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