23/7/2017 决定夯实基础,回顾指针的知识点,该博文转载于CSDN博主百家晓东,小部分修改,外加编译图引证 正文如下: 以下面的情况说明: int a = 5; int* p = &a; 结论: *p 是指针变量p指向的变量a的存储内容,*p = 5; p 是指针变量的存储内容,因为p指向a,所以p的存储内容(存储空间的值)就是系统为变量a分配的存储地址: &p 是对指针变量p进行取地址的操作. 分析: int a = 5; int* p = &a; int* q
当我们讨论指针时,通常假设它是一种可以用 void * 指针来表示的东西,在 x86_64 平台下是 8 个字节大小.例如,下面是来自 维基百科中关于 x86_64 的文章 的摘录: Pushes and pops on the stack are always in 8-byte strides, and pointers are 8 bytes wide. 从 CPU 的角度来看,指针无非就是内存的地址,所有的内存地址在 x86_64 平台下都是由 64 位来表示,所以假设它是 8 个字节是
A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null. Return a deep copy of the list. 这道链表的深度拷贝题的难点就在于如何处理随机指针的问题,由于每一个节点都有一个随机指针,这个指针可以为空,也可以指向链表的任意一个节点,如果我们在每生成一个新节点给其随机指
Follow up for problem "Populating Next Right Pointers in Each Node". What if the given tree could be any binary tree? Would your previous solution still work? Note: You may only use constant extra space. For example,Given the following binary tr
Given a binary tree struct TreeLinkNode { TreeLinkNode *left; TreeLinkNode *right; TreeLinkNode *next; } Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL. Initially, al