由于exa提供了悬赏,我在这里发布一个回答。
在Ross Ridge的评论和对clang的一些尝试后,我意识到这是一个有点愚蠢的问题。
让我们假设我们有以下程序,它使用内联汇编直接调用write()
函数。
#include <stdio.h>
int main(void)
{
char *buf = "test\n";
ssize_t n;
asm volatile (
"movl $0x00000002, %%edi\n" /* first argument == stderr */
"movl $0x00000006, %%edx\n" /* third argument == number of bytes */
"movl $1, %%eax\n" /* syscall number == write on amd64 linux */
"syscall\n"
: "=A"(n) /* %rax: return value */
: "S"(buf)); /* %rsi: second argument == address of data to write */
return n;
}
我们可以使用
gcc
或者
clang
进行编译,得到的结果大致相同。
$ gcc -o syscall.gcc syscall.c
$ clang -o syscall.clang syscall.c
$ ./syscall.gcc
test
$ ./syscall.clang
test
如果我们希望查看生成此代码所使用的精确LLVM指令,我们只需使用
-emit-llvm
标志即可。如您所见,有一行
call i64 asm sideeffect
,其中包含完整的内联汇编字符串。
$ clang -S -emit-llvm syscall.c
$ cat syscall.ll
; ModuleID = 'syscall.c'
target datalayout = "e-m:e-i64:64-f80:128-n8:16:32:64-S128"
target triple = "x86_64-pc-linux-gnu"
@.str = private unnamed_addr constant [6 x i8] c"test\0A\00", align 1
; Function Attrs: nounwind uwtable
define i32 @main() #0 {
%1 = alloca i32, align 4
%buf = alloca i8*, align 8
%n = alloca i64, align 8
store i32 0, i32* %1
store i8* getelementptr inbounds ([6 x i8]* @.str, i32 0, i32 0), i8** %buf, align 8
%2 = load i8** %buf, align 8
%3 = call i64 asm sideeffect "movl $$0x00000002, %edi\0Amovl $$0x00000006, %edx\0Amovl $$1, %eax\0Asyscall\0A", "=A,{si},~{dirflag},~{fpsr},~{flags}"(i8* %2) #1, !srcloc !1
store i64 %3, i64* %n, align 8
%4 = load i64* %n, align 8
%5 = trunc i64 %4 to i32
ret i32 %5
}
attributes #0 = { nounwind uwtable "less-precise-fpmad"="false" "no-frame-pointer-elim"="true" "no-frame-pointer-elim-non-leaf" "no-infs-fp-math"="false" "no-nans-fp-math"="false" "stack-protector-buffer-size"="8" "unsafe-fp-math"="false" "use-soft-float"="false" }
attributes #1 = { nounwind }
!llvm.ident = !{!0}
!0 = metadata !{metadata !"Ubuntu clang version 3.5-1ubuntu1 (trunk) (based on LLVM 3.5)"}
!1 = metadata !{i32 134, i32 197, i32 259, i32 312}
int 0x80
和设置适当的寄存器。 - Zach Riggle