Rust 是否优化按值传递临时结构？

Question

假设我有一个 Rust 结构向量。结构相当大。当我想插入一个新的时，我编写这样的代码：

my_vec.push(MyStruct {field1: value1, field2: value2, ... });

推送的定义是

fn push(&mut self, value: T)

这意味着值是按值传递的。我想知道 Rust 是否先创建一个临时对象，然后复制到推送函数，或者它是否优化代码，以便不创建和复制临时对象？

Answer 1

让我们来看看。这个程序：

struct LotsOfBytes {
    bytes: [u8; 1024]
}

#[inline(never)]
fn consume(mut lob: LotsOfBytes) {
}

fn main() {
    let lob = LotsOfBytes { bytes: [0; 1024] };
    consume(lob);
}

编译为以下 LLVM IR 代码：

%LotsOfBytes = type { [1024 x i8] }

; Function Attrs: noinline nounwind uwtable
define internal fastcc void @_ZN7consume20hf098deecafa4b74bkaaE(%LotsOfBytes* noalias nocapture dereferenceable(1024)) unnamed_addr #0 {
entry-block:
  %1 = getelementptr inbounds %LotsOfBytes* %0, i64 0, i32 0, i64 0
  tail call void @llvm.lifetime.end(i64 1024, i8* %1)
  ret void
}

; Function Attrs: nounwind uwtable
define internal void @_ZN4main20hf3cbebd3154c5390qaaE() unnamed_addr #2 {
entry-block:
  %lob = alloca %LotsOfBytes, align 8
  %lob1 = getelementptr inbounds %LotsOfBytes* %lob, i64 0, i32 0, i64 0
  %arg = alloca %LotsOfBytes, align 8
  %0 = getelementptr inbounds %LotsOfBytes* %lob, i64 0, i32 0, i64 0
  call void @llvm.lifetime.start(i64 1024, i8* %0)
  call void @llvm.memset.p0i8.i64(i8* %lob1, i8 0, i64 1024, i32 8, i1 false)
  %1 = getelementptr inbounds %LotsOfBytes* %arg, i64 0, i32 0, i64 0
  call void @llvm.lifetime.start(i64 1024, i8* %1)
  call void @llvm.memcpy.p0i8.p0i8.i64(i8* %1, i8* %0, i64 1024, i32 8, i1 false)
  call fastcc void @_ZN7consume20hf098deecafa4b74bkaaE(%LotsOfBytes* noalias nocapture dereferenceable(1024) %arg)
  call void @llvm.lifetime.end(i64 1024, i8* %1)
  call void @llvm.lifetime.end(i64 1024, i8* %0)
  ret void
}

这行代码特别有趣：

call fastcc void @_ZN7consume20hf098deecafa4b74bkaaE(%LotsOfBytes* noalias nocapture dereferenceable(1024) %arg)

如果我理解正确的话，这意味着它consume是用指向 的指针调用的LotsOfBytes，所以是的，rustc 优化了按值传递大结构。