Is it possible to splice buffers in C++ with zero copy?

Question

Suppose I have two buffers:

uint8_t* buf1[100];
uint8_t* buf2[10];
uint8_t* buf3[90];

Where buf1 is full of data and I need to pass 10 bytes of this data to buf2 and the rest to buf3. Is there a way to do this without copying?

If not, is there a high level library (like std::vector) in which splicing is possible without copying?

[std::list::splice](https://en.cppreference.com/w/cpp/container/list/splice) — Aykhan Hagverdili, Dec 07 '19 at 05:07
Those are not buffers with data. They are arrays of pointers that aren't initialized.. — doug, Dec 07 '19 at 05:07
What does it mean to "pass 10 bytes of data" to an array of uint8_t's ? Assuming it doesn't mean copying, perhaps what you want is something like `uint8_t* buf1[100]; uint8_t * buf2 = &buf1[0]; uint8_t * buf3 = &buf[10];`. ? — Jeremy Friesner, Dec 07 '19 at 05:26

L. F. · Accepted Answer · 2019-12-07T08:45:56.083

6

No, this is not possible. The three buffers are distinct objects with different addresses. Therefore, some form of copying is necessary.

You can use a span:

span<uint8_t*, 10> buf2(buf1, 10);
span<uint8_t*, 90> buf3(buf1 + 10, 90);

span is not in the standard library as of C++17, but it is available in the GSL. See What is a "span" and when should I use one?.

edited Dec 07 '19 at 08:45

answered Dec 07 '19 at 05:13

L. F.

1 Answers1