so that you can create the appropriate sized array (or use a std::vector and just keep reading until you reach the end of the stream).

The "flexible array" hack keeps the spatial locality (that is has the allocated memory in a contiguous block to the rest of the structure), which you lose when you are forced to use dynamic memory. There isn't really an elegant way around that (e.g. you could allocate a large buffer, but you would have to make it sufficiently large enough to hold any number of elements you wanted - and if the actual data being read in was smaller than the buffer, there would be wasted space allocated).

Also, before people start suggesting keeping an int* to a separately allocated piece of memory in the struct instead, that is not a satisfactory answer. I want to allocate a single piece of memory to hold both my struct and the array elements. Using a std::vector also falls into the same category.

That is the way you would do it in C++. You can down-vote it all you want, but the fact remains: a non-standard extension is not going to work when you move to a compiler that does not support it. If you keep to the standard (e.g. avoid using compiler-specific hacks), you are less likely to run into these types of issues.