6 ms·

Another option is a struct with a FAM at the end. typedef struct { size_t len; uint8_t data[]; } ByteBuf; Then, allocation becomes ByteBuf

Another option is a struct with a FAM at the end. typedef struct { size_t len; uint8_t data[]; } ByteBuf; Then, allocation becomes ByteBuf *b = malloc(sizeof(*b) + sizeof(uint8_t) * array_size); b->len = array_size; and data is no longer a pointer.

ATsch 6y ago
Well, your ByteBuf is still a pointer. You also now need to dereference it to get the length. It also can't be passed by value, since it's very big. You can also not have multiple ByteBufs pointing at subsections of the same region of memory. Thing is, you rarely want to share just a buffer anyway. You probably have additional state, locks, etc. So what I do is embed my ByteBuf directly into another structure, which then owns it completely: typedef struct { ... ByteBuf mybuffer; ... } SomeThing; So we end up with the same amount of pointers (1), but with some unique advantages.
- saagarjha 6y ago
  sizeof(ByteBuf) == sizeof(size_t), and you can pass it by value; I just don't think you can do anything useful with it because it'll chop off the data.
- kkdwivedi 6y ago
  Right, totally depends on what you're doing. My example is not a good fit for intrusive use cases.
kevin_thibedeau 6y ago
This will an alignment problem on any platform with data types larger than size_t. You'd need an alignas(max_align_t) on the struct. At which point some people are going to be unhappy about the wasteful padding on a memory constrained target.