unsigned

Object-C学习笔记(三)

那年仲夏 提交于 2019-12-14 19:42:48
【推荐】2019 Java 开发者跳槽指南.pdf(吐血整理) >>> 住院好几天了。昨天刚安上无线网卡,写了半天笔记结果破输入法死机白忙活了,还得重搞。 NSArray 是Cocoa的一个类,用于存放对象,这个类不可以直接存放C中数据类型的数据,需要转换成Object-C的对象才可以。 创建一个NSArray对象的方法有很多,这里先说一个比较简单的方法。 + (NSArray) arrayWithObjects:(id) obj, ... count 用于获取NSArray中的对象个数 -(unsigned int) count; objectAtIndex 用于根据索引获取相应的对象,注意如果访问了不存在的索引,不会像PHP一样最多出现个Notice,将会导致程序崩溃,其实这是个好事,因为他会让你及时的修正,避免由于使用非法索引而导致出现难以调试的Bug。 - (id) objectAtIndex:(unsigned int)index; componentsJoinedByString 这个跟PHP中的implode类似,用于根据一个字符串来将数组连接成一个新的字符串。 - (NSString *) componentsJoinedByString:(NSString *)dec; componentsSeparatedByString 这个跟PHP中的explode类似

单片机蜂鸣器的控制程序与驱动电路图

ⅰ亾dé卋堺 提交于 2019-12-14 02:24:44
蜂鸣器从结构区分分为压电式蜂鸣器和电磁式蜂鸣器。压电式为压电陶瓷片发音,电流比较小一些,电磁式蜂鸣器为线圈通电震动发音,体积比较小。按照驱动方式分为有源蜂鸣器和无源蜂鸣器。这里的有源和无源不是指电源,而是振荡源。有源蜂鸣器内部带了振荡源,如图1中所示,给了 BUZZ 引脚一个低电平,蜂鸣器就会直接响。而无源蜂鸣器内部是不带振荡源的,要让他响必须给 500Hz~4.5KHz 之间的脉冲频率信号来驱动它才会响。有源蜂鸣器往往比无源蜂鸣器贵一些,因为里边多了振荡电路,驱动发音也简单,靠电平就可以驱动,而无源蜂鸣器价格比较便宜,此外无源蜂鸣器声音频率可以控制,而音阶与频率又有确定的对应关系,因此就可以做出来“do re mi fa sol la si”的效果,可以用它制作出简单的音乐曲目,比如生日歌、两只老虎等等。 来看一下图1的电路,蜂鸣器电流依然相对较大,因此需要用三极管驱动,并且加了一个 100 欧的电阻作为限流电阻。此外还加了一个 D4 二极管,这个二极管叫做续流二极管。我们的蜂鸣器是感性器件,当三极管导通给蜂鸣器供电时,就会有导通电流流过蜂鸣器。而我们知道,电感的一个特点就是电流不能突变,导通时电流是逐渐加大的,这点没有问题,但当关断时,经“电源-三极管-蜂鸣器-地”这条回路就截断了,过不了任何电流了,那么储存的电流往哪儿去呢,就是经过这个 D4 和蜂鸣器自身的环路来消耗掉了

Is this an unavoidable signed and unsigned integer comparison?

廉价感情. 提交于 2019-12-13 20:04:52
问题 Probably not, but I can't think of a good solution. I'm no expert in C++ yet. Recently I've converted a lot of int s to unsigned int s in a project. Basically everything that should never be negative is made unsigned. This removed a lot of these warnings by MinGW: warning: comparison between signed and unsigned integer expressions [-Wsign-compare] I love it. It makes the program more robust and the code more descriptive. However, there is one place where they still occur. It looks like this:

Ruby Unsigned Right Shift

不问归期 提交于 2019-12-13 18:25:17
问题 I have a snippet of java code: return (int)(seed >>> (48 - bits)); As you can see it uses the unsigned right shift operator (>>>). I'm trying to implement this code in ruby which has no unsigned right shift operator, only a signed right shift operator. As I'm not very familiar with the >>> operator I'm not quite sure on how I'd impelemnt this in ruby. I tried doing a few searches to see if anyone had come across this problem before, but couldn't find anything relevent. Any help would be much

How to add binary that is unsigned char type in C?

拥有回忆 提交于 2019-12-13 11:16:45
问题 Here is the sheet that I need to follow https://imgur.com/a/JuLpQZt Here is my code as of now #include<stdio.h> #include<conio.h> #include<stdlib.h> #include<windows.h> void arithmetic(); int ALU(unsigned char operand1, unsigned char operand2, unsigned char control_signals); int askOperand1(); int askOperand2(); unsigned char operand1; unsigned char operand2; unsigned char control_signals; const int ACC = 16; //ACC = ACCUMULATOR int main() { for(;;) { system("cls"); ALU(operand2,operand2

Why does 1ul << 64 return 1 instead of 0? [duplicate]

穿精又带淫゛_ 提交于 2019-12-13 10:52:00
问题 This question already has answers here : Unexpected output when executing left-shift by 32 bits (2 answers) GCC left shift overflow (4 answers) Closed 3 years ago . Consider the following piece of code: // Simply loop over until 64 is hit. unsigned long x = 0; for (int i = 0; i <= 64; i++) { if (i == 64) { x = 1ul << i; printf("x: %d\n", x); } } We know that unsigned long is 64-bit wide, and left shifting 1 by 64 positions would become 1000...000 (64 zeros behind one), and would have been

UINT_MAX the same as ULONG_MAX in C

对着背影说爱祢 提交于 2019-12-13 05:45:32
问题 While solving exercises from the K&R C book, I stumbled upon the exercise 2.1. At first I got as UINT_MAX as -1 , but then I used the %u placeholder, but now its giving me the same number as ULONG_MAX . In the book in Appendix B, they say that UINT_MAX should be 65535 and ULONG_MAX should be 4294967295 , but when running the exercise, its giving me for both UINT_MAX and ULONG_MAX as 4294967295 . Why is that? 回答1: First of all, the right way to print an unsigned long is not %u but %lu . Second

linux下socketcan代码编写

无人久伴 提交于 2019-12-13 04:54:56
linux下socketcan代码编写 1.相关头文件 #include <stdio.h> #include <stdlib.h> #include <string.h> #include <unistd.h> #include <net/if.h> #include <sys/ioctl.h> #include <sys/socket.h> #include <linux/can.h> #include <linux/can/raw.h> 2.can设备的初始化 /** * function :CAN设备初始化 * parameter :void * return :s (socketcan套接字) * */ int Can_Init(void) { int s,ret; struct sockaddr_can addr; struct ifreq ifr; system("sudo ip link set can0 type can bitrate 100000"); system("sudo ifconfig can0 up"); printf("this is a can send demo\r\n"); //1.Create socket s = socket(PF_CAN, SOCK_RAW, CAN_RAW); if (s < 0) { perror("socket

Converting from unsigned long int to signed int and vice versa

不打扰是莪最后的温柔 提交于 2019-12-13 04:04:21
问题 I would like to pass a signed int to gsl_rng_uniform_int (const gsl_rng * r, unsigned long int n) . The signed int that I'm passing is greater than or equal to zero. The function will return a number between 0 and n , so if I pass it a positive signed int, it will return something that is in the range of a signed int. I then want to store the return value in a signed int. What's the cleanest way to do this with the obvious expected behaviour? I'm on a 64-bit compiler on a 64-bit Linux machine

Is it possible to use jnlp without signing the jars?

て烟熏妆下的殇ゞ 提交于 2019-12-12 12:25:35
问题 Is there any way at all to use jnlp without having to sign the jars involved? (The application is being used in a secure environment so from the security point of view signing is not necessary) 回答1: Yes you can, however the webstart application will be in 'sandbox' mode and cannot access filesystem, network and many other parts of the OS. Simply leave out the security element to accomplis this. security Element Each application is, by default, run in a restricted execution environment,