如何使 GDB 断点仅在到达给定次数后才中断？

Question

我有一个被多次调用的函数，最终出现段错误。

但是，我不想在这个函数上设置断点并在每次调用它后停止，因为我将在这里工作很多年。

我听说我可以在 GDB 中为断点设置一个计数器，每次命中断点时，计数器都会递减，并且只有在计数器达到断点时才会触发= 0。

这准确吗？如果准确，我该怎么做？请给出设置此类断点的 gdb 代码。

Answer 1

阅读 GDB 手册的第 5.1.6 节。您要做的就是首先设置一个断点，然后为该断点编号设置一个“忽略计数”，例如ignore 23 1000。

如果您不知道要忽略断点多少次，并且不想手动计数，以下内容可能会有所帮助：

  ignore 23 1000000   # set ignore count very high.

  run                 # the program will SIGSEGV before reaching the ignore count.
                      # Once it stops with SIGSEGV:

  info break 23       # tells you how many times the breakpoint has been hit, 
                      # which is exactly the count you want

Answer 2

继续

这是一种方便的方法，可以跳过最后一个命中断点 n - 1 次（因此在第 n 个处停止） hit):

main.c

#include <stdio.h>

int main(void) {
    int i = 0;
    while (1) {
        i++; /* Line 6 */
        printf("%d\n", i);
    }
}

用法:

gdb -n -q main.out

GDB 会话:

Reading symbols from main.out...done.
(gdb) start
Temporary breakpoint 1 at 0x6a8: file main.c, line 4.
Starting program: /home/ciro/bak/git/cpp-cheat/gdb/main.out

Temporary breakpoint 1, main () at main.c:4
4           int i = 0;
(gdb) b 6
Breakpoint 2 at 0x5555555546af: file main.c, line 6.
(gdb) c
Continuing.

Breakpoint 2, main () at main.c:6
6               i++; /* Line 6 */
(gdb) c 5
Will ignore next 4 crossings of breakpoint 2.  Continuing.
1
2
3
4
5

Breakpoint 2, main () at main.c:6
6               i++; /* Line 6 */
(gdb) p i
$1 = 5
(gdb)
(gdb) help c
Continue program being debugged, after signal or breakpoint.
Usage: continue [N]
If proceeding from breakpoint, a number N may be used as an argument,
which means to set the ignore count of that breakpoint to N - 1 (so that
the breakpoint won't break until the Nth time it is reached).