我对指针知之甚少。
我遇到了以下程序。 输出似乎正常，但实际上发生的事情我无法理解。
#include<stdio.h> #include<conio.h> void main() { int k; int a[] = {1,2,3}; int *b[3] ; int **c[3]; int ***d[3]; int ****e[3]; int*****f[3]; for (k = 0 ; k <3; k++) { b[k] = a + k; c[k] = b + k ; d[k] = c + k; e[k] = d + k ; f[k] = e + k; } for (k = 0 ; k <3; k++) { printf("%3d", *b[k]); printf("%3d", **c[k]); printf("%3d", ***d[k]); printf("%3d", ****e[k]); printf("%3d\n", *****f[k]); } }I know little about pointers.
I came across the following program. The output seems normal but what is actually going going on I could not figure it out.
#include<stdio.h> #include<conio.h> void main() { int k; int a[] = {1,2,3}; int *b[3] ; int **c[3]; int ***d[3]; int ****e[3]; int*****f[3]; for (k = 0 ; k <3; k++) { b[k] = a + k; c[k] = b + k ; d[k] = c + k; e[k] = d + k ; f[k] = e + k; } for (k = 0 ; k <3; k++) { printf("%3d", *b[k]); printf("%3d", **c[k]); printf("%3d", ***d[k]); printf("%3d", ****e[k]); printf("%3d\n", *****f[k]); } }最满意答案
第一个for循环只是基本的指针算法。 a[]保存int ，后面的每个数组都有一个指针。 b[]是指向int的指针 c[]是指向int指针 等等
所以它在内存中是这样的：
Memory Address: 0x00441234 <---+ 0x00441238 <----+ 0x0044123C <---+ ********** | ********** | ********** | var name: * a (+0) * | * a (+1) * | * a (+2) * | ********** | ********** | ********** | value: * 1 * | * 2 * | * 3 * | ********** | ********** | ********** | | | | | | | +-> 0x00442345 | +->0x00442349 | +->0x0044234D | | ************ | | ************ | | ************ | | * b (+0) * | | * b (+1) * | | * b (+2) * | | ************ | | ************ | | ************ | | *0x00441234* -+ | *0x00441238* --+ | *0x0044123C* --+ | ************ | ************ | ************ | | | | | | | 0x00443345 | 0x00443349 | 0x0044334D | ************ | ************ | ************ | * c (+0) * | * c (+1) * | * c (+2) * | ************ | ************ | ************ +-- *0x00442345* +-*0x00442349* +-*0x0044234D* ************ ************ ************并且D每个元素指向C每个元素，依此类推。 最终的结果是你将每个数组中的每个元素（通过一些指针链）设置回a的元素。 然后在第二个for循环中，你一遍又一遍地打印a[]的元素。
The first for loop is just basic pointer arithmetic. a[] holds ints, each array after that holds a pointer. b[] is a pointer to int c[] is a pointer to pointer to int etc
So it's something like this in memory:
Memory Address: 0x00441234 <---+ 0x00441238 <----+ 0x0044123C <---+ ********** | ********** | ********** | var name: * a (+0) * | * a (+1) * | * a (+2) * | ********** | ********** | ********** | value: * 1 * | * 2 * | * 3 * | ********** | ********** | ********** | | | | | | | +-> 0x00442345 | +->0x00442349 | +->0x0044234D | | ************ | | ************ | | ************ | | * b (+0) * | | * b (+1) * | | * b (+2) * | | ************ | | ************ | | ************ | | *0x00441234* -+ | *0x00441238* --+ | *0x0044123C* --+ | ************ | ************ | ************ | | | | | | | 0x00443345 | 0x00443349 | 0x0044334D | ************ | ************ | ************ | * c (+0) * | * c (+1) * | * c (+2) * | ************ | ************ | ************ +-- *0x00442345* +-*0x00442349* +-*0x0044234D* ************ ************ ************And each element of D points to each element of C, and so on. The end result being you're setting each element in each of the arrays (via some chain of pointers) back to the elements of a. And then in the second for loop you're printing the elements of a[] over and over again.
更多推荐
发布评论