Can Rust optimise away the bit-wise copy during move of an object someday?

问题

Consider the snippet

struct Foo {
    dummy: [u8; 65536],
}

fn bar(foo: Foo) {
    println!("{:p}", &foo)
}

fn main() {
    let o = Foo { dummy: [42u8; 65536] };
    println!("{:p}", &o);
    bar(o);
}

A typical result of the program is

0x7fffc1239890
0x7fffc1229890

where the addresses are different.

Apparently, the large array dummy has been copied, as expected in the compiler's move implementation. Unfortunately, this can have non-trivial performance impact, as dummy is a very large array. This impact can force people to choose passing argument by reference instead, even when the function actually "consumes" the argument conceptually.

Since Foo does not derive Copy, object o is moved. Since Rust forbids the access of moved object, what is preventing bar to "reuse" the original object o, forcing the compiler to generate a potentially expensive bit-wise copy? Is there a fundamental difficulty, or will we see the compiler someday optimise away this bit-wise copy?

回答1:

Given that in Rust (unlike C or C++) the address of a value is not considered to matter, there is nothing in terms of language that prevents the elision of the copy.

However, today rustc does not optimize anything: all optimizations are delegated to LLVM, and it seems you have hit a limitation of the LLVM optimizer here (it's unclear whether this limitation is due to LLVM being close to C's semantics or is just an omission).

So, there are two avenues of improving code generation for this:

• teaching LLVM to perform this optimization (if possible)
• teaching rustc to perform this optimization (optimization passes are coming to rustc now that it has MIR)

but for now you might simply want to avoid such large objects from being allocated on the stack, you can Box it for example.

来源：https://stackoverflow.com/questions/38571270/can-rust-optimise-away-the-bit-wise-copy-during-move-of-an-object-someday