blocks|key|1343315|text|int*+arr[8];+//+An+array+of+int+pointers.
int+(*arr)[8];+//+A+pointer+to+an+array+of+integers|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1343316|第三个与第一个相同。|unstyled|1343317|一般规则是operator+precedence。随着函数指针的出现，它可能会变得更加复杂。|offset|length|1343318|entityMap|0|LINK|mutability|MUTABLE|url|http://unixwiz.net/techtips/reading-cdecl.html^0|0|0|5|J|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|T|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|F|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

<pre><code>int* arr[8]; // An array of int pointers.
int (*arr)[8]; // A pointer to an array of integers
</code></pre>

The third one is same as the first.

The general rule is <a href="http://unixwiz.net/techtips/reading-cdecl.html" rel="noreferrer">operator precedence</a>. It can get even much more complex as function pointers come into the picture.

blocks|key|1343365|text|按照K&R的建议，使用cdecl程序。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1343366|$+cdecl
Type+`help'+or+`?'+for+help
cdecl>+explain+int*+arr1[8];
declare+arr1+as+array+8+of+pointer+to+int
cdecl>+explain+int+(*arr2)[8]
declare+arr2+as+pointer+to+array+8+of+int
cdecl>+explain+int+*(arr3[8])
declare+arr3+as+array+8+of+pointer+to+int
cdecl>|code-block|syntax|javascript|1343367|它也以另一种方式工作。|1343368|cdecl>+declare+x+as+pointer+to+function(void)+returning+pointer+to+float
float+*(*x)(void+)|1343369|entityMap|0|LINK|mutability|MUTABLE|url|http://cdecl.org/^0|B|5|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

Use the <a href="http://cdecl.org/" rel="noreferrer">cdecl</a> program, as suggested by K&amp;R.



<pre class="lang-none prettyprint-override"><code>$ cdecl
Type `help' or `?' for help
cdecl&gt; explain int* arr1[8];
declare arr1 as array 8 of pointer to int
cdecl&gt; explain int (*arr2)[8]
declare arr2 as pointer to array 8 of int
cdecl&gt; explain int *(arr3[8])
declare arr3 as array 8 of pointer to int
cdecl&gt;
</code></pre>

It works the other way too.

<pre class="lang-none prettyprint-override"><code>cdecl&gt; declare x as pointer to function(void) returning pointer to float
float *(*x)(void )
</code></pre>

blocks|key|1558893|text|int+*a[4];+//+Array+of+4+pointers+to+int

int+(*a)[4];+//a+is+a+pointer+to+an+integer+array+of+size+4

int+(*a[8])[5];+//a+is+an+array+of+pointers+to+integer+array+of+size+5+|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1558894|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>int *a[4]; // Array of 4 pointers to int

int (*a)[4]; //a is a pointer to an integer array of size 4

int (*a[8])[5]; //a is an array of pointers to integer array of size 5 
</code></pre>

blocks|key|1343373|text|typedef+int+(*PointerToIntArray)[];
typedef+int+*ArrayOfIntPointers[];|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1343374|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>typedef int (*PointerToIntArray)[];
typedef int *ArrayOfIntPointers[];
</code></pre>

blocks|key|1343585|text|根据经验，右一元运算符(如[]、()等)优先于左运算符。因此，int+*(*ptr)()[];应该是一个指向函数的指针，该函数返回一个指向整数的指针数组(尽快获得正确的运算符)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1343586|entityMap^0|D|2|G|2|V|G|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|O|8|@]|D|@]|E|$]]]|G|$]]

As a rule of thumb, right unary operators (like <code>[]</code>, <code>()</code>, etc) take preference over left ones. So, <code>int *(*ptr)()[];</code> would be a pointer that points to a function that returns an array of pointers to int (get the right operators as soon as you can as you get out of the parenthesis)

blocks|key|1343480|text|我想我们可以用一个简单的规则..|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1343481|example+int+*+(*ptr)()[];
start+from+ptr+|code-block|syntax|javascript|1343482|“ptr是指向”go+go+..its“”的指针)“now+go+left+its+a”"(“out+out+go+right+"()”so“指向不带任何参数的函数”go+left“并返回指向整数数组”go+left“的指针”go+right“”|offset|length|style|CODE|1343483|entityMap^0|0|0|1|3|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]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I think we can use the simple rule ..

<pre><code>example int * (*ptr)()[];
start from ptr 
</code></pre>

" <code>ptr</code> is a pointer to " 
go towards right ..its ")" now go left its a "("
come out go right "()" so
" to a function which takes no arguments " go left "and returns a pointer " go right "to 
an array" go left " of integers "

blocks|key|1343666|text|这里有一个有趣的网站，它解释了如何在C：http://www.unixwiz.net/techtips/reading-cdecl.html中读取复杂类型|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1343667|entityMap|0|LINK|mutability|MUTABLE|url|http://www.unixwiz.net/techtips/reading-cdecl.html^0|K|1E|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

Here's an interesting website that explains how to read complex types in C:
<a href="http://www.unixwiz.net/techtips/reading-cdecl.html" rel="nofollow noreferrer">http://www.unixwiz.net/techtips/reading-cdecl.html</a>

blocks|key|1559162|text|我猜第二个声明会让很多人感到困惑。这里有一个理解它的简单方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1559163|让我们有一个整数数组，即int+B[8]。|offset|length|style|CODE|1559164|我们还有一个变量A，它指向B。现在，A处的值是B，即(*A)+==+B。因此，A指向整数数组。在您的问题中，arr类似于A。|1559165|类似地，在int*+(*C)+[8]中，C是指向整数指针数组的指针。|1559166|entityMap^0|0|C|8|0|Q|9|0|5|D|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@$D|P|E|Q|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|R|8|@$D|S|E|T|F|G]]|9|@]|A|$]]|$1|J|3|K|5|6|7|U|8|@$D|V|E|W|F|G]]|9|@]|A|$]]|$1|L|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|M|$]]

I guess the second declaration is confusing to many. Here's an easy way to understand it.

Lets have an array of integers, i.e. <code>int B[8]</code>.

Let's also have a variable A which points to B. Now, value at A is B, i.e. <code>(*A) == B</code>. Hence A points to an array of integers. In your question, arr is similar to A.

Similarly, in <code>int* (*C) [8]</code>, C is a pointer to an array of pointers to integer.

blocks|key|1343736|text|int+*arr1[5]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1343737|在此声明中，arr1是一个由5个指向整数的指针组成的数组。原因:方括号的优先级高于*(取消引用运算符)。在这种类型中，行数是固定的(这里是5)，但列数是可变的。|unstyled|offset|length|style|CODE|1343738|int+(*arr2)[5]|1343739|在此声明中，arr2是指向包含5个元素的整数数组的指针。原因:在这里，()括号的优先级高于[]。在这种类型中，行数是可变的，但列数是固定的(这里是5)。|1343740|entityMap^0|0|6|4|0|0|6|4|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|R|8|@$G|S|H|T|I|J]]|9|@]|A|$]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$B|C]]|$1|M|3|N|5|F|7|V|8|@$G|W|H|X|I|J]]|9|@]|A|$]]|$1|O|3|-4|5|F|7|Y|8|@]|9|@]|A|$]]]|P|$]]

<pre><code>int *arr1[5]
</code></pre>

In this declaration, <code>arr1</code> is an array of 5 pointers to integers.
Reason: Square brackets have higher precedence over * (dereferncing operator).
And in this type, number of rows are fixed (5 here), but number of columns is variable.

<pre><code>int (*arr2)[5]
</code></pre>

In this declaration, <code>arr2</code> is a pointer to an integer array of 5 elements.
Reason: Here, () brackets have higher precedence than []. 
And in this type, number of rows is variable, but the number of columns is fixed (5 here).

blocks|key|1559035|text|在指向一个整数的指针中，如果指针是递增的，那么它将进入下一个整数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1559036|在指针数组中，如果指针递增，则跳转到下一个数组|1559037|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|$]]

In pointer to an integer if pointer is incremented then it goes next integer.

in array of pointer if pointer is incremented it jumps to next array

What is the difference between the following declarations:

<pre><code>int* arr1[8];
int (*arr2)[8];
int *(arr3[8]);
</code></pre>

What is the general rule for understanding more complex declarations?

C pointer to array/array of pointers disambiguation

下面的声明有什么区别：int* arr1[8];int (*arr2)[8];int *(arr3[8]);理解更复杂的声明的一般规则是什么？

问C指针指向数组/指针数组消除歧义
EN

回答 10

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问C指针指向数组/指针数组消除歧义EN

回答 10

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问C指针指向数组/指针数组消除歧义
EN