printf上的分段错误 - NASM 64位Linux

时间:2014-09-05 20:53:38

标签: linux assembly x86-64 nasm calling-convention

我尝试使用scanf输入四个浮点数,将它们存储到堆栈中,然后使用vmovupd将它们复制到寄存器中以供使用。我的问题是当我尝试输出这4个数字时,程序在printf处出现故障。

我认为它与堆栈有关,但我尝试过多次弹出(多次指示一次)无济于事。我对汇编编码还不熟悉所以使用gdb对我来说有点太高级了。

您会注意到我添加了一个名为debug的文件。它允许我查看寄存器和堆栈(这就是为什么有dumpstack指令。)这是由我的教授提供的,它确实帮助了一些但显然不够(或者我可能是只是遗漏了一些东西)。

这里是.cpp

#include <iostream>

using namespace std;

extern "C" double ComputeElectricity();

int main()
{
    cout << "Welcome to electric circuit processing by Chris Tarazi." << endl;
    double returnValue = ComputeElectricity();
    cout << "The driver received this number: " << returnValue << endl; 
    return 0;
}

这里是ASM代码:

%include "debug.inc"
extern printf
extern scanf
global ComputeElectricity

;---------------------------------Declare variables-------------------------------------------

segment .data

greet db "This progam will help you analyze direct current circuits configured in parallel.", 10, 0
voltage db "Please enter the voltage of the entire circuit in volts: ", 0
first db "Enter the power consumption of device 1 (watts): ", 0
second db "Enter the power consumption of device 2 (watts): ", 0
third db "Enter the power consumption of device 3 (watts): ", 0
fourth db "Enter the power consumption of device 4 (watts): ", 0
thankyou db "Thank you. The computations have completed with the following results.", 10, 0
circuitV db "Curcuit total voltage: %1.18lf v", 10, 0
deviceNum db "Device number:                1                    2                    3                    4", 10, 0
power db "Power (watts): %1.18lf %1.18lf %1.18lf %1.18lf", 10, 0
current db "Current (amps): %1.18lf %1.18lf %1.18lf %1.18lf", 10, 0
totalCurrent db "Total current in the circuit is %1.18lf amps.", 10, 0
totalPower db "Total power in the circuit is %1.18lf watts.", 10, 0

bye db "The analyzer program will now return total power to the driver.", 10, 0

string db "%s", 0
floatfmt db "%lf", 0
fourfloat db "%1.18lf %1.18lf %1.18lf %1.18lf", 0

;---------------------------------Begin segment of executable code------------------------------

segment .text

dumpstack 20, 10, 10

ComputeElectricity:

;dumpstack 30, 10, 10

;---------------------------------Output greet message------------------------------------------

    mov qword rax, 0
    mov rdi, string 
    mov rsi, greet
    call printf

;---------------------------------Prompt for voltage--------------------------------------------

    mov qword rax, 0
    mov rdi, string
    mov rsi, voltage
    call printf

;---------------------------------Get  voltage--------------------------------------------------

    push qword 0
    mov qword rax, 0
    mov rdi, floatfmt
    mov rsi, rsp
    call scanf
    vbroadcastsd ymm15, [rsp]
    pop rax

;---------------------------------Prompt for watts 1--------------------------------------------

    mov qword rax, 0
    mov rdi, string
    mov rsi, first
    call printf

;---------------------------------Get watts 1---------------------------------------------------

    push qword 0
    mov qword rax, 0
    mov rdi, floatfmt
    mov rsi, rsp
    call scanf

;---------------------------------Prompt for watts 2--------------------------------------------

    mov qword rax, 0
    mov rdi, string
    mov rsi, second         
    call printf 

;---------------------------------Get watts 2---------------------------------------------------

    push qword 0
    mov qword rax, 0
    mov rdi, floatfmt
    mov rsi, rsp
    call scanf

;---------------------------------Prompt for watts 3--------------------------------------------

    mov qword rax, 0
    mov rdi, string
    mov rsi, third      
    call printf 

;---------------------------------Get watts 3---------------------------------------------------

    push qword 0
    mov qword rax, 0
    mov rdi, floatfmt
    mov rsi, rsp
    call scanf

;---------------------------------Prompt for watts 4--------------------------------------------

    mov qword rax, 0
    mov rdi, string
    mov rsi, fourth 
    call printf 

;---------------------------------Get watts 4---------------------------------------------------

    push qword 0
    mov qword rax, 0
    mov rdi, floatfmt
    mov rsi, rsp
    call scanf

    ;dumpstack 50, 10, 10

;---------------------------------Move data into correct registers------------------------------

    vmovupd ymm14, [rsp]                ; move all 4 numbers from the stack to ymm14

    pop rax
    pop rax
    pop rax
    pop rax

    ;dumpstack 55, 10, 10       

    vextractf128 xmm10, ymm14, 0        ; get lower half
    vextractf128 xmm11, ymm14, 1        ; get upper half

;---------------------------------Move data into low xmm registers------------------------------

    movsd xmm1, xmm11                   ; move ymm[128-191] (3rd value) into xmm1
    movhlps xmm0, xmm11                 ; move from highest value from xmm11 to xmm0

    movsd xmm3, xmm10
    movhlps xmm2, xmm10

    ;showymmregisters 999

;---------------------------------Output results-------------------------------------------------

    ;dumpstack 60, 10, 10

    mov rax, 4
    mov rdi, fourfloat
    push qword 0
    call printf
    pop rax

ret

1 个答案:

答案 0 :(得分:5)

问题在于您的堆栈使用情况。

首先,ABI文档授权rspcall之前与进行16字节对齐。

由于call将在堆栈上推送一个8字节的返回地址,因此需要将rsp调整为16加8的倍数才能返回16字节对齐。 16 * n + 8 包含任何push指令或对RSP的其他更改,而不仅仅是sub rsp, 24。这是段错误的直接原因,因为printf将使用对齐的SSE指令,这些指令会对未对齐的地址造成错误。

如果你解决了这个问题,那么你的堆栈仍然是不平衡的,因为你继续推动值但从不弹出它们。很难理解你想用堆栈做什么。

通常的方法是在你的函数开头(序言)为本地人分配空间,并在结尾(结语)释放它。如上所述,此数量(包括任何推送)应该是16加8的倍数,因为函数条目上的RSP (在调用者call之后)距离8个字节一个16字节的边界。


在glibc的大多数版本中,当{!= 0}时,printf只关心16字节堆栈对齐。(因为这意味着有FP args,所以它将所有XMM寄存器转储到堆栈中,因此它可以为%f次转化编制索引。)

如果您使用未对齐的堆栈调用它,即使它恰好在您的系统上运行,它仍然是一个错误;未来的glibc版本可以包含依赖于16字节堆栈对齐的代码,即使没有FP args也是如此。例如,scanf已经在未对齐的堆栈上崩溃,即使在大多数GNU / Linux发行版上都使用AL=0