blocks|key|555190|text|这是因为arr现在是一个指针，它可以指向单个int或函数不知道的1000个int数组。您必须将数组的大小传递给函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|555191|在main中，arr被声明为int5，因此其大小可以由编译器计算。|555192|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

This is because arr is now a pointer and it could point to a single int or an array of 1000 ints the function just does not know. You will have to pass the size of the array into the function.

In main arr is declared as an int[5] and so the size can be calculated by the compiler.

blocks|key|771758|text|这是因为已知大小的数组和指向未知大小的数组的指针之间的差异。sizeof是在编译时完成的，编译器不可能提前知道动态创建的内存区域的大小。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|771759|entityMap^0|U|6|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

It's because of the difference between an array of known size and a pointer to an array of unknown size. <code>sizeof</code> is done at compile time and it's not possible for the compiler to tell the size of a dynamically created memory region in advance.

blocks|key|555317|text|C数组不会将自己的大小存储在任何地方，所以只有在编译时知道大小的情况下，sizeof才能按照预期的方式工作。编译器将malloc()视为任何其他函数，因此sizeof无法判断arr指向数组的第一个元素，更不用说它有多大了。如果需要知道数组的大小，则需要显式地将其作为单独的参数传递给函数，或者使用包含指向数组及其大小的指针的结构。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|555318|entityMap^0|10|6|1M|8|25|6|2F|3|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|G|$]]

C arrays don't store their own sizes anywhere, so <code>sizeof</code> only works the way you expect if the size is known at compile time. <code>malloc()</code> is treated by the compiler as any other function, so <code>sizeof</code> can't tell that <code>arr</code> points to the first element of an array, let alone how big it is. If you need to know the size of the array, you need to explicitly pass it to your function, either as a separate argument, or by using a struct containing a pointer to your array and its size.

blocks|key|555402|text|我不理解第一个问题(请发布一些代码)，但对于第二个问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|555403|sizeof(*arr)+;+++//+is+sizeof(int)+as+arr+is+of+type+"int+*"
sizeof(arr[0])+;+//+is+sizeof(int),+as+arr[0]+is+the+first
+++++++++++++++++//+++++item+of+an+array+of+int|code-block|syntax|javascript|555404|事实是，*arr和arr[0]的意思完全相同，但表达方式不同。实际上，假设n是一个索引：*(arr+%2B+n)与arr[n]相同。|offset|length|style|CODE|555405|entityMap^0|0|0|4|4|9|6|18|A|1J|6|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|N|$]]

I didn't understand the first question (post some code, please), but for the second:

<pre><code>sizeof(*arr) ; // is sizeof(int) as arr is of type "int *"
sizeof(arr[0]) ; // is sizeof(int), as arr[0] is the first
 // item of an array of int
</code></pre>

Fact is, <code>*arr</code> and <code>arr[0]</code> mean exactly the same thing, but said differently. In fact, let say n is an index: <code>*(arr + n)</code> is the same than <code>arr[n]</code>.

blocks|key|771998|text|您应该了解静态数组和动态数组之间的区别。静态数组是像int、float、double等类型，它们是不同的。甚至|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|771999|int+a[10];|code-block|syntax|javascript|772000|具有不同于的类型|772001|int+b[11];|772002|在编译时，静态数组中元素的数量是已知的，sizeof运算符返回它们占用的字节数。|772003|对于已初始化的指针，无论是指向某个变量，还是指向已分配的内存，在运行时它们所指向的位置都是显而易见的，并且编译器无法确定将来该数组的大小。因此，sizeof的指针是4(例如64位系统中的8)。|772004|请注意，sizeof运算符在编译时工作，而不是在运行时。|772005|如果您需要知道已分配内存的大小(通过malloc)，则别无选择，只能将数组的大小保存在另一个变量中，例如，在执行malloc之后。|772006|entityMap^0|Q|3|U|5|10|6|0|0|0|0|K|6|0|20|6|0|4|6|0|I|6|1K|6|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$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]]|$1|K|3|L|5|6|7|16|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|18|8|@$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1B|8|@$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|S|3|T|5|6|7|1E|8|@$9|1F|A|1G|B|C]]|D|@]|E|$]]|$1|U|3|V|5|6|7|1H|8|@$9|1I|A|1J|B|C]|$9|1K|A|1L|B|C]]|D|@]|E|$]]|$1|W|3|-4|5|6|7|1M|8|@]|D|@]|E|$]]]|X|$]]

You should understand the difference between static and dynamic arrays. Static arrays are types like <code>int</code>, <code>float</code>, <code>double</code> etc. They are different. Even

<pre><code>int a[10];
</code></pre>

has a different type from

<pre><code>int b[11];
</code></pre>

At compile time, the number of elements in static arrays are known and the <code>sizeof</code> operator returns the number of bytes they occupy.

With pointers that were initialized, either to point to some variable, or to an allocated memory, it is at run-time that it would be apparent where they are pointing to and the compiler cannot determine what would be the size of that array in the future. Therefore, the <code>sizeof</code> a pointer gives you 4 (or 8 in 64bits systems for example).

Note that <code>sizeof</code> operator works at compile-time and not at run-time.

If you need to know the size of an allocated memory (through <code>malloc</code>), you have no choice but to save the size of the array in another variable, for example right after you did the <code>malloc</code>.

a simple question that bugs me.
Say I have an array defined in main like so <code>int arr[5]</code>. Now, if I'm still inside main and I set <code>int i = sizeof(arr)/sizeof(arr[0])</code> then I is set to be 5, but if I pass the array as a function parameter and do the exact same calculation in this function, I get a different number. Why is that? At first I thought its because in a function <code>arr</code> is a pointer, but as far as I know <code>arr</code> is a pointer inside main too!

Also, if I do something very similar only I initialize the array dynamically, I get weird results:

<pre><code>int *arr = (int*) malloc(sizeof(int) * 5);
int length = sizeof(*arr) / sizeof(arr[0]);
printf("%d\n",length);
</code></pre>

Here the output is <code>1</code>. Any ideas why?
Thanks in advance!

size of array in c

编译器

一个困扰我的简单问题。假设我在main中定义了一个数组，比如int arr[5]。现在，如果我仍然在main中，并且我设置了int i = sizeof(arr)/sizeof(arr[0])，那么我被设置为5，但是如果我将数组作为函数参数传递，并在这个函数中进行完全相同的计算，我会得到一个不同的数字。为什么会这样呢？...

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