blocks|key|5718552|text|从你的十六进制地址-|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5718553|The+address+of+firstInt+is:+0x7fff5fbff7cc|code-block|syntax|javascript|5718554|地址的大小是6个字节长。但是unsignedint的大小是4个字节。当您试图使用%25u打印地址时，它将导致未定义的行为。|offset|length|style|CODE|5718555|因此，始终使用%25p打印地址。|5718556|entityMap^0|0|0|E|B|14|2|0|7|2|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@$I|Y|J|Z|K|L]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|P|$]]

From your hexadecimal address-

<pre><code>The address of firstInt is: 0x7fff5fbff7cc
</code></pre>

The size of the address is 6 bytes long. But Size of <code>unsignedint</code> is 4 bytes. When you trying to print the address using <code>%u</code>, It will cause undefined behaviour.

So always print the address using <code>%p</code>.

blocks|key|5718615|text|看来你在修一台64位的机器。你的指针是64位长|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5718616|两者(&firstInt和pointerFirstInt)是完全相同的。但显示方式不同。"%25p"知道指针是64位，并以十六进制形式显示它们。"%25u"显示十进制数，并假定为32位。所以只显示了一部分。|offset|length|style|CODE|5718617|如果将1606416332转换为十六进制，则如下所示：0x5FBFF7CC。您可以看到，这是64位地址的下半部。|5718618|编辑:进一步解释：|5718619|由于printf是var-arg函数，所以您提供给它的所有参数都放在堆栈上。因为你在这两种情况下都放了8个字节。由于Pcs机使用的是小终端，所以首先将较低的字节放在上面。printf函数解析字符串并到达%25[DatatypeSpecifier]点，并根据DatatypeSpecifier所需的数据类型从堆栈读取多少字节。因此，对于"%25u"，它只读取4个字节，而忽略其他字节。由于您编写的是"%25u"而不是"%25x"，所以它以十进制而不是十六进制形式显示值。|5718620|entityMap^0|0|3|9|D|F|18|4|1Y|4|0|3|A|R|A|0|0|2T|K|4L|4|5E|4|5L|4|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Q|8|@$D|R|E|S|F|G]|$D|T|E|U|F|G]|$D|V|E|W|F|G]|$D|X|E|Y|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|Z|8|@$D|10|E|11|F|G]|$D|12|E|13|F|G]]|9|@]|A|$]]|$1|J|3|K|5|6|7|14|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|15|8|@$D|16|E|17|F|G]|$D|18|E|19|F|G]|$D|1A|E|1B|F|G]|$D|1C|E|1D|F|G]]|9|@]|A|$]]|$1|N|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]]|O|$]]

it seems that you are working on an 64bit machine. so your pointer is 64bit long

both (<code>&amp;firstInt</code> and <code>pointerFirstInt</code>) are exactly same. but are displayed differently.
<code>"%p"</code> knows that pointers are 64bit and displays them in hexadecimal. <code>"%u"</code> shows decimal number and assumes 32bit. so only a part is shown.

if you convert <code>1606416332</code> to hexadecimal it looks like: <code>0x5FBFF7CC</code>. you see that this is the lower half of the 64bit address.

edit:
further explanations:

since printf is a var-arg function all the parametes you give to it were put on the stack. since you put 8 byte on it in both cases. since Pcs using little endian the lower bytes are put on it first.
the printf function parses the string and comes to an <code>%[DatatypeSpecifier]</code> point and reads as many bytes from stack as the datatype that is refered by DatatypeSpecifier requires. so in case of <code>"%u"</code> it only reads 4 bytes and ignores the other bytes. Since you wrote <code>"%u"</code> and not <code>"%x"</code> it displays the value in decimal and not in hexadecimal form.

<pre><code>int firstInt =10;
int *pointerFirstInt = &amp;firstInt;


printf("The address of firstInt is: %u", &amp;firstInt);
printf("\n");
printf("The address of firstInt is: %p", pointerFirstInt);
printf("\n");
</code></pre>

The above code returns the following:

<pre><code>The address of firstInt is: 1606416332
The address of firstInt is: 0x7fff5fbff7cc
</code></pre>

I know that <code>0x7fff5fbff7cc</code> is in hexadecimal, but when i attempt to convert that number to decimal it does not equal <code>1606416332</code>. Why is this? Shouldn't both return the same memory address?

Difference between printing pointer address and ampersand 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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int firstInt =10;int *pointerFirstInt = &firstInt;printf("The address of firstInt is: %u", &firstInt);printf("\n");printf("The address of firstInt is: %p", poin...

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