blocks|key|3756877|text|您可能会看到222，因为用于abc中的本地数组的内存已被用于其他东西--函数xyz的堆栈。你把一个地址传递到那个记忆里。再进行几次函数调用，*a可能包含一些其他值。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3756878|应该在函数执行后超出范围。为什么不发生这种事？
|blockquote|3756879|3756880|这个变量已经超出了范围。使用函数外部的地址是不正确的代码:使用指向函数返回的本地数据的指针是未定义的行为。|3756881|entityMap^0|E|3|12|3|1Y|2|0|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]|$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|U|8|@]|D|@]|E|$]]|$1|I|3|-4|5|6|7|V|8|@]|D|@]|E|$]]|$1|J|3|K|5|6|7|W|8|@]|D|@]|E|$]]|$1|L|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|M|$]]

You might see 222 because the memory that was used for the local array in <code>abc</code> has been used for something else - the stack for the function <code>xyz</code>. And you're passing around an address to that memory. Make a few more function calls and <code>*a</code> may contain some other value.

<blockquote>
 should go out of scope after the execution of the function. why is that not happening ?
</blockquote>

The variable has gone out of scope. Using that address outside the function is incorrect code: using a pointer to local data returned from a function is undefined behavior.

blocks|key|3757011|text|2:它正在发生，整个程序都有未定义的行为。这不是一个正确的程序，也没有什么意义去思考如果和但是。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|3757012|entityMap^0|E|6|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

2: It is happening, and the entire program has undefined behaviour. It is not a correct program, and there's little point musing about ifs and buts.

blocks|key|3733839|text|变量a和b是自动变量；在其他函数中使用它们的入口是一种未定义的行为。任何事情都可能发生:您不能期望输出(例如，优化编译器可以删除一些非法代码)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3733840|entityMap^0|2|1|4|1|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

The variables <code>a</code> and <code>b</code> are automatic variables; using their adress in a other function is an undefined behavior. Anything can happen : you can't expect an output (eg, an optimizing compiler can delete some illegal code).

blocks|key|3734018|text|若要返回指针，必须是指向动态分配变量或静态或全局变量的指针。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3734019|将指针返回到堆栈变量将使您有指向堆栈的指针，在调用新方法时将重用该指针。|3734020|在您的示例中，重用堆栈变量用于另一个数组，并覆盖在调用第一个方法时存储的旧值。|3734021|再次尝试调用printf，您将看到不同的输出，因为对printf的第一次调用更改了堆栈内容。|offset|length|style|CODE|3734022|entityMap^0|0|0|0|6|6|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|P|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|Q|8|@$H|R|I|S|J|K]]|9|@]|A|$]]|$1|L|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|M|$]]

to return a pointer it must be a pointer to dynamically allocated variable or static or global variable.

returning a pointer to a stack variable will cause you to have pointer to the stack which will be reused when you call a new method.

it happened in your case to reuse the stack variable for another array and overwrite the old value stored when you called the first method.

try to call <code>printf</code> again you will see different output because the first call to printf changed the stack content.

blocks|key|3733948|text|为什么不发生这种事？
|type|blockquote|depth|inlineStyleRanges|entityRanges|data|3733949|unstyled|3733950|它确实发生了，只是形式上的。未定义的行为是没有义务崩溃或行为不当的-他们“任何事情都可能发生”意味着它也可以运行似乎没有任何错误。I+just+answered+a+similar+question.|offset|length|style|BOLD|3733951|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/12463241/not-null-terminating-a-c-style-string/12463277#12463277^0|0|0|L|C|1T|Z|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|-4|5|C|7|S|8|@]|9|@]|A|$]]|$1|D|3|E|5|C|7|T|8|@$F|U|G|V|H|I]]|9|@$F|W|G|X|1|Y]]|A|$]]|$1|J|3|-4|5|C|7|Z|8|@]|9|@]|A|$]]]|K|$L|$5|M|N|O|A|$P|Q]]]]

<blockquote>
 why is that not happening ?
</blockquote>

It does happen, just formally. Undefined behavior is not obligated to crash or to misbehave - they "anything might happen" means it can also run seemingly without any error. <a href="https://stackoverflow.com/questions/12463241/not-null-terminating-a-c-style-string/12463277#12463277">I just answered a similar question.</a>

blocks|key|3733978|text|函数abc和xyz分别将一个地址传回给本地创建的数组。随后的调用正在混合以前使用的内存(并将其传回给您)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3733979|这些称为自动局部变量。|3733980|您需要将这些数组声明为静态的，或者以不同的方式分配内存。|3733981|entityMap^0|2|3|6|3|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Q|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|R|8|@]|D|@]|E|$]]|$1|J|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|K|$]]

The functions <code>abc</code> and <code>xyz</code> are each passing back an address to a locally created array. Subsequent calls are mashing the memory that was previously used (and passed back to you). 

These are called automatic local variables. 

You'll need to declare those arrays as static or allocate the memory in a different way.

<blockquote>
 Possible Duplicate: 
 <a href="https://stackoverflow.com/questions/6441218/can-a-local-variables-memory-be-accessed-outside-its-scope">Can a local variable&#39;s memory be accessed outside its scope?</a> 
</blockquote>



I recently came across the following code:

<pre><code>#include &lt;stdio.h&gt;

int* abc () {
 int a[3] = {1,10,100};
 return a;
}
int* xyz () {
 int b[1] = {222};
 return b;
}
int main() {
 int *a, *b;
 a = abc();
 b = xyz();
 printf("%d\n", *a);
 return 0;
}
</code></pre>

the output is <code>222</code>. <code>'a'</code> is pointing to the array declared inside the <code>xyz()</code>.

my question is:

<ol>
<li>why is a pointing to the array declared inside <code>xyz()</code>.</li>
<li>the array declared inside the function <code>xyz()</code> should go out of scope after the execution of the function. why is that not happening ?</li>
</ol>

strange behavior in c

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没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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可能重复：最近我发现了以下代码：#include <stdio.h>int* abc () { int a[3] = {1,10,100}; return a;}int* xyz () { int b[1] = {222}; return b;}int main() { int *a, *b; a = abc(); b = xyz(); printf("%d\n",

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