printf 上的分段错误 - NASM 64 位 Linux

我尝试使用输入四个浮点数scanf，将它们存储到堆栈中，然后使用vmovupd将它们复制到寄存器以供使用。我的问题是，当我尝试输出这 4 个数字时，程序段错误位于printf.

我认为这是堆栈的问题，但我尝试多次弹出（一次多条指令）但无济于事。我对汇编编码还是新手，所以使用gdb对我来说有点太先进了。

你会注意到我已经包含了一个名为debug。它允许我查看寄存器和堆栈（这就是为什么有一个dumpstack）这是我的教授提供的，它确实帮助了一些人，但显然还不够（或者也许我只是错过了一些东西）。

这是.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;
}

这是ASM code:

%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

问题出在你的堆栈使用上。

首先，ABI 文档要求rsp16 字节对齐before a call.

Since a call将把一个8字节的返回地址压入堆栈，你需要调整rsp16 的倍数加 8 以恢复 16 字节对齐。这16 * n + 8 is 包括 any pushRSP 的说明或其他更改，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SSE指令将因未对齐的地址而出错。

如果你解决这个问题，你的堆栈仍然不平衡，因为你不断推入值但从未弹出它们。很难理解你想用堆栈做什么。

通常的方法是在函数开头（序言）为局部变量分配空间，并在函数末尾（尾声）释放空间。如上所述，该金额（包括任何推送）应为 16 加 8 的倍数，因为 RSP 功能entry（在呼叫者之后call) 距 16 字节边界 8 字节。

在大多数 glibc 版本中，printf当 AL != 0 时，只会关心 16 字节堆栈对齐。（因为这意味着存在 FP 参数，因此它将所有 XMM 寄存器转储到堆栈，以便可以为它们建立索引%f转换。）

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could即使没有 FP 参数，也包含依赖于 16 字节堆栈对齐的代码。例如，scanf即使使用未对齐的堆栈，也已经崩溃了AL=0在大多数 GNU/Linux 发行版上。