Ser*_*kov 11 linux gdb coredump
我在分析核心文件时使用的是真正的gdb脚本我尝试取消引用指针并获取"源代码命令文件中的错误:无法访问地址处的内存"然后我的gdb脚本停止.我想要的只是继续执行我的gdb脚本而不停止.可能吗?
这是一个测试程序和测试gdb脚本,用于演示我的问题.在这种情况下,指针具有NULL值,但在实际情况下,指针将不具有null无效值.
这是测试C程序:
#include <stdio.h>
struct my_struct {
int v1;
int v2;
};
int main()
{
my_struct *p;
printf("%d %d\n", p->v1, p->v2);
return 0;
}
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这是一个测试gdb脚本:
>cat analyze.gdb
p p->v1
q
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这是问题的证明(我想从gdb这里得到这个错误信息然后去处理quit命令):
>gdb -silent a.out ./core.22384 -x ./analyze.gdb
Reading symbols from /a.out...done.
[New Thread 22384]
Core was generated by `./a.out'.
Program terminated with signal 11, Segmentation fault.
#0 0x0000000000400598 in main () at main.cpp:11
11 printf("%d %d\n", p->v1, p->v2);
./analyze.gdb:1: Error in sourced command file:
Cannot access memory at address 0x0
Missing separate debuginfos, use: debuginfo-install glibc-2.12-1.80.el6.x86_64
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>cat ./analyze.v2.gdb
python
def my_ignore_errors(arg):
try:
gdb.execute("print \"" + "Executing command: " + arg + "\"")
gdb.execute (arg)
except:
gdb.execute("print \"" + "ERROR: " + arg + "\"")
pass
my_ignore_errors("p p")
my_ignore_errors("p p->v1")
gdb.execute("quit")
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这是它的工作原理:
>gdb -silent ./a.out -x ./analyze.v2.gdb -c ./core.15045
Reading symbols from /import/home/a.out...done.
[New Thread 15045]
Core was generated by `./a.out'.
Program terminated with signal 11, Segmentation fault.
#0 0x0000000000400598 in main () at main.cpp:11
11 printf("%d %d\n", p->v1, p->v2);
$1 = "Executing command: p p"
$2 = (my_struct *) 0x0
$3 = "Executing command: p p->v1"
$4 = "ERROR: p p->v1"
$5 = "Executing command: quit"
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gdb的命令语言无法在处理命令时忽略错误.
但是,如果您的gdb是使用Python扩展构建的,那么这很容易完成.搜索"ignore-errors"脚本.有了它,你可以:
(gdb)ignore-errors print*foo
...并且将显示打印中的任何错误,但不会中止脚本的其余部分.