我正在学习再次使用汇编语言,到目前为止,我唯一遇到的问题是对C进行调用。我拥有的这本书面向32位,而我正在使用64位。显然,调用约定存在很大差异,并且http://www.x86-64.org/documentation网站已关闭。因此,经过一些挖掘/测试,在C编译虚拟程序并花费3天时间,我想我会发布我的发现,如果它可以帮助其他人。从64位asm调用printf时如何传递参数?
RAX是否需要浮点数? 堆栈填充“阴影空间”16或32位? 这个宏是否适合小程序对齐堆栈?我知道你可以用对齐方式NOP-填充代码,我不确定堆栈帧。
; pf.asm compiled with 'nasm -o pf.o -f elf64 -g -F stabs'
; linked with 'gcc -o pf pf.o'
; 64-bit Bodhi (ubuntu) linux
%include "amd64_abi.mac"
[SECTION .data]
First_string: db "First string.",10,"%s", "%d is an integer. So is %d",10
db "Floats XMM0:%5.7f XMM1:%.6le XMM2:%lg",10,0
Second_String: db "This is the second string... %s's are not interpreted here.",10
db " Neither are %d's nor %f's. 'Cause it is a passed value.", 10, 0
; Just a regular string for insert.
[SECTION .bss]
[SECTION .text]
EXTERN printf
GLOBAL main
main:
_preserve_64AMD_ABI_regs ; Saves RBP, RBX, R12-R15
mov rdi, First_string ; Start of string to be formatted. Null terminated
mov rsi, Second_String ; String addy of first %s in main string. Not interpretted
mov rcx, 0456 ; Second Integer (Register is specific for ordered arguments.)
mov rdx,; First integer (Order of assignment does not matter.)
; Order of Integer/Pointer Registers:
; $1:RDI $2:RSI $3:RDX $4:RCX $5:R8 $6:R9
mov rax,0AABBCCh ; Test value to be stored in xmm0
cvtsi2sd xmm0, rax ; Convert quad to scalar double
mov rax,003333h ; Test value to be stored in xmm1
cvtsi2sd xmm1, rax ; Convert quad to scalar double
cvtsi2sd xmm2, rax ; Convert quad to scalar double
divsd xmm2, xmm0 ; Divide scalar double
sub rsp, 16 ; Allocates 16 byte shadow memory
_prealign_stack_to16 ; Move to the lower end 16byte boundry (Seg-Fault otherwise)
; mov rax, 3 ; Count of xmm registers used for floats. ?!needed?!
Before_Call:
call printf ; Send the formatted string to C-printf
_return_aligned_stack ; Returns RSP to the previous alignment
add rsp, 16 ; reallocate shadow memory
_restore_64AMD_ABI_regs_RET
; Ends pf.asm
; amd64_abi.mac
; Aligns stack (RSP) to 16 byte boundry, padding needed amount in rbx
%macro _preserve_64AMD_ABI_regs 0
push rbp
mov rbp, rsp
push rbx
push r12
push r13
push r14
push r15
%endmacro
%macro _restore_64AMD_ABI_regs_RET 0
pop r15
pop r14
pop r13
pop r12
pop rbx
mov rsp, rbp
pop rbp
ret
%endmacro
%macro _prealign_stack_to16 0
mov rbx, 0Fh ; Bit mask for low 4-bits 10000b = 16 :: 01111b = 15b
and rbx, rsp ; get bits 0-3 into rbx
sub rsp, rbx ; remove them from rsp, rounding down to multiple of 16 (10h)
%endmacro
; De-aligns stack (RSP)from 16 byte boundry using saved rbx offset
%macro _return_aligned_stack 0
add rsp, rbx
%endmacro
输出: 第一个字符串。 这是第二个字符串...%s在这里没有解释。 %d和%f都不是。因为这是一个传递的价值。 123是一个整数。所以是456 花车XMM0:11189196.0000000 XMM1:1.310700e + 04 XMM2:0.0011714
资源: System V的ABI v0.96:http://www.uclibc.org/docs/psABI-x86_64.pdf(这是不提供x86-64.org网站已关闭) 汇编语言一步步。 Jeff Duntemann第12章 Intel 64位指令集。 http://www.intel.com/content/www/us/en/processors/architectures-software-developer-manuals.html
最明显的方法先用C编写代码,然后用C编译器生成程序集。它永远不会错。 – 2013-05-07 15:55:43