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CSAPP: 位操作实现基本运算

本小妞迷上赌 提交于 2019-12-02 09:09:23
目录 实验要求 实现代码 1、pow2plus1 2、pow2plus4 3、bitXor 4、tmin 5、isTmax 6、allOddBits 7、negate 8、isAsciiDigit 9、conditional 10、isLessOrEqual 11、logicalNeg 12、howManyBits 13、floatScale2 14、floatFloat2Int 15、floatPower2 @(位操作实现简单函数) 实验要求 给出15个函数,规定了实现每个函数需要的逻辑和算术操作符(规定数量)。 只能使用规定的操作符! ˜ & ˆ | + << >> 不能使用循环或者条件语句 不能使用超过8位的常数(ff) 实现代码 1、pow2plus1 /* * pow2plus1 - returns 2^x + 1, where 0 <= x <= 31 */ int pow2plus1(int x) { /* exploit ability of shifts to compute powers of 2 */ return (1 << x) + 1; } 2、pow2plus4 /* * pow2plus4 - returns 2^x + 4, where 0 <= x <= 31 */ int pow2plus4(int x) { /* exploit

根据位操作将八位十六进制转化为浮点数

回眸只為那壹抹淺笑 提交于 2019-12-02 06:44:52
1 #include<stdio.h> 2 #include<stdlib.h> 3 //*********************************************// 4 //****name: 八位十六进制数转浮点数输出 ***// 5 //****length: 32位 ***// 6 //****formula: x[真值] = ((-1)^s)*1.M*(2^E) ***// 7 //****time: 2019-10-24 ***// 8 //**********************************************// 9 10 11 #define NUMBER_SIGN 1 //数符 12 #define FRAME_SHIFT 8 //移码 13 #define CODING 32 //编码 14 #define DIDITAL_CODING 23 //数码 15 16 float HextoFloat(unsigned int hex) 17 { 18 int i = 1;//提供2的n次方 19 unsigned int ihex = (hex << NUMBER_SIGN); 20 unsigned int decimal_bin = 0; 21 unsigned int dacade = 0; 22 double

Why does << 32 not result in 0 in javascript?

情到浓时终转凉″ 提交于 2019-12-02 06:32:48
问题 This is false: (0xffffffff << 31 << 1) === (0xffffffff << 32) It seems like it should be true. Adding >>> 0 anywhere does not change this. Why is this and how can I correctly write code that handles << 32 ? 回答1: The shift operators always effectively has a right operand in the range 0-31. From the Mozilla docs: Shift operators convert their operands to 32-bit integers in big-endian order and return a result of the same type as the left operand. The right operand should be less than 32, but if

How to read specific bits of an unsigned int

ε祈祈猫儿з 提交于 2019-12-02 05:57:03
问题 I have an uint8_t and I need to read/write to specific bits. How would I go about doing this. Specifically what I mean is that I need to write and then later read the first 7 bits for one value and the last bit for another value. edit: forgot to specify, I will be setting these as big endian 回答1: You're looking for bitmasking . Learning how to use C's bitwise operators: ~ , | , & , ^ and so on will be of huge help, and I recommend you look them up. Otherwise -- want to read off the least

学习mysql_day2

孤街醉人 提交于 2019-12-02 04:58:39
创建数据库 create database mysql_demo1 charset=utf8; MariaDB [ ( none ) ] > create database mysql_demo1 charset = utf8 ; Query OK , 1 row affected ( 0.00 sec ) 使用数据库 use mysql_demo1; MariaDB [ ( none ) ] > use mysql_demo1 ; Database changed 创建表参数 unsigned  整型无符号 signed  整型有符号 primary key  详解参考:http: //c.biancheng.net/view/2440.html foreign key 详解参考:http: //c.biancheng.net/view/2441.html unique key 详解参考:http: //c.biancheng.net/view/2445.html auto_increment 自增字段 not null  不为空 default  默认值 int   int 类型 char  存储字符串 varchar   存储字符串( char 是属于固定长度的字符类型,而 varchar 是属于可变长度的字符类型。) tinyint 型的字段如果不设置 UNSIGNED

Android原生(Native)C开发之二 framebuffer篇

蓝咒 提交于 2019-12-02 03:34:01
Android原生(Native)C开发之二 framebuffer篇 如对Android原生(Natvie)C开发还任何疑问,请参阅http://emck.avaw.com/?p=205 虽然现在能通过交叉环境编译程序,并push到Android上执行,但那只是console台程序,是不是有些单调呢?下面就要看如何通过Linux的 framebuffer 技术在Android上画图形,关于Linux的framebuffer技术,这里就不再详细讲解了,请大家google一下。 操作framebuffer的主要步骤如下: 1、打开一个可用的FrameBuffer设备; 2、通过mmap调用把显卡的物理内存空间映射到用户空间; 3、更改内存空间里的像素数据并显示; 4、退出时关闭framebuffer设备。 下面的这个例子简单地用framebuffer画了一个渐变的进度条,代码 framebuf.c 如下: #include <unistd.h> #include <stdio.h> #include <fcntl.h> #include <linux/fb.h> #include <sys/mman.h> inline static unsigned short int make16color(unsigned char r, unsigned char g, unsigned

converting an array of signed bytes to unsigned bytes [duplicate]

烈酒焚心 提交于 2019-12-02 03:22:22
问题 This question already has answers here : Can we make unsigned byte in Java (15 answers) Closed 2 years ago . I have a array of bytes. bytes[] = [43, 0, 0, -13, 114, -75, -2, 2, 20, 0, 0] I want to convert it to unsigned bytes in Java. this is what I did: created a new array and copy the values with & 0xFF: this.bytes = new byte[bytes.length]; for (int i=0;i<bytes.length;i++) this.bytes[i] = (byte) (bytes[i] & 0xFF); but the values stay negative in the new array as well. what am I doing wrong?

How to read specific bits of an unsigned int

ε祈祈猫儿з 提交于 2019-12-02 03:18:50
I have an uint8_t and I need to read/write to specific bits. How would I go about doing this. Specifically what I mean is that I need to write and then later read the first 7 bits for one value and the last bit for another value. edit: forgot to specify, I will be setting these as big endian You're looking for bitmasking . Learning how to use C's bitwise operators: ~ , | , & , ^ and so on will be of huge help, and I recommend you look them up. Otherwise -- want to read off the least significant bit? uint8_t i = 0x03; uint8_t j = i & 1; // j is now 1, since i is odd (LSB set) and set it? uint8

域名解析为ip地址

旧城冷巷雨未停 提交于 2019-12-02 02:57:13
int GetIpByDomainName(const char *DName,unsigned char* ipaddr) { struct hostent *phost; if(NULL == DName) { //DBG_ERR("DOMAIN NAME IS NULL\r\n"); return APP_FAIL; } if((phost = gethostbyname(DName)) == NULL) { // DBG_ERR("DOMAIN NAME CHANGE IP FAILE\r\n"); return APP_FAIL; } sprintf((char*)ipaddr,"%u.%u.%u.%u", ((unsigned char)phost->h_addr_list[0][0]), ((unsigned char)phost->h_addr_list[0][1]), ((unsigned char)phost->h_addr_list[0][2]), ((unsigned char)phost->h_addr_list[0][3]) ); return APP_OK; } 来源: CSDN 作者: wangjidong198612 链接: https://blog.csdn.net/wangjidong198612/article/details