assembly - 如何读取这段Rust代码的汇编代码？

Question

我正在尝试读取一段 Rust 汇编代码，但实际上，它比 C/C++ 编译器生成的 ASM 代码更难读取。那么，如何分析下面这段Rust代码的ASM代码呢？

fn main() { 
    let closure = |x| println!("{}", x);
    let x: fn(x: i32) -> () = closure; 
    println!("{}", x as i32);
}

相应的汇编代码如下，带有一些注释(我只粘贴了主要部分，完整版本请使用此固定链接:https://play.rust-lang.org/?version=nightly&mode=release&edition=2018&gist=e7ba4844f1ce6e881912dc074152988d):

playground::main: # @playground::main
# %bb.0:
    subq    $72, %rsp
    leaq    core::ops::function::FnOnce::call_once(%rip), %rax
    movl    %eax, 4(%rsp)
    leaq    4(%rsp), %rax
    movq    %rax, 8(%rsp)
    movq    core::fmt::num::imp::<impl core::fmt::Display for i32>::fmt@GOTPCREL(%rip), %rax
    movq    %rax, 16(%rsp)
    leaq    .L__unnamed_2(%rip), %rax  # the contents of rdx come from .L__unnamed_2(%rip), how to evaluate this part?
    movq    %rax, 24(%rsp)  # the contents of rdi come from rax.
    movq    $2, 32(%rsp)
    movq    $0, 40(%rsp)
    leaq    8(%rsp), %rax
    movq    %rax, 56(%rsp)
    movq    $1, 64(%rsp)
    leaq    24(%rsp), %rdi  # rdi should be the register holding the value passed to println!.
    callq   *std::io::stdio::_print@GOTPCREL(%rip)
    addq    $72, %rsp
    retq
                                        # -- End function

main:                                   # @main
# %bb.0:
    subq    $8, %rsp
    movq    %rsi, %rcx
    movslq  %edi, %rdx
    leaq    playground::main(%rip), %rax
    movq    %rax, (%rsp)
    leaq    .L__unnamed_1(%rip), %rsi
    movq    %rsp, %rdi
    callq   *std::rt::lang_start_internal@GOTPCREL(%rip)
                                        # kill: def $eax killed $eax killed $rax
    popq    %rcx
    retq
                                        # -- End function

.L__unnamed_1:
    .quad   core::ptr::drop_in_place<std::rt::lang_start<()>::{{closure}}>
    .quad   8                               # 0x8
    .quad   8                               # 0x8
    .quad   std::rt::lang_start::{{closure}}
    .quad   std::rt::lang_start::{{closure}}
    .quad   core::ops::function::FnOnce::call_once{{vtable.shim}}

.L__unnamed_3:

.L__unnamed_4:
    .byte   10

.L__unnamed_2:
    .quad   .L__unnamed_3
    .zero   8
    .quad   .L__unnamed_4
    .asciz  "\001\000\000\000\000\000\000"

而且，我正在尝试找出 Rust 编译器如何处理闭包函数指针和普通函数。因此，在这里我尝试使用闭包作为示例，但似乎找不到任何与变量“x”的使用相对应的有效汇编代码。

Answer 1

没有实际调用闭包，因此没有生成调用代码，但 x 变量的使用实际上是在您的帖子中未包含的函数中，该函数的名称具有误导性 core::ops::function::FnOnce::call_once 在 ASM 输出中，但在同一个 Playground 示例的 LLVM 输出中具有更加困惑的名称 @_ZN4core3ops8function6FnOnce9call_once17hefa1aa47132c4122E 。这是闭包的实际内容(println!("{}", x))

core::ops::function::FnOnce::call_once: # @core::ops::function::FnOnce::call_once
# %bb.0:
 # allocate a bunch of stack space for variables and print arguments
 subq   $72, %rsp

 # %edi has the value of x passed in to the closure, which we store in a new stack allocated variable
 movl   %edi, 4(%rsp)

 # we then load the address of that variable into another variable
 leaq   4(%rsp), %rax
 movq   %rax, 8(%rsp)

 # the following is mostly populating the std::fmt::Arguments struct which is passed to print
 movq   core::fmt::num::imp::<impl core::fmt::Display for i32>::fmt@GOTPCREL(%rip), %rax
 movq   %rax, 16(%rsp)
 leaq   .L__unnamed_2(%rip), %rax
 movq   %rax, 24(%rsp)
 movq   $2, 32(%rsp)
 movq   $0, 40(%rsp)

 # the address of the address of x is loaded into the arguments struct here
 leaq   8(%rsp), %rax
 movq   %rax, 56(%rsp)

 # finish populating the arguments and then call print
 movq   $1, 64(%rsp)
 leaq   24(%rsp), %rdi
 callq  *std::io::stdio::_print@GOTPCREL(%rip)
 addq   $72, %rsp
 retq

playground main 函数是创建闭包的地方，但它实际上并未被调用，并且与上面的函数一样，主要填充复杂的 std::fmt::Arguments 结构

playground::main: # @playground::main
# %bb.0:
subq    $72, %rsp

 # this creates the closure by storing a pointer to the closure's function
leaq    core::ops::function::FnOnce::call_once(%rip), %rax
movl    %eax, 4(%rsp)

 # this stores the closure in main's `x` variable (line 3 of the example)
leaq    4(%rsp), %rax
movq    %rax, 8(%rsp)

 # populate the std::fmt::Arguments struct
movq    core::fmt::num::imp::<impl core::fmt::Display for i32>::fmt@GOTPCREL(%rip), %rax
movq    %rax, 16(%rsp)
leaq    .L__unnamed_2(%rip), %rax  # the contents of rdx come from .L__unnamed_2(%rip), how to evaluate this part?
movq    %rax, 24(%rsp)  # the contents of rdi come from rax.
movq    $2, 32(%rsp)
movq    $0, 40(%rsp)

 # store the closure (stored in `x`) in the std::fmt::Arguments struct
leaq    8(%rsp), %rax
movq    %rax, 56(%rsp)

 # finish populating and call print
movq    $1, 64(%rsp)
leaq    24(%rsp), %rdi  # rdi should be the register holding the value passed to println!.
callq   *std::io::stdio::_print@GOTPCREL(%rip)
addq    $72, %rsp
retq

从 LLVM 输出中，std::fmt::Arguments 定义为 %"std::fmt::Arguments"= type { [0 x i64], { [0 x { [0 x i8] ]*, i64 }]*, i64 }, [0 x i64], { i64*, i64 }, [0 x i64], { [0 x { i8*, i64* }]*, i64 }, [0 x i64] } 而且我不了解太多的内部细节，所以我不确定为什么它引用静态内存区域 .L__unnamed_2 但深入研究 std::fmt::Arguments可能会提供更多线索