blocks|key|2484193|text|更改指针的内容(相对于所指向的对象)不会影响指向同一对象的其他指针。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2484194|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Changing the contents of a pointer (as opposed to the object being pointed to) will not affect other pointers to the same object.

blocks|key|2962689|text|指针B的位置将保持不变。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2962690|因为指针A和指针B是不同的指针，所以它们有不同的内存address.That，这意味着它们是相互独立的，尽管它们都指向相同的内存地址。|2962691|但是当您试图更改指针A的内容时要小心，此操作可能会导致指针B成为野生指针！|offset|length|style|BOLD|2962692|entityMap^0|0|0|W|4|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|N|8|@$F|O|G|P|H|I]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|K|$]]

Pointer B's location will stay the same.

Because pointer A and pointer B are different pointer,they have different memory address.That means they are independent each other although both of them point to the same memory address.

But you should be careful when you try to change pointer A's content,this operation may cause pointer B be a wild pointer by accident!

blocks|key|695447|text|您可以看到以下代码的输出：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|695448|#include+<iostream>

int+main(int+argc,+char+**argv){
++++int+a+=+3,+d+=+4;
++++int+*b+=+&a,+*c+=+&a;
++++std::cout+<<+b+<<+",+"+<<+c+<<+std::endl;
++++b+=+&d;
++++std::cout+<<+b+<<+",+"+<<+c+<<+std::endl;
}|code-block|syntax|javascript|695449|以下是代码的输出：|695450|0x7ffca3788e40,+0x7ffca3788e40
0x7ffca3788e44,+0x7ffca3788e40|695451|显然，将b指向的地址更改为，与指针c指向的地址无关。|offset|length|style|CODE|695452|但是当你使用references时，你可能会有一个类似的现象。引用是另一个对象的别名的实体。假设你有|695453|int+&ref+=+a;
ref+=+5;|695454|现在，更改变量ref还会使变量a的值发生变化，等于5。|695455|entityMap^0|0|0|0|0|4|1|H|1|0|6|A|0|0|7|3|F|1|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|12|8|@$M|13|N|14|O|P]|$M|15|N|16|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|17|8|@$M|18|N|19|O|P]]|9|@]|A|$]]|$1|S|3|T|5|D|7|1A|8|@]|9|@]|A|$E|F]]|$1|U|3|V|5|6|7|1B|8|@$M|1C|N|1D|O|P]|$M|1E|N|1F|O|P]]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1G|8|@]|9|@]|A|$]]]|X|$]]

You can see the output of the below code :

<pre><code>#include &lt;iostream&gt;

int main(int argc, char **argv){
 int a = 3, d = 4;
 int *b = &amp;a, *c = &amp;a;
 std::cout &lt;&lt; b &lt;&lt; ", " &lt;&lt; c &lt;&lt; std::endl;
 b = &amp;d;
 std::cout &lt;&lt; b &lt;&lt; ", " &lt;&lt; c &lt;&lt; std::endl;
}
</code></pre>

The following is the output of the code :

<pre><code>0x7ffca3788e40, 0x7ffca3788e40
0x7ffca3788e44, 0x7ffca3788e40
</code></pre>

Clearly, changing the address <code>b</code> points to, has nothing to do with the address pointed by pointer <code>c</code>.

But you can have a near-similar phenomena when you use <code>references</code>. A reference is an entity that is an alias for another object. Suppose you have

<pre><code>int &amp;ref = a;
ref = 5;
</code></pre>

Now, changing the variable <code>ref</code> will also bring a change in the value of variable <code>a</code>, equating it to 5.

blocks|key|2962675|text|指针保存内存中一段数据的内存地址。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2962676|指针变量的值是此地址。|2962677|int+targetA+=+1;
int+targetB+=+2;

int+*pointerA+=+&targetA;
int+*pointerB+=+&targetA;

printf("%25p,+%25p\n",+pointerA,+pointerB);
//+=>+0x7fff5c78e8f8,+0x7fff5c78e8f8

pointerB+=+&targetB;

printf("%25p,+%25p\n",+pointerA,+pointerB);
//+=>+0x7fff5c78e8f8,+0x7fff5c78e8f4|code-block|syntax|javascript|2962678|通过为指针分配不同的地址/位置，您只需更改指针指向的内容，其他变量就不会受到影响。|2962679|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|$]]

A pointer holds the memory address of a piece of data in memory.

The value of the pointer variable is this address.

<pre><code>int targetA = 1;
int targetB = 2;

int *pointerA = &amp;targetA;
int *pointerB = &amp;targetA;

printf("%p, %p\n", pointerA, pointerB);
// =&gt; 0x7fff5c78e8f8, 0x7fff5c78e8f8

pointerB = &amp;targetB;

printf("%p, %p\n", pointerA, pointerB);
// =&gt; 0x7fff5c78e8f8, 0x7fff5c78e8f4
</code></pre>

By assigning a different address / location to the pointer you just change what it points to, other variables are not affeted.

blocks|key|693615|text|让我们看一下这个代码片段。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|693616|#include+<iostream>
#include+<string>

using+namespace+std;

int+main+()
{
++int+*+myp1,+*+myp2;
++int+myvalue;
++myp1+=+&myvalue;
++myp2+=+&myvalue;
++myvalue+=+10;
++cout+<<+"first+address+is+"+<<+myp1+<<+'\n';
++cout+<<+"second+address+is+"+<<+myp2+<<+'\n';|code-block|syntax|javascript|693617|在这里，输出是相同的(当我运行它时，我得到的第一个地址是0x759ba27bb1d8，第二个地址是0x759ba27bb1d8)。|693618|++int+myvalue2;
++myp1+=+&myvalue2;
++myvalue2+=+20;
++cout+<<+"first+address+is+"+<<+myp1+<<+'\n';
++cout+<<+"second+address+is+"+<<+myp2+<<+'\n';
++return+0;|693619|}在这里，地址是不同的，但是myp2指向与上面相同的地址。|693620|这是有意义的；你有两个独立的东西指向同一个地址。为什么要改变一件事来改变另一件事？|693621|一个(可能是非常糟糕的类比)是:考虑到你和世界另一端的某个人在指向天空(比如说，在天空中的同一点)。然后，你指着地面。仅仅因为你指着地面，并不意味着另一个人会停止指着天空。|693622|希望这能有所帮助。|693623|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

Let's take a look at this code snippet. 

<pre><code>#include &lt;iostream&gt;
#include &lt;string&gt;

using namespace std;

int main ()
{
 int * myp1, * myp2;
 int myvalue;
 myp1 = &amp;myvalue;
 myp2 = &amp;myvalue;
 myvalue = 10;
 cout &lt;&lt; "first address is " &lt;&lt; myp1 &lt;&lt; '\n';
 cout &lt;&lt; "second address is " &lt;&lt; myp2 &lt;&lt; '\n';
</code></pre>

Here, the outputs are the same (when I run it, I get first address is 0x759ba27bb1d8
second address is 0x759ba27bb1d8).

<pre><code> int myvalue2;
 myp1 = &amp;myvalue2;
 myvalue2 = 20;
 cout &lt;&lt; "first address is " &lt;&lt; myp1 &lt;&lt; '\n';
 cout &lt;&lt; "second address is " &lt;&lt; myp2 &lt;&lt; '\n';
 return 0;
</code></pre>

}
Here, the addresses are different, but myp2 points to the same address as above.

And it makes sense; you have two separate things pointing to the same address. Why would changing what one thing points to change the other?

A (possibly very bad analogy) would be: consider that you and someone on the other end of the world are pointing at the sky (say, at the same point in the sky). Then, you point at the ground. Just because you pointed at the ground, doesn't mean the other guy will stop pointing at the sky.

Hope that helps.

I can't seem to find the answer on here, but I'm sorry if this is a duplicate. Here's my question: When I have two pointers to the same location, then I change the address of one (let's say pointer A), will I (by accident) be changing the address of the other pointer (pointer B)? Or will pointer B's location stay the same?

Two pointers to the same location?

翻译质量差，导致语言生硬或混乱。

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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我似乎在这里找不到答案，但如果这是重复的，我很抱歉。下面是我的问题:当我有两个指向同一位置的指针时，我会更改其中一个的地址(例如指针A)，我会(偶然地)更改另一个指针(指针B)的地址吗？还是指针B的位置保持不变？

问两个指向同一地点的指针？
EN

回答 5

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问两个指向同一地点的指针？EN

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问两个指向同一地点的指针？
EN