在没有优化的程序上使用gdb调试，但在当前上下文中没有用于局部变量的符号 [英] Debugging with gdb on a program with no optimization but still there is no symbol in the current context for local variables

问题描述

我一直有这个问题，虽然现在，但总是似乎推迟了问这个问题，因为它看起来像我做错了...但现在我觉得不然...采取这个代码：

  #include< string> 
 #include< iostream> 
 #include< algorithm> 
 int main（int argc，char ** argv）
 {
 if（argc< 2）
 {
 std :: cout< usage：< argv [0]< < string> << std :: endl; 
 return 1; 
} 
 std :: string str = argv [1]; 
 std :: reverse（str.begin（），str.end（））; 
 std :: cout<< str<< std :: endl; 
 return 0; 
} 
  

使用命令编译：

  g ++ -std = c ++ 11 -Wall main.cpp -o main -O0 -ggdb3 
  

我正在使用一个最近的中继线版本的gcc，我的树干被约在9月23日左右ish ...还要注意，我不编译优化！

现在，我开始调试，如下：

  gdb --quiet  - args ./main string 
  

我在第12行设置了一个断点>

  b 12 
  

然后我运行程序

 运行
   
 
然后我尝试打印出字符串，看看是什么。
  str 
  
这个，亲爱的朋友，对我来说看起来很奇怪：
输出的上一个命令是：
 在当前上下文中没有符号str。 
 
快速检查本地变量不会显示字符串
  info locals 
  
 get是
  std :: __ ioinit = {static _S_refcount = 2，static _S_synced_with_stdio = true} 
  
所以我想知道，我是错的，或者是编译器或调试器在故障...这是一个很痛苦屁股的问题为我很长一段时间...所以感谢甚至阅读这个问题...）
 
 
 
编辑：现在已经清楚，我的gcc构建错误，我想知道是否有任何人遇到一个错误报告或任何其他情况下，似乎有一个类似于这个问题...我也将尝试检查与最近的一个gdb构建，确保没有一个问题，我当前的调试器（与ubuntu一起）...这是有意义吗？
 
 
 
 EDIT2：所以在编译gdb v7.5 ，我得到相同的结果，除了没有当地人在所有...我想这意味着它是一个gcc的问题，谢谢大家... 
解决方案
不，即使--quiet它为我工作。可能您的设置有问题。
 〜/ tmp $ g ++ -Wall tmp.cpp -o tmp -O0 -ggdb3 
〜/ tmp $ gdb --quiet --args ./tmp string 
从/xxxxxxxx/tmp...done读取符号。 
（gdb）b 12 
断点1在0x400c95：文件tmp.cpp，第12行。
（gdb）run 
启动程序：/ xxxxxxxx / tmp string 
断点1，main（argc = 2，argv = 0x7fffffffdc58）at tmp.cpp：12 
 12 std :: reverse（str.begin（），str.end（））; 
（gdb）print str 
 $ 1 =string
（gdb）info locals 
 str =string
（gdb）
 c $ c> 
 
I've been having this problem for while now, but always seem to put off asking this question because it seems like I am doing something wrong... but right now I feel otherwise... taken this code:
#include <string>
#include <iostream>
#include <algorithm>
int main(int argc, char** argv)
{
  if(argc < 2)
  {
    std::cout << "usage: " << argv[0] << " <string>" << std::endl;
    return 1;
  }
  std::string str = argv[1];
  std::reverse(str.begin(), str.end());
  std::cout << str << std::endl;
  return 0;
}
Compiled with the command:
g++ -std=c++11 -Wall main.cpp -o main -O0 -ggdb3
I am using a very recent trunk version of gcc, I the trunk was taken around September 23rd ish... Also note that I am not compiling with optimization!


Now anyway, I start debugging, like this:
gdb --quiet --args ./main string
I set a break point at line 12 (the reverse algorithm)
b 12
then I run the program
run
then I try to print out the string, to see what it is
print str
And this, my dear friends, is what seems strange to me:
The output of that previous command is:
No symbol "str" in current context.
a quick check to the local variables doesn't show the string either
info locals
all I get is
std::__ioinit = {static _S_refcount = 2, static _S_synced_with_stdio = true}
so I'm wondering, am I at fault, or is the compiler or debugger at fault... this has been a pretty pain in the ass problem for a long time for me... so thanks for even reading this question... :)

EDIT: now that it has become clear that there is something wrong with my gcc build, I'm wondering if anyone has come across a bug report or any other case where there seems to a be a problem similar to this... I will also try checking with a recent build of gdb to make sure that there is definitely not a problem with my current debugger (that comes with ubuntu)... does that make sense?

EDIT2: So after compiling gdb v7.5, I got relatively the same result except there were no locals present at all... I guess that means it's a gcc issue, thanks everyone...
解决方案
No, even with --quiet it works for me. Maybe there's something wrong with your setup.
~/tmp $ g++ -Wall tmp.cpp -o tmp -O0 -ggdb3
~/tmp $ gdb --quiet --args ./tmp string
Reading symbols from /xxxxxxxx/tmp...done.
(gdb) b 12
Breakpoint 1 at 0x400c95: file tmp.cpp, line 12.
(gdb) run
Starting program: /xxxxxxxx/tmp string
Breakpoint 1, main (argc=2, argv=0x7fffffffdc58) at tmp.cpp:12
12        std::reverse(str.begin(), str.end());
(gdb) print str
$1 = "string"
(gdb) info locals
str = "string"
(gdb)


                        
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