size_t in Go without using cgo

Asked May 23 '18 at 05:24

Active May 23 '18 at 06:29

Viewed 2,572 times

This is a C struct that I want to port into Go struct:

struct InputBuffer_t {
   char* buffer;
   size_t buffer_length;
   ssize_t input_length;
};

But there is a way to declare buffer_length variable without using C.size_t cgo pointer in Go.

This is a concern for portability. If I write the Go struct in this way, is it will be portable?

type InputBuffer struct {
   Buffer        string
   BufferLength uint32
   InputLength  uint32
};

edited May 23 '18 at 06:29

asked May 23 '18 at 05:24

Shajal Ahamed

1

It's portable. Although your variable names are not idiomatic Go :) – Jonathan Hall May 23 '18 at 06:17
1

Notice that in 64 bit systems, size_t is not uint32 but uint64, and ssize_t is equivalent to int64 – Not_a_Golfer May 23 '18 at 07:15
4

If you want a type that changes based on the bitness of the target system, just use `int`. – Jonathan Hall May 23 '18 at 07:41

0 Answers0