是什么（* P）[8]和* P [8] C之间的区别？

本文介绍了是什么（* P）[8]和* P [8] C之间的区别？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 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 :-)