为什么gdb在从函数返回时停在与"i b"不同的行？

Question

这是我试图调试的程序:

#include <stdio.h>
int i = 5;

int main(void)
{
    int x = 3;

    display(x);
    return 0;
}


void display(int x)
{
for ( i=0; i<x; ++i ) {
    printf("i is %d.\n", i);
}
}

此代码来自http://www.dirac.org/linux/gdb/05-Stepping_And_Resuming.php#breakpointsandwatchpoints.这是问题所在:

(gdb) break display 
Breakpoint 1 at 0x40051e: file try5.c, line 15.
(gdb) run
Starting program: /home/ja/gdb/learning/try5 

Breakpoint 1, display (x=3) at try5.c:15
(gdb) frame 1
#1  0x000000000040050c in main () at try5.c:8
(gdb) break 
Breakpoint 2 at 0x40050c: file try5.c, line 8.
(gdb) c
Continuing.
i is 0.
i is 1.
i is 2.

Breakpoint 2, main () at try5.c:9
(gdb) i b
Num     Type           Disp Enb Address            What
1       breakpoint     keep y   0x000000000040051e in display at try5.c:15
    breakpoint already hit 1 time
2       breakpoint     keep y   0x000000000040050c in main at try5.c:8
    breakpoint already hit 1 time
(gdb) c
Continuing.

Program exited normally.
(gdb) q

Debugger finished

它应该在main()的第8 行停止,但它在第9行停在main()处.对我来说这是误导.我认为它应该停在第9行,因为这就是'break'命令所做的 - 在下一条指令处设置一个断点.但为什么"信息断点"说断点设置在第8行？

Answer 1

正如您所看到的，断点放置在正确的位置，因为它在从函数返回后确实中断了。如果进行反汇编，您还会看到断点放置在正确的指令处（在本例中为 0x00401192）：

b display
r
f 1
b
disassemble $pc
...
   0x0040118d <+29>:    call   0x401199 <display>
=> 0x00401192 <+34>:    mov    $0x0,%eax
   0x00401197 <+39>:    leave

i b
...
2       breakpoint     keep y   0x00401192 in main at try5.c:8

但它显示错误的行号。首先我认为这可能与函数返回有关，所以我在显示调用后添加了额外的指令，但它仍然显示错误的行。

这对我来说看起来是一个错误。