blocks|key|1811678|text|以下假设:您希望创建一个长度为1(稍后可能更长)的int指针数组。并且您希望数组中的单个指针指向名为“value”的局部变量。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1811679|然后：|1811680|int*+arr[]+=+{+0+};+//+or+{+NULL+},+if+you+prefer
//+++++++%5E++creates+an+array;+either+you+specify+size+explicitly+or+it+will+be+deduced
//++++++++++from+initializer,+as+in+this+case|code-block|syntax|javascript|1811681|如果将数组传递给函数，则数组会自动衰减为指向第一个元素的指针。所以：|1811682|add_value_to_array(arr,+&value);
//+++++++++++++++++%5E+decays+to+pointer
//++++++++++++++++++++++++%5E+here+you+still+need+a+pointer;+you+can+create+it+directly,+though|1811683|剩下的小问题是：arr衰减到指针的类型是int**。在您的函数中需要相同的类型：|offset|length|style|CODE|1811684|void+add_value_to_array(int**+array,+int*+value)
//++++++++++++++++++++++++++%5E+one+asterisk+less
{
++++*array
//++%5E+this+one+is+NOW+correct:+array+points+to+the+first+element+of+arr
//++++dereferencing+gives+you+exactly+this+element,+i.+e.+the+pointer

++++++++=+/*+&+*/value;
//+++++++++++%5E+must+be+dropped:+you+already+have+a+pointer+and+you+assign+it+to
//+++++++++++++another+one,+which+is+fine.
}|1811685|请注意，指针只是内存中某处变量的地址。稍微简化一下：|1811686|int+n+=+10;
int*+p0+=+&n;
int*+p1+=+&n;+//+p1+and+p0+point+both+to+the+same+variable+n
int*+p2+=+p0;+//+now+again+p2+points+to+n
*p0+=+12;+++++//+change+the+value+pointed+to
printf("%25d\n",+*p2);+//+will+print+12,+as+both+pointers+point+to+the+same+address.|1811687|函数参数在这方面没有区别，它们只是普通变量。如果被指针对象是一个数据值或指针本身，它就不起作用：|1811688|int+n+=+10;
int+m+=+12;
int*+p+=+&n;
int**+p1+=+&p;+//+pointer+to+pointer+to+int
int**+p2+=+p1;+//+both+p1+and+p2+point+to+p
*p1+=+&m;++++++//+re-assign+the+POINTER
printf("%25d\n",+**p2);+//+again+prints+12:
++++++++++++++++++++++//+p1+and+p2+both+point+to+p+which+was+reset+to+m,+though...|1811689|entityMap^0|0|0|0|0|0|8|3|K|5|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|15|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|16|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|17|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|18|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|19|8|@$O|1A|P|1B|Q|R]|$O|1C|P|1D|Q|R]]|9|@]|A|$]]|$1|S|3|T|5|F|7|1E|8|@]|9|@]|A|$G|H]]|$1|U|3|V|5|6|7|1F|8|@]|9|@]|A|$]]|$1|W|3|X|5|F|7|1G|8|@]|9|@]|A|$G|H]]|$1|Y|3|Z|5|6|7|1H|8|@]|9|@]|A|$]]|$1|10|3|11|5|F|7|1I|8|@]|9|@]|A|$G|H]]|$1|12|3|-4|5|6|7|1J|8|@]|9|@]|A|$]]]|13|$]]

Following assumption: You want to create an array of length 1 (maybe greater length later) of pointers to int. And you want that single pointer in the array to point to the local variable named 'value'.

Then:

<pre><code>int* arr[] = { 0 }; // or { NULL }, if you prefer
// ^ creates an array; either you specify size explicitly or it will be deduced
// from initializer, as in this case
</code></pre>

Arrays automatically decay to pointer to first element, if you pass them to a function. So:

<pre><code>add_value_to_array(arr, &amp;value);
// ^ decays to pointer
// ^ here you still need a pointer; you can create it directly, though
</code></pre>

Little problem left: <code>arr</code> decaying to pointer is of type <code>int**</code>. You need the same type in your function:

<pre><code>void add_value_to_array(int** array, int* value)
// ^ one asterisk less
{
 *array
// ^ this one is NOW correct: array points to the first element of arr
// dereferencing gives you exactly this element, i. e. the pointer

 = /* &amp; */value;
// ^ must be dropped: you already have a pointer and you assign it to
// another one, which is fine.
}
</code></pre>

Be aware that pointers simply are addresses of variables somewhere in memory. A bit simplified:

<pre><code>int n = 10;
int* p0 = &amp;n;
int* p1 = &amp;n; // p1 and p0 point both to the same variable n
int* p2 = p0; // now again p2 points to n
*p0 = 12; // change the value pointed to
printf("%d\n", *p2); // will print 12, as both pointers point to the same address.
</code></pre>

Function parameters do not differ in this respect, they are just ordinary variables. And it doesn't play a role if the pointee is a data value or a pointer itself:

<pre><code>int n = 10;
int m = 12;
int* p = &amp;n;
int** p1 = &amp;p; // pointer to pointer to int
int** p2 = p1; // both p1 and p2 point to p
*p1 = &amp;m; // re-assign the POINTER
printf("%d\n", **p2); // again prints 12:
 // p1 and p2 both point to p which was reset to m, though...
</code></pre>

blocks|key|1597389|text|感谢你们的回答，这帮助我找到了bug+:我必须通过地址而不是值来传递我的value_ptr。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1597390|以下是我更正后的代码：|1597391|void+add_value_to_array(int+***+array,+int+**+value_ptr)+{
++++*array+=+value_ptr;
}
int+main()+{

++++int+**+arr+=+{+0+};
++++int+value+=+5;
++++int+*+value_ptr+=++&value;

++++add_value_to_array(&arr,+&value_ptr);
++++//arr+=+&value_ptr;
++++printf("looool\n");
++++printf("looool\n");
++++printf("looool\n");
++++printf("looool\n");
++++printf("%25p\n",+arr[0]);

}|code-block|syntax|javascript|1597392|感谢大家的帮助！|1597393|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Thank you all for your answers, this helped me find the bug :
I had to pass my value_ptr by address and not by value. 
Here is my corrected code :

<pre class="lang-c prettyprint-override"><code>void add_value_to_array(int *** array, int ** value_ptr) {
 *array = value_ptr;
}
int main() {

 int ** arr = { 0 };
 int value = 5;
 int * value_ptr = &amp;value;

 add_value_to_array(&amp;arr, &amp;value_ptr);
 //arr = &amp;value_ptr;
 printf("looool\n");
 printf("looool\n");
 printf("looool\n");
 printf("looool\n");
 printf("%p\n", arr[0]);

}
</code></pre>

Thank you all for your help !

I am trying to make a simple piece of c code using pointers work but memory is being overwritten unexpectedly.

I want to create an array of pointers to integers then create a pointer to an integer and assign it's address to an array.

Therefore, the array will point to a pointer to an integer. I did this in a function called <code>add_value_to_array()</code>.

Here is my code :

<pre class="lang-c prettyprint-override"><code>void add_value_to_array(int *** array, int * value) {
 *array = &amp;value;
}
int main() {

 int ** arr = { 0 };
 int value = 5;
 int * value_ptr = &amp;value;

 add_value_to_array(&amp;arr, value_ptr);
 //arr = &amp;value_ptr;
 printf("looool\n");
 printf("looool\n");
 printf("looool\n");
 printf("looool\n");
 printf("%p\n", arr[0]);

}
</code></pre>

What I want is :

<pre><code>arr -&gt; value_ptr -&gt; 5
(arr = &amp;value_ptr
*array = value_ptr
value_ptr = &amp;value
*value_ptr = 5
**arr = 5)
</code></pre>

however, this is what I have right after <code>add_value_to_array()</code> is called, but memory is overwritten when I call the <code>printfs()</code>, I get garbage values in my <code>arr</code> variable.

Even weirder, if I do directly <code>arr = &amp;value_ptr</code> instead of calling <code>add_value_to_array</code>, things go as expected and memory is not overwritten by the <code>printfs()</code>.

So it seems that memory is allocated differently if I use a function or if I do I do things outside of it.

What is happening that I am seeing this behavior?

Assigning address of pointer to pointer-to-pointer

我试图使用指针来编写一段简单的c代码，但是内存被意外地覆盖了。我想创建一个指向整数的指针数组，然后创建一个指向整数的指针，并将其地址分配给一个数组。因此，数组将指向一个指向整数的指针。我在一个名为add_value_to_array()的函数中做到了这一点。下面是我的代码：void add_value_to_array...

问将指针的地址分配给指针到指针
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