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调试 nasm 应用程序时 GDB 关闭一行

[英]GDB is off by a line when debugging a nasm application

I just encountered a weird issue with gdb , though I'm not sure, if I'm missing something.我刚刚遇到了gdb的一个奇怪问题,但我不确定是否遗漏了什么。 Let's suppose I have these two files:假设我有这两个文件:

somefile.asm : somefile.asm

01 section .text
02 
03 funca:
04  mov eax, 5
05  mov ebx, 5
06  cmp eax, ebx
07  je aisequal
08  mov ecx, 13
09  mov edx, 19
10  ret
11
12  aisequal:
13  mov ecx, 17
14  mov edx, 21
15  ret

and somefile_test.asm :somefile_test.asm

01 %include "somefile.asm"
02
03 section .text
04  global _start
05
06 _start:
07  xor eax, eax
08  xor ebx, ebx
09  xor ecx, ecx
10  xor edx, edx
11  call funca
12
13  mov eax, 1
14  mov ebx, 0
15  int 0x80

I compile and link it using我编译并链接它使用

nasm -f elf -g -F dwarf somefile_test.asm 
ld -m elf_i386 -o somefile_test.out somefile_test.o

And then debug my application using gdb :然后使用gdb调试我的应用程序:

gdb somefile_test.out

I now set a breakpoint in the imported file:我现在在导入的文件中设置断点:

GNU gdb (GDB) 10.1
(gdb) b somefile.asm:5
Breakpoint 1 at 0x8049000: file somefile.asm, line 5.
(gdb) r
Starting program: /<bla>/somefile_test.out

Breakpoint , funca () at somefile.asm:5
5       mov ebx, 5

Now appearantly, the execution stopped at the correct position.现在看来,执行在正确的 position 处停止。 The next line to be executed would be 5, which is mov ebx, 5 .要执行的下一行是 5,即mov ebx, 5 However , the last line should've been mov eax, 5 which should have already been executed.但是,最后一行应该是mov eax, 5应该已经执行了。 It was not:不是:

(gdb) i r eax
eax            0x0                 0

It gets even weirder:它变得更加奇怪:

(gdb) si
6       cmp eax, ebx
(gdb) i r eax ebx
eax            0x5                 5
ebx            0x0                 0

Now, eax is set, but ebx is not (yet).现在, eax已设置,但ebx尚未设置。 If I execute the next line, it is set:如果我执行下一行,则设置为:

(gdb) si
7       je aisequal
(gdb) i r eax ebx
eax            0x5                 5
ebx            0x5                 5

However, I'd expect the program to jump to line 12 (aisequal) now, but it doesn't:但是,我希望程序现在跳转到第 12 行(aisequal),但它没有:

(gdb) si
8       mov ecx, 13

On the next instruction, it suddenly goes to the right line:在下一条指令中,它突然转到正确的行:

(gdb) si
14      mov edx, 21
(gdb) i r eax ebx edx
eax            0x5                 5
ebx            0x5                 5
edx            0x0                 0

And so on:等等:

(gdb) si
15      ret
(gdb) i r eax ebx ecx edx
eax            0x5                 5
ebx            0x5                 5
ecx            0x11                17
edx            0x0                 0

If I put all my code in a single file, everything works as expected:如果我将所有代码放在一个文件中,一切都会按预期工作:

01 section .text
02  global _start
03
04 _start:
05  xor eax, eax
06  xor ebx, ebx
07  xor ecx, ecx
08  xor edx, edx
09  call funca
10
11  mov eax, 1
12  mov ebx, 0
13  int 0x80
14
15 funca:
16  mov eax, 5
17  mov ebx, 5
18  cmp eax, ebx
19  je aisequal
20  mov ecx, 13
21  mov edx, 19
22  ret
23
24  aisequal:
25  mov ecx, 17
26  mov edx, 21
27  ret
GBU gdb (GDB) 10.1
(gdb) b 16
Breakpoint 1 at 0x8049019: file singlefile.asm, line 16.
(gdb) r
Starting program: /<bla>/singlefile.out 

Breakpoint 1, funca () at singlefile.asm:16
16      mov eax, 5
(gdb) i r eax ebx ecx edx
eax            0x0                 0
ebx            0x0                 0
ecx            0x0                 0
edx            0x0                 0
(gdb) si
17      mov ebx, 5
(gdb) i r eax ebx ecx edx
eax            0x5                 5
ebx            0x0                 0
ecx            0x0                 0
edx            0x0                 0
(gdb) si
18      cmp eax, ebx
(gdb) si
19      je aisequal
(gdb) si
25      mov ecx, 17
(gdb) si
26      mov edx, 21
(gdb) i r eax ebx ecx edx
eax            0x5                 5
ebx            0x5                 5
ecx            0x11                17
edx            0x0                 0
(gdb) si
aisequal () at singlefile.asm:27
27      ret
(gdb) i r eax ebx ecx edx
eax            0x5                 5
ebx            0x5                 5
ecx            0x11                17
edx            0x15                21
(gdb) si
_start () at singlefile.asm:11
11      mov eax, 1

Now I've only picked up gdb two days ago, so I'm not that familiar with it.现在两天前才拿到gdb ,所以不是很熟悉。 Can someone explain to me what's happening?有人可以向我解释发生了什么吗? Is this a bug or am I missing something?这是一个错误还是我错过了什么?

I am using我在用

nasm 2.15.05-1
binutils 2.35.1-1
gdb 10.1-4
gcc 10.2.0-4

on Linux 5.9.14-arch1-1 #1 SMP PREEMPT Sat, 12 Dec 2020 14:37:12 +0000 x86_64 GNU/LinuxLinux 5.9.14-arch1-1 #1 SMP PREEMPT Sat, 12 Dec 2020 14:37:12 +0000 x86_64 GNU/Linux

Apparently, this is a regression in nasm.显然,这是 nasm 的回归

Reported it to them.向他们报告了。

This looks like a bug in nasm .这看起来像nasm中的错误。 It didn't reproduce for me using nasm-2.14 .它没有为我使用nasm-2.14重现。

GDB will only display source info that the compiler/assembler tells it. GDB 只会显示编译器/汇编器告诉它的源信息。 If the assembler puts out incorrect info, then GDB will display that incorrect info and can't do anything about it.如果汇编器输出了不正确的信息,那么 GDB 将显示不正确的信息并且不能做任何事情。

To verify that the problem is in nasm , run objdump -dS somefile_test.o and compare the assembly and source listing.要验证问题出在nasm中,请运行objdump -dS somefile_test.o并比较程序集和源代码列表。 If they are also off by one, it's a bug in nasm .如果它们也相差一个,那是nasm中的一个错误。

Here is what I see:这是我看到的:

somefile_test.o:     file format elf32-i386


Disassembly of section .text:

00000000 <funca>:
section .text

funca:
 mov eax, 5
   0:   b8 05 00 00 00          mov    $0x5,%eax
 mov ebx, 5
   5:   bb 05 00 00 00          mov    $0x5,%ebx
 cmp eax, ebx
   a:   39 d8                   cmp    %ebx,%eax
 je aisequal
   c:   74 0b                   je     19 <aisequal>

Note how instructions and source perfectly line up.请注意说明和源代码是如何完美排列的。

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