bit

Suppose I have 0.625 as a floating point is 0b.101 , so if I want the first two bits of that as an integer i.e. 0b10 = 2 , how can I achieve this in python? I've tried taking the number to a power of 2 and casting to an int, so if I want n bits I do int(0.625*(2**n)) . But that is not working for me. The problem occurs when I have a number greater than 1 so 24.548838022726972 will give me 392 rather than 12 for the first four bits. ( 24.548838022726972 = 0b11000.100011001... ) If you want the n most significant bits, one way to start is to use math.frexp to normalise your number to lie in the

总结下linux kernel switch driver。 这里的switch driver是为监听CPU gpio口状态变化的，switch可以理解为gpio口状态变化。 switch driver是因android引入的。 下面以耳机接入为例说下androd系统中耳机接入检测的过程 kernel/arch/arm64/boot/dts/xxx.dts switch_gpio { compatible = "mstar,switch-gpio"; switch-name = "h2w"; switch-gpio = <66>; switch-inverse = <0>; }; 上面的switch-gpio是想要监听的CPU gpio index，用来告诉switch driver要监听的gpio口 解析上述dts switch_gpio section是在kernel switch driver中： drivers/switch/switch_gpio.c static int gpio_switch_probe(struct platform_device *pdev) { struct gpio_switch_platform_data *pdata = pdev->dev.platform_data; struct gpio_switch_data *switch

To avoid relying on the wireless tools I want to get the essid directly from the device with ioctl, in C this wouldn't be a problem, but in Ruby it's quite different. The problem is following struct from wireless.h that is used as input/reply of ioctl: struct iw_point { void __user *pointer; /* Pointer to the data (in user space) */ __u16 length; /* number of fields or size in bytes */ __u16 flags; /* Optional params */ }; The pointer part must be a valid address of a memory area, followed by the length in bytes, followed by a flag field. I tried with Array#pack and the bit-struct gem, but

import random print('*'*30+'欢迎进入大富翁游戏'+'*'*30) username = input('请输入用户名：') print('请充值购买金币后方可游戏，充值金额必须是100的整数倍') bit_conunt = 0 while True: my_money= int(input('请输入充值金额：')) if my_money > 0 and my_money % 100 == 0: bit_conunt = (my_money//100)*30 print('充值成功，您的充值金额是%d,账户金币为%d个。'%(my_money,bit_conunt),'请开始你的游戏！') break else: print('充值金额不符合规定，请输入正确金额进行充值。') while 1==1: big_small1 = random.randint(1,6) big_small2 = random.randint(1,6) guess_big_small = int(input('请输入你猜的数值（2-12之间）：')) if big_small1 + big_small2 >= guess_big_small : bit_conunt = bit_conunt -2 print('你输了，您的剩余金币为：{}'.format(bit_conunt)

EnumSet EnumSet是与枚举类型一起使用的专用 Set 集合，EnumSet 中所有元素都必须是枚举类型。与其他Set接口的实现类HashSet/TreeSet(内部都是用对应的HashMap/TreeMap实现的)不同的是，EnumSet在内部实现是位向量(稍后分析)，它是一种极为高效的位运算操作，由于直接存储和操作都是bit，因此EnumSet空间和时间性能都十分可观，足以媲美传统上基于 int 的“位标志”的运算，重要的是我们可像操作set集合一般来操作位运算，这样使用代码更简单易懂同时又具备类型安全的优势。注意EnumSet不允许使用 null 元素。试图插入 null 元素将抛出 NullPointerException，但试图测试判断是否存在null 元素或移除 null 元素则不会抛出异常，与大多数collection 实现一样，EnumSet不是线程安全的，因此在多线程环境下应该注意数据同步问题，ok~，下面先来简单看看EnumSet的使用方式。 EnumSet用法 创建EnumSet并不能使用new关键字，因为它是个抽象类，而应该使用其提供的静态工厂方法，EnumSet的静态工厂方法比较多，如下： 创建一个具有指定元素类型的空EnumSet。 EnumSet<E> noneOf(Class<E> elementType) /

社团作业=_= 开发版上的LED灯负极连接在PB5口，正极串联一510Ω电阻后与3.3V相连 若开发板不带LED灯则需要自行连接，务必串联一个合适的电阻防止LED灯烧坏 零、一个有趣的延时函数 来自于开发板配套资料当中的例程，第一次看到的时候觉得耳目一新，代码如下： void Delay(u32 count) { u32 i = 0; for (; i < count; i++) ; } 当中的u32类型是在stm32f10x.h当中的一个宏定义，对应uint32_t，表示32位无符号型整数，在我的开发板当中就是unsigned int类型。 因为STM32的主频比电脑CPU慢得多，因此可以通过这种循环的方式来达到延时的效果 一、库函数版本 1.初始化 以下是初始化PB5端口的代码 // 定义一个类型GPIO_InitTypeDef,名字叫做GPIO_InitStructure的结构体 GPIO_InitTypeDef GPIO_InitStructure; // PORTB时钟使能 RCC_APB2PeriphClockCmd(RCC_APB2Periph_GPIOB, ENABLE); // 配置结构体GPIO_InitStructure GPIO_InitStructure.GPIO_Pin = GPIO_Pin_5; // 设置GPIO端口号为5 GPIO

This question already has an answer here: How is an integer stored in memory? 4 answers I have a question about int32 storage (c#) . 32 bits means that the biggest number for int is 2^32. 2^32 = 4294967296 , if you divide it by 2 you get the maximum value for an int32 : 4294967296 / 2 = -2147483648 to 2147483648 So I thought half of the bits are for negative figures and the other for positive. But that cant be true because 2^16 = 65536 . Now my question : How is this actually set up in memory? Im really curious about your answers. Only one bit is used for sign (negative or positive) Int32

I'm trying to check whether or not a number has the second bit flag (ie 0000 0010). My code is as follows: int flags = Integer.parseInt(fields[1]); String strflags = Integer.toBinaryString(flags); flags = Integer.parseInt(strflags); int secondBitTest = Integer.parseInt("00000010", 2); if((flags & secondBitTest) == 2) { System.out.println("YES"); } However I think I might be doing this wrong, since when I try to input 147 nothing is returned. You can check if any bit is set using this code that I found here . if (x & (1<<n) != 0) { //n-th bit is set } else { //n-th bit is not set } x is the