Rust Strings and C variadic functions

Question

I need pass a vector of Rust string to a C variadic function. But I can't figure out what is the expected(CString, [u8]..) format.

References:

API reference: isc_event_block
My extern C declaration: ibase
How I'm calling: que_events
C example: api16
API implementation: isc_event_block

My api16 example code version:

strcpy (ids[0], "evento");
length = (short)  isc_event_block((char **) &event_buffer, (char **) &result_buffer, 1, ids[0], 0);
printf("event_buffer: '%s' %d\n", event_buffer, sizeof(event_buffer));
printf("result_buffer: '%s' %d\n", result_buffer, sizeof(result_buffer));

Result of version of api16:

event_buffer: 'evento' 8
result_buffer: '' 8

My Rust code:

let mut event_buffer: Vec<u8> = Vec::with_capacity(256);
let mut result_buffer: Vec<u8> = Vec::with_capacity(256);
let mut len = 0;
let names = "evento".to_string();

unsafe {
   len = self.ibase.isc_event_block()(
      event_buffer.as_mut_ptr() as *mut _,
      result_buffer.as_mut_ptr() as *mut _,
      names.len() as u16,
      names.as_ptr()
   );
   event_buffer.set_len(len as usize);
   result_buffer.set_len(len as usize);
}

println!("{:?} {:?}", len, names);
println!("{:x?} {:x?}", event_buffer, result_buffer);
println!("{:?} {:?}", String::from_utf8_lossy(&event_buffer.clone()), String::from_utf8_lossy(&result_buffer.clone()));

Result of my Rust code:

12 ["evento"]
[e0, 4f, 51, 28, a8, 7f, 0, 0, 0, 0, 0, 0] [0, 50, 51, 28, a8, 7f, 0, 0, 0, 0, 0, 0]
"�OQ(�\u{7f}\0\0\0\0\0\0" "\0PQ(�\u{7f}\0\0\0\0\0\0"

I already tried use CString or CStr, like here.

What am I doing wrong?

Aiden4 · Accepted Answer · 2022-12-16 09:08:39Z

3

You're doing multiple things wrong in the rust version. For the first two arguments, you're intended to pass a pointer to a location that can hold a single pointer to a byte buffer. Instead, you're passing in a pointer to a byte buffer, so a pointer is getting written into that buffer, which isn't what you want.

Secondly, the id_count parameter corresponds to the number of strings you're passing as the variadic parameters, rather than the length of a single variadic string, meaning your c code just reads a bunch of uninitialized memory, which definitely isn't what you want. Additionally, that string does need to be null-terminated, and it isn't in your example, you do need CString. What you really want is something like this:

use std::ffi::{c_char, c_long, c_ushort, CStr, CString};
use std::ptr;
use std::slice;
use std::str;
fn main() {
    let mut event_buffer = ptr::null_mut();
    let mut result_buffer = ptr::null_mut();
    let names = CString::new("evento").unwrap();
    let len = unsafe {
        isc_event_block(
            &mut event_buffer,
            &mut result_buffer,
            1,
            names.as_ptr() as *mut c_char,
        )
    };
    debug_assert!(!event_buffer.is_null() && !result_buffer.is_null());
    let event_slice = unsafe { slice::from_raw_parts(event_buffer.cast(), len as usize) };
    let result_slice = unsafe { slice::from_raw_parts(event_buffer.cast(), len as usize) };
    let event_str = str::from_utf8(event_slice).unwrap();
    let result_str = str::from_utf8(result_slice).unwrap();
    println!("event: {event_str}");
    println!("result: {result_str}");
}

Playground

Adding a simple stub to print the string passed in, and write a couple of strings into the buffers, I got this:

"evento"
event: some events
result: some events

edited Dec 16, 2022 at 9:08

answered Dec 16, 2022 at 4:15

Aiden4

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3 Comments

Batels Over a year ago

"Secondly, the id_count parameter corresponds to the number of strings" My fail. In my original code I use the names as a vec<String> instead a single String. I forgot to change here.

Batels Over a year ago

For a single CString your answer works like a charm, but how I can use Vec<CString> in this place?

Aiden4 Over a year ago

@Batels You'll have to use libffi or similar to call it with a dynamic number of arguments. If you only have a small number of different expected arg counts, manually writing out each of the possibilities by hand is an option, too.

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