blocks|key|2939836|text|因为“字符串”指针的大小是8个字节。下面是一些使用具有适当“大小”的sizeof()的示例。size_of()这个词有时是用来欺骗不习惯使用它的人的。在这种情况下，指针的大小为8字节。下面是一个典型的32位系统的表示形式.|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2939837|sizeof+(char)+++=+1
sizeof+(double)+=+8
sizeof+(float)++=+4
sizeof+(int)++++=+4
sizeof+(long)+++=+4
sizeof+(long+long)++=+8
sizeof+(short)++=+2
sizeof+(void+*)+=+4

sizeof+(clock_t)++++=+4
sizeof+(pid_t)++=+4
sizeof+(size_t)+=+4
sizeof+(ssize_t)++++=+4
sizeof+(time_t)+=+4|code-block|syntax|javascript|2939838|来源|2939839|您忽略了如何确定字符串正在消失(char数组)。它可能被传递给一个函数，您需要将显式长度作为变量传递，或者在某个地方跟踪它。使用sizeof()不会告诉你这个。|2939840|看看我的前一个问题关于这一点，你会看到，甚至我缺乏最初的理解。|2939841|entityMap|0|LINK|mutability|MUTABLE|url|http://www.pccl.demon.co.uk/C/sizeof.html|1|https://stackoverflow.com/questions/2051462/question-about-why-string-is-truncated-on-first-instance-of-0^0|Y|8|1A|9|0|0|0|2|0|0|1S|8|0|4|5|1|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|16|8|@]|D|@$9|17|A|18|1|19]]|E|$]]|$1|M|3|N|5|6|7|1A|8|@$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|O|3|P|5|6|7|1D|8|@]|D|@$9|1E|A|1F|1|1G]]|E|$]]|$1|Q|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]|Y|$5|T|U|V|E|$W|Z]]]]

Because the size of the "string" pointer is 8 bytes. Here are some examples of using <code>sizeof()</code> with their appropriate "size". The term <code>size_of()</code> is sometimes deceiving for people not used to using it. In your case, the size of the pointer is 8 bytes.. below is a representation on a typical 32-bit system.

<pre><code>sizeof (char) = 1
sizeof (double) = 8
sizeof (float) = 4
sizeof (int) = 4
sizeof (long) = 4
sizeof (long long) = 8
sizeof (short) = 2
sizeof (void *) = 4

sizeof (clock_t) = 4
sizeof (pid_t) = 4
sizeof (size_t) = 4
sizeof (ssize_t) = 4
sizeof (time_t) = 4
</code></pre>

<a href="http://www.pccl.demon.co.uk/C/sizeof.html" rel="nofollow noreferrer">Source</a>

You are leaving out how you are determining your string is disappearing (char array). It is probably being passed to a function, which you need to pass the explicit length as a variable or track it somewhere. Using <code>sizeof()</code> won't tell you this.

See my <a href="https://stackoverflow.com/questions/2051462/question-about-why-string-is-truncated-on-first-instance-of-0">previous question</a> about this and you'll see even my lack of initial understanding.

blocks|key|1416353|text|在C89中，sizeof运算符在编译时只找到以字节为单位的变量的大小(在本例中为8字节的void指针)。它的工作方式与您期望的在普通数组上的工作方式一样，因为它们的大小在编译时是已知的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1416354|char+arr[100];+//+sizeof+arr+==+100
char+*p+=+arr;+//+sizeof+p+==+4+(or+8+on+64-bit+architectures)
char+*p+=+malloc(100);+//+sizeof+p+==+4+(or+8).+Still!|code-block|syntax|javascript|1416355|要知道堆分配内存的大小，您需要手动跟踪它，sizeof不会帮助您。|1416356|entityMap^0|6|6|18|4|0|0|L|6|0^^$0|@$1|2|3|4|5|6|7|O|8|@$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|L|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

In C89, <code>sizeof</code> operator only finds the size of a variable in bytes at compile time (in this case a <code>void</code> pointer of 8 bytes). It works the way you'd expect it to work on plain arrays, because their size is known at compile time.

<pre><code>char arr[100]; // sizeof arr == 100
char *p = arr; // sizeof p == 4 (or 8 on 64-bit architectures)
char *p = malloc(100); // sizeof p == 4 (or 8). Still!
</code></pre>

To know the size of heap-allocated memory, you need to keep track of it manually, <code>sizeof</code> won't help you.

blocks|key|2939906|text|sizeof返回给它的指针(void+*)的大小，而不是您分配的内存的大小。如果以后要使用，则必须将内存大小存储在单独的变量中。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2939907|entityMap^0|0|6|E|6|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|$]]

<code>sizeof</code> returns the size of the pointer (<code>void *</code>) that you gave it, not the size of the memory you allocated. You would have to store the size of the memory in a separate variable if you want to use it later.

blocks|key|1237569|text|你不能。正如所指出的，您可以得到void+*错误位置的大小，但这并不能告诉您多少。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1237570|不能让获得被恶意攻击的*mallocated的大小。也就是说，没有函数调用来返回100+(在您的示例中)--除非您编写自己的内存管理例程。|BOLD|1237571|最简单的就是在某个地方记住它..。也许..。|1237572|stuct+{
++void+*data;
++unsigned+int+size
+}+myStructureWhichRemebersItsSize;

myStructureWhichRemebersItsSize+*someData;
someData.data+=+malloc(100);++//+and+check+for+NULL
someData.size+=+100;|code-block|syntax|javascript|1237573|entityMap^0|G|6|0|0|3|B|B|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@$9|V|A|W|B|H]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|I|3|J|5|6|7|Z|8|@]|D|@]|E|$]]|$1|K|3|L|5|M|7|10|8|@]|D|@]|E|$N|O]]|$1|P|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|Q|$]]

You cannot. As pointed out, you can get the size of the <code>void *</code> mallocated, but that does not tell you much.

You cannot get the size of the malloed <code>*mallocated</code>. That is to say, there is no function call to return 100 (in your example) - unless you write your own memory management routines. 

Simplest is just to remember it somewhere ... maybe ....

<pre><code>stuct {
 void *data;
 unsigned int size
 } myStructureWhichRemebersItsSize;

myStructureWhichRemebersItsSize *someData;
someData.data = malloc(100); // and check for NULL
someData.size = 100;
</code></pre>

blocks|key|2939970|text|void*是内存中某个位置的类型，如果您不知道它包含了什么。应该避免这种情况。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2939971|char*是指向内存中包含char的某个位置的值的类型。标识内存中的位置需要8个字节。|2939972|sizeof告诉您一个特定类型需要多少字节。不是用malloc分配了多少内存，而是有多少内存编译器知道类型应该占用多少。将sizeof应用于值通常被认为是糟糕的样式，正如这里有人提到的那样，有时会在C99中调用智能行为。|2939973|char[100]包含100个字符的值的类型。char[100]+a;是堆栈上100+char的字符串。|2939974|char(*)[100]是指向包含100个字符的值的指针的类型。char(*b)[100];使b指向100个字符，可能在堆上。你可能想要的|2939975|char+(*s)[100]+=+malloc(+sizeof(+char[100]+)+);
printf(+"%25u+byte+pointer+to+%25u+bytes+of+size+%25u+elements\n",
++sizeof+s,+sizeof+*s,+sizeof+**s+);|code-block|syntax|javascript|2939976|entityMap^0|0|5|0|0|5|D|4|0|0|6|P|6|1P|6|0|0|9|N|C|17|4|0|0|C|W|E|1B|1|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|19|8|@$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]|$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|L|3|M|5|6|7|1G|8|@$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]]|D|@]|E|$]]|$1|N|3|O|5|P|7|1N|8|@]|D|@]|E|$Q|R]]|$1|S|3|-4|5|6|7|1O|8|@]|D|@]|E|$]]]|T|$]]

<code>void*</code> is the type of a location in memory if you don't know what it contains. It should be avoided.

<code>char*</code> is the type of a value which points to some location in memory which holds a <code>char</code>. Identifying a location in memory takes eight bytes.

<code>sizeof</code> tells you how many bytes a particular type takes. Not how many were allocated with <code>malloc</code> but just how much memory compiler knows the type should take. Applying <code>sizeof</code> to values is often considered bad style, and as someone else mentioned here, sometimes invokes smart behavior in C99.

<code>char[100]</code> the type of a value which holds 100 chars. <code>char[100] a;</code> is a string of 100 <code>char</code>s on the stack.

<code>char(*)[100]</code> is the type of a pointer to a value that holds 100 chars. <code>char(*b)[100];</code> makes <code>b</code> point to 100 chars, possibly on the heap. What you probably want is

<pre><code>char (*s)[100] = malloc( sizeof( char[100] ) );
printf( "%u byte pointer to %u bytes of size %u elements\n",
 sizeof s, sizeof *s, sizeof **s );
</code></pre>

<blockquote>
 Possible Duplicate: 
 <a href="https://stackoverflow.com/questions/1533519/newbie-questions-about-malloc-and-sizeof">newbie questions about malloc and sizeof</a> 
</blockquote>



I am trying to read strings into a program. When I noticed that the strings were sometimes being corrupted, I tried the following code: 

<pre><code> void *mallocated = malloc(100);
 printf("sizeof(mallocated) = %d\n", sizeof(mallocated));
</code></pre>

According to my program, the size of <code>mallocated</code> was <code>8</code>, even though I allocated 100 bytes for it. Because of this, whenever I try to store a string longer than 8 bytes, everything after the 8th byte will sometimes disappear. Why is this happening, and how can I prevent it?

Using sizeof() on malloc'd memory

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。

文章

问答

视频

教程

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云智能顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

聚焦“写作效率、视觉美观与运行性能”三方面进行全面升级，为您提供更高效、稳定的创作环境

社区富文本&Markdown编辑器全新改版上线，欢迎大家体验!

诚挚邀请您参与本次调研，分享您的真实使用感受与建议。您的反馈至关重要，感谢您的支持与参与！

社区新版编辑器体验调研

可能重复：我正在尝试将字符串读入程序中。当我注意到字符串有时被破坏时，我尝试了以下代码： void *mallocated = malloc(100); printf("sizeof(mallocated) = %d\n", sizeof(mallocated));根据我的程序，mallocated的大小是8，尽管我为...

问在malloc内存上使用sizeof()
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问在malloc内存上使用sizeof()EN