如何避免`#[global_allocator]的冲突？

Question

假设我有三个板条箱：foo、bar和baz。

Cargo.toml

[workspace]
members = [
    "foo",
    "bar",
    "baz",
]

[profile.dev]
panic = "abort"

[profile.release]
panic = "abort"

foo/Cargo.toml

[package]
name = "foo"
version = "0.1.0"
edition = "2018"

[lib]
name = "foo"
crate-type = ["staticlib"]

[dependencies]
bar = { path = "../bar", default-features = false }

foo/src/lib.rs

#![no_std]

extern crate bar as _;

use core::alloc::{GlobalAlloc, Layout};

#[global_allocator]
static ALLOCATOR: Allocator = Allocator;

struct Allocator;
unsafe impl GlobalAlloc for Allocator {
    unsafe fn alloc(&self, _: Layout) -> *mut u8 {
        todo!()
    }

    unsafe fn dealloc(&self, _: *mut u8, _: Layout) {
        todo!()
    }
}

bar/Cargo.toml

[package]
name = "bar"
version = "0.1.0"
edition = "2018"

[lib]
name = "bar"

[features]
default = ["heap"]
heap = []

bar/src/lib.rs

#![no_std]
#![feature(alloc_error_handler)]

use core::alloc::GlobalAlloc;
use core::alloc::Layout;

extern crate alloc;

#[cfg(feature = "heap")]
#[global_allocator]
static ALLOCATOR: Allocator = Allocator;

struct Allocator;
unsafe impl GlobalAlloc for Allocator {
    unsafe fn alloc(&self, _: Layout) -> *mut u8 {
        todo!()
    }

    unsafe fn dealloc(&self, _: *mut u8, _: Layout) {
        todo!()
    }
}

#[panic_handler]
fn panic(_: &core::panic::PanicInfo<'_>) -> ! {
    todo!();
}

#[alloc_error_handler]
fn alloc_fail(_: Layout) -> ! {
    todo!();
}

baz/Cargo.toml

[package]
name = "baz"
version = "0.1.0"
edition = "2018"

[lib]
name = "baz"
crate-type = ["staticlib"]

[dependencies]
bar = { path = "../bar" }

baz/src/lib.rs

#![no_std]

extern crate bar as _;

在这里，bar提供全局分配程序，foo和baz都依赖于bar。然而，foo不允许bar的全局分配器使用它的原始分配器，因此foo依赖于bar，而没有默认的特性。

在每个目录(foo/、bar/和baz/)上运行foo/成功。但是，在项目根目录上运行命令会导致错误：

%cargo clippy
    Checking bar v0.1.0 (/tmp/tmp.wlBjM4wNFx/bar)
    Checking baz v0.1.0 (/tmp/tmp.wlBjM4wNFx/baz)
    Checking foo v0.1.0 (/tmp/tmp.wlBjM4wNFx/foo)
error: the `#[global_allocator]` in this crate conflicts with global allocator in: bar

error: could not compile `foo` due to previous error

实际上，由于foo使用其原始全局分配程序并禁用bar的全局分配程序，因此不存在全局分配程序冲突。如何避免此错误？

版本

rustup 1.24.3 (ce5817a94 2021-05-31)
cargo 1.56.0-nightly (b51439fd8 2021-08-09)
rustc 1.56.0-nightly (0035d9dce 2021-08-16)

Answer 1

这是工作区的属性。bjorn3说论祖利普

如果foo、bar和baz都被编译在一起，那么如果foo或baz依赖于堆特性，就会为bar全局启用堆特性。Cargo将结合每个依赖的机箱所需的所有特性，并使用这些特性编译依赖关系一次。

因此，bar是用heap特性编译的，因为baz需要它，在编译foo时也启用了它的特性。因此，foo和bar的全局分配器同时启用，导致错误。