break * main VS break main（）在GDB中

Question

break *main和break main()之间的区别是什么？ 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，那就是：

(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，那就是：

(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) 

为什么他们不同？本质上的区别是什么？

当我看一个数组，如果我用break main()，它会出现：

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

，但如果我用break *main，就不会出现，为什么呢？

Answer 1

，是什么本质的区别

不同的是，b *main断裂上的main的第一条指令，而b main断裂第一指令后功能序言。

在我的生成（g++ -g t.cc，使用gcc 4.8.4-2ubuntu1~14.04.3和gdb 7.9），将信号源的反汇编如下所示：

(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. 

和设置b *main与b main生产：

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

我无法重现您观察到的问题：

(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;