是什么（* P）[8]和* P [8] C之间的区别？ [英] What is the difference between (*p)[8] and *p[8] in C?

问题描述

这两个声明如下：

  INT（* P）[8];
为int * P [8];
 

解决方案

首先是一个的单的指针指向8整数数组，而第二个是一个的阵列的8指针，各自为整数。

如果你只踢了 CDECL ，这是美妙的这样的事情：

  PAX $ CDECL
键入'帮助'或'？求助CDECL＆GT;解释INT（* P）; [8]
申报数p作为指针为int数组8CDECL＆GT;解释为int * P [8];
申报数p作为指针数组8为intCDECL＆GT;解释的char *（* FP []）（整型，浮点*）;
声明FP为指针数组功能（INT，指针浮）
    返回字符指针
 

实际上，有一个顺时针/螺旋规则你可以用它来做到这一点在你的脑袋，但我没有担心，自从我发现 CDECL ，出于同样的原因我不再转换任意大的32位数字从十进制在我的脑海诅咒任何更多的 - 我的可以，如果我有的到，但它是如此一个工具更容易： - ）

The two declarations are as follows:

int (*p)[8];
int *p[8];

解决方案

The first is a single pointer to an array of 8 integers, while the second is an array of 8 pointers, each to an integer.

If you just kick up cdecl, it's wonderful for this sort of thing:

pax$ cdecl
Type `help' or `?' for help

cdecl> explain int (*p)[8];
declare p as pointer to array 8 of int

cdecl> explain int *p[8];
declare p as array 8 of pointer to int

cdecl> explain char*(*fp[])(int,float*);
declare fp as array of pointer to function (int, pointer to float)
    returning pointer to char

There's actually a clockwise/spiral rule you can use to do this in your head but I haven't had to worry about that since I discovered cdecl, for the same reason I no longer convert large arbitrary 32 bit numbers from decimal to hex in my head any more - I can if I have to but it's so much easier with a tool :-)

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