blocks|key|1343597|text|参数x已经包含来自main的变量x的地址，因此它必须编写为|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1343598|void+change_number(int+*x)
{

++int+*pointer_x;
++pointer_x+=+x;
++*x+=+*x+%2B+3;
++printf("In+sub+program:+x+=+%25d\n",+*pointer_x);

}|code-block|syntax|javascript|1343599|entityMap^0|2|1|9|4|G|1|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|L|$]]

The parameter <code>x</code> already contains address of the variable <code>x</code> from <code>main</code> so it have to be written as

<pre><code>void change_number(int *x)
{

 int *pointer_x;
 pointer_x = x;
 *x = *x + 3;
 printf("In sub program: x = %d\n", *pointer_x);

}
</code></pre>

blocks|key|1198840|text|编写void+change_number(int+*x)时，x作为int+*接收。因此，x指向int，*x指向int。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1198841|因此，您需要更改以下内容：|1198842|pointer_x+=+x;
*x+=+*x+%2B+3;
printf("In+sub+program:+x+=+%25d\n",+*pointer_x);|code-block|syntax|javascript|1198843|现在这个打印正确了。但是要恢复值，只需在末尾添加以下一行：|1198844|*x+=+*x+-+3;|1198845|entityMap^0|2|Q|U|1|X|5|18|1|1B|3|1F|2|1J|3|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|17|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|18|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|19|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|1A|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|R|$]]

When you write <code>void change_number(int *x)</code>, <code>x</code> is received as an <code>int *</code>. So, <code>x</code> points to <code>int</code> and <code>*x</code> is <code>int</code>.

So you'll need to change the following:

<pre><code>pointer_x = x;
*x = *x + 3;
printf("In sub program: x = %d\n", *pointer_x);
</code></pre>

Now this prints correctly. But to restore the value, just add this line at the end:

<pre><code>*x = *x - 3;
</code></pre>

blocks|key|1343646|text|您粘贴的代码应该无法在行pointer_x+=+&x;上编译，因为x和pointer_x都是int*类型|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1343647|使用指针变量上的address-of运算符提供指针到指针-在本例中，&x生成一种类型的int**。|1343648|此外，行x+=+x+%2B+3通过3*sizeof(int)字节在内存中推进指针位置，而不是修改原始的int变量。|1343649|也许你打算写*x+=+*x+%2B+3呢？|1343650|void+change_number(int+*x)
{
++++int+*pointer_x;
++++pointer_x+=+x;
++++*x+=+*x+%2B+3;
++++printf("In+sub+program:+x+=+%25d\n",+*pointer_x);
}|code-block|syntax|javascript|1343651|entityMap^0|C|F|X|1|Z|9|1A|4|0|Y|2|17|5|0|4|9|F|D|1D|3|0|6|B|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|1D|8|@$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|L|3|M|5|N|7|1G|8|@]|D|@]|E|$O|P]]|$1|Q|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|R|$]]

The code you've pasted should fail to compile on the line <code>pointer_x = &amp;x;</code> as both <code>x</code> and <code>pointer_x</code> are type <code>int*</code>

Using the address-of operator on a pointer variable gives you a pointer-to-pointer - in this case, <code>&amp;x</code> yields a type of <code>int**</code>

In addition, the line <code>x = x + 3</code> advances a pointer location in memory by <code>3*sizeof(int)</code> bytes, it's not modifying the original <code>int</code> variable.

Perhaps you intended to write <code>*x = *x + 3</code> instead?

<pre><code>void change_number(int *x)
{
 int *pointer_x;
 pointer_x = x;
 *x = *x + 3;
 printf("In sub program: x = %d\n", *pointer_x);
}
</code></pre>

I'm working with the program that scans a number in the main program. After that this program calls for a function change_number(), and as an argument gives the numbers memory address. After this program should add 3 to the number in the sub-program, print it out in the subprogram and restore that new value. However, when trying to print the number out in the subprogram change_number(), it prints out it's memory address. My understanding is that a program should return integers value when referring to the pointer with * -notation or just by inserting a variables name. Another compiler which i have tried says the following error message, and does not even compile, either with x -notation or with *pointer_x -notation: 

<ul>
<li>"You are trying to initialize a variable to a value that is of the wrong type. 
code.c:20: warning: assignment from incompatible pointer type".</li>
</ul>

I don't understand because my pointer is introduced as an integer, just like the integer itself. Here is the code:

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

void change_number(int *x);

int main()
{

 int x;
 printf("Give number x: ");
 scanf("%d", &amp;x);
 printf("In main program: x = %d\n", x);
 change_number(&amp;x);
 printf("In main program: x = %d\n", x);

 return 0;

}

void change_number(int *x)
{

int *pointer_x;
pointer_x = &amp;x;
x = x + 3;
printf("In sub program: x = %d\n", *pointer_x);

}
</code></pre>

Pointer returning memory address?

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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我正在用这个程序扫描主程序中的一个数字。之后，这个程序调用一个函数change_number()，并作为参数给出数字内存地址。在此程序之后，应将子程序中的数字加3，然后在子程序中打印出来，并恢复该新值。但是，当试图在子程序change_number()中打印出数字时，它会打印出它的内存地址。我的理解是，当程序引用带有*...

问指针返回内存地址？
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