Question

是否可以让gdb调试器与gcc样式的嵌套函数一起使用？如果是这样，需要哪些版本的gcc / gdb？

Answer 1

看起来还没有。它有一个相应的错误http://sourceware.org/bugzilla/show_bug.cgi?id=8300。

Answer 2

看来gcc / gdb已经掌握了它。对于早期版本，它只是不起作用，但这是一个简单会话的记录：

jla@jla-desktop$ gdb --version
GNU gdb (GDB) Fedora (7.2-51.fc14)
Copyright (C) 2010 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-redhat-linux-gnu".
For bug reporting instructions, please see:
<http://www.gnu.org/software/gdb/bugs/>.
~
jla@jla-desktop$ gcc --version
gcc (GCC) 4.5.1 20100924 (Red Hat 4.5.1-4)
Copyright (C) 2010 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

~
jla@jla-desktop$ cat hello.c
#include<stdio.h>

void main(void)
{
    char* hello="hello";
    int sub(int a){
        int b;
        int c;
        b=a;
        a++;
        c=a;
        return a+b+c;
    }

    printf("%s(%i)\n", hello, sub(3));
}

~
jla@jla-desktop$ gcc -ggdb3 hello.c -o hello
~
jla@jla-desktop$ gdb hello
GNU gdb (GDB) Fedora (7.2-51.fc14)
Copyright (C) 2010 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-redhat-linux-gnu".
For bug reporting instructions, please see:
<http://www.gnu.org/software/gdb/bugs/>...
Reading symbols from /home/jla/hello...done.
(gdb) b main
Breakpoint 1 at 0x80483f1: file hello.c, line 5.
(gdb) list
1   #include<stdio.h>
2   
3   void main(void)
4   {
5       char* hello="hello";
6       int sub(int a){
7           int b;
8           int c;
9           b=a;
10          a++;
(gdb) r
Starting program: /home/jla/hello 

Breakpoint 1, main () at hello.c:5
5       char* hello="hello";
Missing separate debuginfos, use: debuginfo-install glibc-2.13-2.i686
(gdb) info locals
hello = 0x7aeff4 "|\355z"
(gdb) s
15      printf("%s(%i)\n", hello, sub(3));
(gdb) s
sub (a=3) at hello.c:9
9           b=a;
(gdb) info locals
b = 134518388
c = 0
(gdb) p a
$1 = 3
(gdb) p b
$2 = 134518388
(gdb) p c
$3 = 0
(gdb) where
#0  sub (a=3) at hello.c:9
#1  0x08048405 in main () at hello.c:15
(gdb)

可以gdb调试gcc样式的嵌套C函数吗？

2 个答案: