Linux 如何以编程方式导致 C/C++ 中的核心转储
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How to programmatically cause a core dump in C/C++
提问by hhafez
I would like to force a core dump at a specific location in my C++ application.
我想在我的 C++ 应用程序中的特定位置强制进行核心转储。
I know I can do it by doing something like:
我知道我可以通过执行以下操作来做到这一点：
int * crash = NULL;
*crash = 1;
But I would like to know if there is a cleaner way?
但我想知道是否有更清洁的方法？
I am using Linux by the way.
顺便说一下，我正在使用 Linux。
采纳答案by paxdiablo
Raising of signal number 6 (SIGABRTin Linux) is one way to do it (though keep in mind that SIGABRT is not requiredto be 6 in all POSIX implementations so you may want to use the SIGABRTvalue itself if this is anything other than quick'n'dirty debug code).
提高信号编号 6（SIGABRT在 Linux 中）是一种方法（但请记住，SIGABRT在所有 POSIX 实现中不需要为 6，因此SIGABRT如果这不是 quick'n '脏调试代码）。
#include <signal.h>
: : :
raise (SIGABRT);
Calling abort()will also cause a core dump, and you can even do this withoutterminating your process by calling fork()followed by abort()in the child only - see this answerfor details.
调用abort()也会导致核心转储，您甚至可以在不终止进程的情况下执行此操作，方法是仅在子进程中调用fork()后跟abort()-有关详细信息，请参阅此答案。
回答by Suvesh Pratapa
As listed in the signal manpage, any signal with the action listed as 'core' will force a core dump. Some examples are:
如信号联机帮助页中所列，任何将操作列为“核心”的信号都将强制进行核心转储。一些例子是：
SIGQUIT 3 Core Quit from keyboard
SIGILL 4 Core Illegal Instruction
SIGABRT 6 Core Abort signal from abort(3)
SIGFPE 8 Core Floating point exception
SIGSEGV 11 Core Invalid memory reference
Make sure that you enable core dumps:
确保启用核心转储：
ulimit -c unlimited
回答by Jonathan Leffler
#include <stdlib.h> // C
//#include <cstdlib> // C++
void core_dump(void)
{
abort();
}
回答by Eugene Yokota
回答by smcameron
Invoke
调用
abort();
Related, sometimes you'd like a back trace without an actual core dump, and allow the program to continue running: check out glibc backtrace() and backtrace_symbols() functions: http://www.gnu.org/s/libc/manual/html_node/Backtraces.html
相关的，有时你想要一个没有实际核心转储的回溯，并允许程序继续运行：检查 glibc backtrace() 和 backtrace_symbols() 函数：http: //www.gnu.org/s/libc/手册/html_node/Backtraces.html
回答by ephemient
A few years ago, Google released the coredumperlibrary.
几年前，谷歌发布了coredumper库。
Overview
The coredumper library can be compiled into applications to create core dumps of the running program -- without terminating. It supports both single- and multi-threaded core dumps, even if the kernel does not natively support multi-threaded core files.
Coredumper is distributed under the terms of the BSD License.
Example
This is by no means a complete example; it simply gives you a feel for what the coredumper API looks like.
#include <google/coredumper.h> ... WriteCoreDump('core.myprogram'); /* Keep going, we generated a core file, * but we didn't crash. */
概述
coredumper 库可以编译到应用程序中以创建正在运行的程序的核心转储——无需终止。它支持单线程和多线程核心转储，即使内核本身不支持多线程核心文件。
Coredumper 根据 BSD 许可条款分发。
例子
这绝不是一个完整的例子；它只是让您了解 coredumper API 的样子。
#include <google/coredumper.h> ... WriteCoreDump('core.myprogram'); /* Keep going, we generated a core file, * but we didn't crash. */
It's not what you were asking for, but maybe it's even better :)
这不是您要的，但也许更好:)
回答by sigjuice
#include <assert.h>
.
.
.
assert(!"this should not happen");
回答by rka444
Sometimes it may be appropriate to do something like this:
有时做这样的事情可能是合适的：
int st = 0;
pid_t p = fork();
if (!p) {
signal(SIGABRT, SIG_DFL);
abort(); // having the coredump of the exact copy of the calling thread
} else {
waitpid(p, &st, 0); // rip the zombie
}
// here the original process continues to live
One problem with this simple approach is that only one thread will be coredumped.
这种简单方法的一个问题是只有一个线程会被核心转储。
回答by Zonk
Another way of generating a core dump:
生成核心转储的另一种方法：
$ bash
$ kill -s SIGSEGV $$
Just create a new instance of the bash and kill it with specified signal. The $$is the PID of
the shell. Otherwise you are killing your current bash and will be logged out, terminal closed or disconnected.
只需创建一个新的 bash 实例并用指定的信号杀死它。的$$是外壳的PID。否则，您将杀死当前的 bash，并将被注销、终端关闭或断开连接。
$ bash
$ kill -s SIGABRT $$
$ bash
$ kill -s SIGFPE $$
回答by karthik339
#include <stdio.h>
#include <stdlib.h>
int main()
{
printf("\n");
printf("Process is aborting\n");
abort();
printf("Control not reaching here\n");
return 0;
}
use this approach wherever you want :)
在任何地方使用这种方法:)
