Question

I am facing problem While using watch in gdb . I am trying to keep a watch on variable m in my code . But for some reason i am getting the following message no symbol m in current context. I have kept a break point at line 7 so that scope of m is known .

    steps performed by me :-
    1>g++ -g a.cpp
    2>gdb a.out
    3>(gdb)break 7
    4>(gdb)watch m

我的方案如下:

    # include<iostream>
    # include<stdio.h>
    using namespace std;

    int main()
    {

      int m=10;
      char *abc = (char *)"ritesh";
      cout << abc << endl ;
      m=11; 
      m=13;
      abc=NULL;
      cout << *abc <<endl;

     return 0;
    }

我也看到。我如何使用“监视”的银? 但是,这无助于我。一个人能够解释这一问题,即它正面临的问题。以下是与我方联会有关的资料

    ritesh@ubuntu:~$ gdb a.out 
    GNU gdb (Ubuntu/Linaro 7.3-0ubuntu2) 7.3-2011.08
    Copyright (C) 2011 Free Software Foundation, Inc.
    License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>
    This is free software: you are free to change and redistribute it.
    There is NO WARRANTY, to the extent permitted by law.  Type "show copying"
    and "show warranty" for details.
    This GDB was configured as "i686-linux-gnu".
    For bug reporting instructions, please see:
    <http://bugs.launchpad.net/gdb-linaro/>...
    Reading symbols from /home/ritesh/a.out...done.

Answer 1

当你把方案装上夸张时,它还没有运行。但是,你试图看到一个文号,该编号将开始在职能中“有效”——<代码>main()功能——并将在你从该职能返回时“消失”。

例如,在本法典中

void func() {
  int b = 1;
  ++b;
  cout << b << endl;
}

int main() {
  int a = 1;
  func();
  cout << a << endl;
}

在开始执行一项方案之前,可在<条码>a上查询,并在执行项目进入<条码>/条码>之前对<条码>b的价值进行观察。

友情链接