break*main VS在GDB中打破main()

Question

break *main和break main()本质上有什么区别？例如:

    #include <iostream>
    using namespace std;
    int main()
    {
        int x=30;
        int y=40;
        x=y;
        return 0;
   }

当我使用 break *main和watch x,它是这样的:

(gdb) b *main
Breakpoint 1 at 0x400674: file aa.cpp, line 4.
(gdb) r
Starting program: /root/dd/aa.out 
Breakpoint 1, main () at aa.cpp:4
4       {
(gdb) n
5               int x=30;
(gdb) watch x
Hardware watchpoint 2: x
(gdb) c
Continuing.
Hardware watchpoint 2: x

Old value = 0
New value = 30
main () at aa.cpp:6
6               int y=40;
(gdb) c
Continuing.
Hardware watchpoint 2: x

Old value = 30
New value = 40
main () at aa.cpp:8
8               return 0;
(gdb) 

但是当我使用break main()和watch x,它是这样的:

(gdb) b main()
Breakpoint 1 at 0x400678: file aa.cpp, line 5.
(gdb) r
Starting program: /root/dd/aa.out 
Breakpoint 1, main () at aa.cpp:5
5               int x=30;
(gdb) watch x
Hardware watchpoint 2: x
(gdb) c
Continuing.
Hardware watchpoint 2: x

Old value = 0
New value = 40
main () at aa.cpp:8
8               return 0;
(gdb) 

他们为什么不同？那本质上有什么区别？

当我观看阵列时,如果我使用break main()它,它将出现:

Watchpoint 2 deleted because the program has left the block in
which its expression is valid.

但如果我使用break *main,它将不会出现,为什么？

Answer 1

而本质上有什么不同

区别在于b *main第一条指令的中断main,而在函数序言之后b main的第一条指令中断.

在我的构建(g++ -g t.cc使用gcc 4.8.4-2ubuntu1~14.04.3和gdb 7.9)中,对源代码的反汇编如下所示:

(gdb) disas main
Dump of assembler code for function main():
   0x00000000004006cd <+0>: push   %rbp
   0x00000000004006ce <+1>: mov    %rsp,%rbp
   0x00000000004006d1 <+4>: movl   $0x1e,-0x8(%rbp)
   0x00000000004006d8 <+11>:    movl   $0x28,-0x4(%rbp)
   0x00000000004006df <+18>:    mov    -0x4(%rbp),%eax
   0x00000000004006e2 <+21>:    mov    %eax,-0x8(%rbp)
   0x00000000004006e5 <+24>:    mov    $0x0,%eax
   0x00000000004006ea <+29>:    pop    %rbp
   0x00000000004006eb <+30>:    retq
End of assembler dump.

设置b *mainvs. b main产生:

(gdb) b *main
Breakpoint 1 at 0x4006cd: file t.c, line 4.
(gdb) b main
Breakpoint 2 at 0x4006d1: file t.c, line 5.

我无法重现你观察到的问题:

(gdb) r
Starting program: /tmp/a.out

Breakpoint 1, main () at t.c:4
4       {
(gdb) c
Continuing.

Breakpoint 2, main () at t.c:5
5           int x=30;
(gdb) p x
$1 = 0
(gdb) watch x
Hardware watchpoint 3: x
(gdb) c
Continuing.
Hardware watchpoint 3: x

Old value = 0
New value = 30
main () at t.c:6
6           int y=40;