blocks|key|3069531|text|尝试并可靠地取消引用指针是不可能的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3069532|这是因为应用程序处理内存的方式在不同的编译器之间可能会有所不同，在具有不同选项的相同编译器上也会有所不同(调试/释放模式的处理方式不同)。|3069533|您可以做的是使用一个信号显式地引发段错误：|3069534|#include+<signal.h>

int+main()
{
++++raise(SIGSEGV);
}|code-block|syntax|javascript|3069535|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|O|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|P|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

It impossible to try and reliable do it dereferencing pointers. 
This is because how the application handles memory can vary from compiler to compiler also across the same compiler with different options (debug/release mode handled differently).
What you can do is explicitly raise the segfault using a signal:
<pre><code>#include &lt;signal.h&gt;

int main()
{
 raise(SIGSEGV);
}
</code></pre>

blocks|key|3282605|text|我的一句话：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3282606|*(char+*)0+=+0;|code-block|syntax|javascript|3282607|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

My one line flavor:

<pre><code>*(char *)0 = 0;
</code></pre>

blocks|key|321216|text|有史以来最短的段故障：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|321217|*(int*)0=0;|code-block|syntax|javascript|321218|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Shortest segfault ever:

<pre><code>*(int*)0=0;
</code></pre>

blocks|key|3069508|text|您所做的这两件事都会在C%2B%2B中产生“未定义的行为”(实际上，调用main()是被明确禁止的)。不能保证它们会导致seg错误-这将取决于很多东西，但主要是你正在运行的平台，你没有指定。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3069509|事实上，任何建议的导致seg故障的方法都可能工作，也可能不工作。这是因为seg错误几乎总是与C%2B%2B的未定义行为的概念相关联。|3069510|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Both of the things you do will produce "undefined behaviour" in C++ (well, calling main() is actually explicitly forbidden). There is no guarantee that they will cause seg faults - this will depend on a lot of things, but mostly the platform you are running on, which you haven't specified.

In fact, any suggested method of causing a seg fault may or may not work. This is because a seg fault is almost always associated with C++'s concept of undefined behaviour.

blocks|key|3282548|text|int+*hello+=+NULL;
printf(*hello);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3282549|或者你可以定义一个结构(比如HelloWorld结构)|unstyled|3282550|HelloWorld+*myWorld+=+NULL;
myWorld->world+=+"hello";|3282551|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>int *hello = NULL;
printf(*hello);
</code></pre>

or you can define a struct (say HelloWorld struct)

<pre><code>HelloWorld *myWorld = NULL;
myWorld-&gt;world = "hello";
</code></pre>

blocks|key|3069522|text|对于第一个示例，编译器可能会将字符串放在可写内存中，因此在尝试更改它时不会出现seg错误。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3069523|对于第二种情况，编译器可能在优化调用，也可能在优化调用本身(因为它是一个tail+call)，这意味着堆栈实际上不会随着每次调用的返回地址而增长，因此您可以无限递归。|offset|length|3069524|但正如Neil在他的帖子中提到的那样，这些事情都会导致“未定义的行为”，因此在运行时不需要代码来生成seg错误。|3069525|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Tail_call^0|0|10|9|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Q|8|@]|9|@$D|R|E|S|1|T]]|A|$]]|$1|F|3|G|5|6|7|U|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|I|$J|$5|K|L|M|A|$N|O]]]]

For the first example, the compiler probably put the string in writable memory, so there is no seg fault when trying to change it.

For the second, the compiler may be optimizing the call away or may be optimizing the call itself away to just a jump (since it is a <a href="http://en.wikipedia.org/wiki/Tail_call" rel="nofollow noreferrer">tail call</a>), meaning the stack isn't actually growing with return addresses for each call, so you could recurse indefinitely.

But as Neil mentioned in his post, any of these things results in "undefined behavior", so the code is not required at runtime to generate a seg fault.

blocks|key|320201|text|char+*+ptr+=+0;
*ptr+=+1;

Segmentation+fault|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|320202|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>char * ptr = 0;
*ptr = 1;

Segmentation fault
</code></pre>

blocks|key|320224|text|有很多方法可以生成一个段错误。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|320225|比如取消引用一个错误的指针：|320226|char+*s+=+(char+*)0xDEADBEEF;
*s+=+'a';|code-block|syntax|javascript|320227|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Lots of ways to generate a segfault.

Like dereferencing a bad pointer:

<pre><code>char *s = (char *)0xDEADBEEF;
*s = 'a';
</code></pre>

blocks|key|3069518|text|维基百科的文章实际上列出了三种方法(其中一种是空指针引用)；你为什么不尝试一下呢？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3069519|至于为什么你尝试的两个例子没有，嗯，正如其他人所指出的那样，正确的答案是它是未定义的行为，所以任何事情都可能发生(包括不分段故障)。但有人可能会推测，第一种形式没有失败的原因是因为你的编译器或系统在内存保护方面松懈。对于第二种情况，尾递归感知的编译器可以优化无限递归的main循环，而不会导致堆栈溢出。|offset|length|style|CODE|3069520|entityMap^0|0|3Q|4|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

The Wikipedia article actually lists three methods (one of which is a null pointer dereference); why didn't you try that one?

As for why the two examples you tried didn't, well, the correct answer as others have noted is that it's undefined behavior, so anything could happen (which includes not segfaulting). But one could speculate that the reason that the first form didn't fail for you is because your compiler or system is lax about memory protection. As for the second case, a tail-recursive-aware compiler conceivably could optimize the infinitely recursive <code>main</code> loop and not end up overflowing the stack.

blocks|key|323206|text|如果你尝试连接两个常量...你会得到一个..。至少是一种简单的方法..。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|323207|strcat("a"，"b")；|323208|=)|323209|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

If you try to concatenate two constants... you'll get one... at least is a simple way...

strcat("a", "b");

=)

OK for whatever reason I'm having trouble causing a seg fault. I want to produce one so that I can use <code>gdb</code> to see how to debug one. I have tried both examples from <a href="http://en.wikipedia.org/wiki/Segmentation_fault" rel="noreferrer">the Wikipedia</a> article yet neither work.

The first one:

<pre><code>char *s = "Hello World!";
*s = 'H';
</code></pre>

And the second example:

<pre><code>int main(void) 
{
 main();
}
</code></pre>

EDIT: I'm using Ubutnu 9.10 and <code>g++</code> as my compiler. Can anyone show me some code that is guaranteed to segfault?

Why can't I cause a seg fault?

编译器

好的，不管是什么原因，我在引发seg错误时遇到了麻烦。我想生成一个，这样我就可以使用gdb来查看如何调试一个。我已经尝试了文章中的两个示例，但都不起作用。第一个：char *s = "Hello World!";*s = 'H';第二个例子：int main(void) {    main();}EDIT:我使用Ubunu9.10和g++作为我的编译器。有没有人能给我看一些保证会出现segfaul

问为什么我不能导致seg错误？
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