Rust内联汇编模板语法

Question

我有以下代码test.s:

call $8 + $8

运行nasm test.s编译成功.我希望以下等效的Rust代码能够成功编译,但事实并非如此.

在test.rs:

#![feature(asm)]
fn main() {
    unsafe {
        asm! (
            "call $0 + $0"
            :
            : "i" (8)
            : "memory"
            : "volatile"
        )
    }
}

产量rustc test.rs:

test.rs:4:9: 10:11 error: <inline asm>:1:12: error: invalid token in expression
        call $8 + $8
                  ^

Answer 1

这对我有用:

#![feature(asm)]
fn main() {
    unsafe {
        asm!( "call ${0:c} + ${0:c}"
            :
            : "i"(8)
            : "memory"
            : "volatile"
        )
    }
}

这是LLVM参考中的相关文档.通过查看输出objdump,我们可以验证我们的内联汇编已经发出:

0000000000005190 <_ZN4main20hc3048743ecd04f53eaaE>:
    5190:   e8 7b ae ff ff          callq  10 <_ZN10sys_common11thread_info11THREAD_INFO5__KEY20h20efb688859d2c0dRhsE+0x10>
    5195:   c3                      retq   
    5196:   66 2e 0f 1f 84 00 00    nopw   %cs:0x0(%rax,%rax,1)
    519d:   00 00 00 

更新:这是从内联汇编直接调用函数的示例:

#![feature(asm)]

fn called_even_if_mangled() {
    println!("Just a regular function minding its own business");
}

fn main() {
    unsafe {
        asm!( "call ${0:c}"
            :
            : "i"(called_even_if_mangled)
            : "memory"
            : "volatile", "alignstack"
            // Omit "alignstack" and you'll get a segfault because of a
            // misaligned SSE load on some initialization code regarding
            // stdin.
        )
    }
}

但是你永远不应该做这样的事情,除非你有一个非常好的和令人信服的论据(例如,因为你正在写一个JIT).我不得不花一个小时调试一个神秘的段错误,直到我意识到我还必须放入alignstack选项部分.

你被警告了.