问题描述
这是我的汇编代码和我的主要子程序.这是我的宏和常量:
Here's my assembly code and my main subroutine.Here are my macros and constants:
.text
fmt: .string "x \t\t ln(x)\n"
sfmt: .string "%.10lf \t %.10lf\n"
error: .string "Error"
filename: .string "input.bin"
.data
LIM: .double 0r1.0E-13
zero: .double 0r0.0
one: .double 0r1.0
half: .double 0r0.5
define(i_r,w19)
define(j_r,w20)
define(n_r,w21)
define(fd_r,w22)
define(ln_x,d8)
define(cur_term,d24)
define(n_read,x25)
define(x_j,d26)
BUF_SIZE = 98*8
AT_FDCWD = -100
O_RDONLY = 0
buf_s = 16
.bss
x_arr: .skip BUF_SIZE
fp .req x29
lr .req x30
.balign 4
.global main
这是我的主要子程序:
main:
stp fp,lr,[sp,-16]!
mov fp,sp
ldp fp,lr,[sp],16
ret
然而，我已经使用了 gdb，它只指出 SIGSEGV 信号来自 main() 中的 0x0000000000420358.我怎样才能缩小这个信号来自主要"的位置?P.S 我只知道 GDB 的基础.
I already used gdb however, it only points out that the SIGSEGV signal came from 0x0000000000420358 in main(). How can I narrow down where in 'main' this signal comes from?P.S I only know the basics of GDB.
GDB 资料:(更新)
GDB Stuff:(update)
(gdb) x/i $pc
=> 0x420358: .inst 0x00000000 ; undefined
我不知道这是否有帮助，但这是有效的 C 版本.我正在将它转换为汇编，因为那是我需要提交的内容.此外，我们不能使用任何类型的转换器，因为那被认为是作弊.
I don't know if this helps but this is the C version THAT WORKS. I am converting it to assembly because thats what I need to hand in. Also we cannot use any types of converter since thats considered cheating.
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <unistd.h>
#include <fcntl.h> //Used for the keyword for flags and other predefined values for the argument on openat,etc.
#define LIM 1.0e-13
#define DOUBSIZE 100 //There are 97 double values in the binary file
#define buf_size 98*8
double x[DOUBSIZE];
int main() {
register int i = 1,j = 0,fd = openat(AT_FDCWD,"input.bin",O_RDONLY); //int fd = openat(int dirfd,const char *pathname (basically a string),int flags,mode_t mode);
register double ln_x = 0.0,cur_term;
//double *x; //(only local variable) //(a local variable so it must be in the stack)only assuming there are 32 double precision values in the binary file
register long n_read = read(fd,&x,buf_size); //reads in 8 bytes(lost the double x[...] in this line since x is now pointing at the buffer
if(fd == -1) {
printf("Error!");
return 0;
}
if(n_read == -1) { //Error checker
printf("Error!");
return 0;
}
printf("x \t\t ln(x)\n"); //The header of the thing to be printed
while(j < (buf_size/8)) { //note that it is implied that EOF = -1 in C
if(x[j] <= 0.5) { //if x is less than or equal to 1/2,go to the next double value(assuming I don't know values in the bin file)
j++;
i = 1;
continue;
}
cur_term = (1.0/i) * (pow((double)((x[j]-1.0)/(x[j])),i));
ln_x += cur_term;
while(cur_term >= LIM) { //continue to accumulate terms until the absolute value of the term is less than 1.0E-13
i++; //follows the pattern of the series.
cur_term = (1.0/i)*(pow((double)((x[j]-1.0)/(x[j])),i)); //since it should start with x[1]
ln_x += cur_term; //adds the new term to previous ln(x) value
}
printf("%.10lf \t %.10lf\n",x[j],ln_x); //prints the current value of ln(x) and x
j++; //manages which x double value will be used next
i = 1;
ln_x = 0.0;
}
close(fd);
return 0;
}
推荐答案
原来你的 main 在 .bss 部分，而不是 .text 它所属的位置，因此它只能包含全零字节.(而且它不会是可执行的).
Turns out your main is in the .bss section, not .text where it belongs, so it can only contain all-zero bytes. (And it won't be executable).
GDB 通常只想反汇编 .text 部分中的代码，所以这也解释了 GDB 的奇怪之处.
GDB normally only wants to disassemble code in the .text section so that also explains GDB being weird.
这就是为什么您应该将代码简化为 MCVE(最小/完整/可验证示例)以使其尽可能小，同时仍然包含问题.
This is why you should reduce your code to a MCVE (Minimal / complete / verifiable example) to make it as small as possible while still containing the problem.
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