Mixed-language-link-time-optimization.md

This can be done with or without the runtime enabled, but the example here will disable it in order to demonstrate a very simple optimization enabled by LTO.

main.c

#include <stdint.h>
#include <stdlib.h>

uint32_t *foo();

int main(void) {
    free(foo());
    return 0;
}

foo.rs

#[no_std];
#[allow(ctypes)];

extern {
    fn malloc(size: uint) -> *mut u8;
    fn free(ptr: *mut u8);
    fn abort() -> !;
}

#[lang = "exchange_malloc"]
unsafe fn exchange_malloc(size: uint) -> *mut u8 {
    let ptr = malloc(size);
    if ptr == 0 as *mut u8 {
        abort()
    }
    ptr
}

#[lang = "exchange_free"]
unsafe fn exchange_free(ptr: *mut u8) {
    free(ptr)
}

#[no_mangle]
extern "C" fn foo() -> ~u32 { ~5 }

Build LLVM bytecode from the Rust code:

$ rustc foo.rs --emit-llvm --lib -O

The --lib flag is necessary even if Rust defines main, to avoid treating the Rust part as the entire application.

Compile together the C and Rust code with clang:

$ clang foo.bc main.c -flto -O3 -o main

Verify that LLVM eliminated the call to foo via inlining and removed the allocation as a dead store:

$ gdb main
(gdb) disassemble main
Dump of assembler code for function main:
   0x00000000004005b0 <+0>:	xor    %eax,%eax
   0x00000000004005b2 <+2>:	retq   
End of assembler dump.

To do this with the Rust runtime on, define a regular main function in Rust and pass the necessary link flags to clang. The necessary switches can be obtained with rustc -Z print-link-args foo.rs. In order for segmented stacks to work, rustc will likely need to be taught to take LLVM bytecode files as input to combine with the current crate.