当前位置:网站首页>指針經典筆試題
指針經典筆試題
2022-07-06 05:13:00 【雲朵c】
筆試題一
int main()
{
int a[5] = {
1,2,3,4,5 };
int* ptr = (int*)(&a + 1);
printf("%d\n", *(a + 1));
printf("%d\n", *(ptr - 1));
return 0;
}
//程序的結果是什麼?
筆試題二
//結構體的大小是20個字節
struct Test
{
int Num;
char *pcName;
short sDate;
char cha[2];
short sBa[4];
}*p;
//假設p的值為0x100000。 如下錶錶達式的值分別為多少?
int main()
{
printf("%p\n", p + 0x1);
printf("%p\n", (unsigned long)p + 0x1);
printf("%p\n", (unsigned int*)p + 0x1);
return 0;
}
筆試題三
int main()
{
int a[4] = {
1, 2, 3, 4 };
int *ptr1 = (int *)(&a + 1);
int *ptr2 = (int *)((int)a + 1);
printf("%x\n", ptr1[-1]);
printf("%x\n", *ptr2);
return 0;
}
筆試題四
#include <stdio.h>
int main()
{
int a[3][2] = {
(0, 1), (2, 3), (4, 5) };
int *p;
p = a[0];
printf( "%d", p[0]);
return 0;
}
筆試題五
int main()
{
int a[5][5];
int(*p)[4];
p = a;
printf("%p\n", &p[4][2] - &a[4][2]);
printf("%d\n", &p[4][2] - &a[4][2]);
return 0;
}
筆試題六
int main()
{
int aa[2][5] = {
1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
int* ptr1 = (int*)(&aa + 1);
int* ptr2 = (int*)(*(aa + 1));
printf("%d\n", *(ptr1 - 1));
printf("%d\n", *(ptr2 - 1));
return 0;
}
筆試題七
#include <stdio.h>
int main()
{
char *a[] = {
"work","at","alibaba"};
char**pa = a;
pa++;
printf("%s\n", *pa);
return 0;
}
筆試題八
int main()
{
char *c[] = {
"ENTER","NEW","POINT","FIRST"};
char**cp[] = {
c+3,c+2,c+1,c};
char***cpp = cp;
printf("%s\n", **++cpp);
printf("%s\n", *--*++cpp+3);
printf("%s\n", *cpp[-2]+3);
printf("%s\n", cpp[-1][-1]+1);
return 0;
}
边栏推荐
- Extension of graph theory
- 内核判断i2c地址上是否挂载外设
- Idea one key guide package
- GAMES202-WebGL中shader的编译和连接(了解向)
- Some common skills on unity inspector are generally used for editor extension or others
- 【OSPF 和 ISIS 在多路访问网络中对掩码的要求】
- Summary of three log knowledge points of MySQL
- Postman pre script - global variables and environment variables
- What are the advantages of the industry private network over the public network? What specific requirements can be met?
- Quelques conseils communs sur l'inspecteur de l'unit é, généralement pour les extensions d'éditeur ou d'autres
猜你喜欢
Codeforces Round #804 (Div. 2)
GAMES202-WebGL中shader的编译和连接(了解向)
Microblogging hot search stock selection strategy
趋势前沿 | 达摩院语音 AI 最新技术大全
Hyperledger Fabric2. Some basic concepts of X (1)
CUDA11.1在线安装
Building intelligent gray-scale data system from 0 to 1: Taking vivo game center as an example
Postman assertion
Rce code and Command Execution Vulnerability
Vite configures the development environment and production environment
随机推荐
Three.js学习-光照和阴影(了解向)
Chip debugging of es8316 of imx8mp
Drive development - the first helloddk
Weng Kai C language third week 3.1 punch in
RT thread analysis log system RT_ Kprintf analysis
麦斯克电子IPO被终止:曾拟募资8亿 河南资产是股东
图论的扩展
用StopWatch 统计代码耗时
Realize a binary read-write address book
GAMES202-WebGL中shader的編譯和連接(了解向)
集合详解之 Collection + 面试题
[classic example] binary tree recursive structure classic topic collection @ binary tree
Talking about the type and function of lens filter
Hyperledger Fabric2. Some basic concepts of X (1)
[noip2008 improvement group] stupid monkey
Codeforces Round #804 (Div. 2)
Finance online homework
MPLS experiment
[NOIP2008 提高组] 笨小猴
nacos-高可用seata之TC搭建(02)