unsigned

Proper Way To Initialize Unsigned Char*

由 北城以北 提交于 2019-12-09 04:38:50
问题 What is the proper way to initialize unsigned char* ? I am currently doing this: unsigned char* tempBuffer; tempBuffer = ""; Or should I be using memset(tempBuffer, 0, sizeof(tempBuffer)); ? 回答1: The second method will leave you with a null pointer. Note that you aren't declaring any space for a buffer here, you're declaring a pointer to a buffer that must be created elsewhere. If you initialize it to "" , that will make the pointer point to a static buffer with exactly one byte—the null

narrowing conversion of long long int to long unsigned int inside {} [-Wnarrowing]

由 情到浓时终转凉″ 提交于 2019-12-09 03:03:29
now the newer error "narrowing conversion of long long int to long unsigned int inside {} [-Wnarrowing]" the source code relative to it: // Implements the joint conditional cummulative density functions for CP-nets // For node j in the lexicographic topological sort as defined by the dag code: // the table gives P( s, t | n, c, j ). // The number of rows in each table is the triangular number of min(c, j). #ifndef TABLES_H #define TABLES_H #include <random> // Mersenne Twister #include <chrono> // Used to initialize MT const unsigned long int factorial[17] = {1, 1, 2, 6, 24, 120, 720, 5040,

C# multi-threaded unsigned increment

由 寵の児 提交于 2019-12-09 02:37:01
问题 I want to increment an unsigned integer from multiple threads. I know about Interlocked.Increment, but it does not handle unsigned integers. I could use lock(), but I would rather not if possible for performance reasons. Is it thread safe just to increment it in the normal way? It would not matter if the occasional increment got lost, as it's only used for statistics. What I don't want is the value to get corrupted. 回答1: You say you don't want to use lock for performance reasons - but have

Unsigned / Signed Arithmetic Problems from A Programmer's Perspective Textbook

由 随声附和 提交于 2019-12-09 01:57:37
问题 int x = random(); int y = random(); unsigned ux = (unsigned) x; unsigned uy = (unsigned) y; For each of the following C expressions, you are to indicate whether or not the expression always yields 1. If it always yields 1, describe the underlying mathematical principles. Otherwise, give an example of arguments that make it yield 0. A. (x<y) == (-x>-y) B. ((x+y)<<4) + y-x == 17*y+15*x C. ~x+~y+1 == ~(x+y) D. (ux-uy) == -(unsigned)(y-x) E. ((x >> 2) << 2) <= x For these questions, I got that

would doing arithmetic operation on a pair of signed and unsigned numbers be legal?

由 ◇◆丶佛笑我妖孽 提交于 2019-12-09 00:19:25
问题 I'm more than half way through learning assembly and I'm familiar with the concept of how signed and unsigned integers are presented in bits, I know that it might seem a weird question of which the answer would be pretty obvious, but I'm wondering if using an arithmetic operation like addition makes sense for a pair of numbers that one of them is considered signed and the other one unsigned, I've thought of multiple examples like below that will yield a correct result: 10000001 (1-byte

C++基本内置类型

由 試著忘記壹切 提交于 2019-12-09 00:11:33
基本内置类型 在C++提供的基本内置类型中包含了 算术类型 和 void类型 。而算术类型又分为 布尔型 ， 字符型 ， 整型 以及 浮点型 : 和很多“现代”语言不同，C++并没有规定其基本内置类型的具体大小（所占的bit数），标准只对以下两点提了要求： 标准虽然没有规定具体的大小，但规定了下限，即基本内置类型的最小位数。 标准规定了相关基本内置类型之间的大小关系，比如char \(\lt\) short \(\leq\) int \(\leq\) long \(\leq\) long long。 类型 含义 最小长度 备注 bool 布尔值 NA char 字符型 8 bits 大小等于一个字节 wchar_t 宽字符型 16 bits char16_t unicode字符型 16 bits 大小固定 char32_t unicode字符型 32 bits 大小固定 short 短整型 16 bits int 整型 16 bits long 长整型 32 bits long long 超长整型 64 bits C++11标准引入 float 单精度浮点型 6位有效数字 double 双精度浮点型 10位有效数字 long double 扩展精度浮点型 10位有效数字 语言细节 char的大小等于一个字节： C++并没有byte类型，一般都是默认char类型占用一个字节

double to unsigned int / char

由 谁都会走 提交于 2019-12-08 19:55:37
问题 I read here, that: According to C99 §6.3.1.4 footnote 50: The remaindering operation performed when a value of integer type is converted to unsigned type need not be performed when a value of real floating type is converted to unsigned type. Thus, the range of portable real floating values is (−1, Utype_MAX+1). Now, I am interested in the subtle difference (this time for C++ 03 !) between: double d1 = 257; double d2 = -2; unsigned char c1 = d1; // undefined, since d1 > 256 unsigned char c2 =

Is there a difference between unsigned and unsigned int [duplicate]

由 六眼飞鱼酱① 提交于 2019-12-08 11:39:38
问题 This question already has answers here : Difference between unsigned and unsigned int in C (5 answers) Closed 2 years ago . In C, is there a difference between unsigned as a type directly and unsigned int ? Update: This is a duplicate of this question, but the other question didn't answer about small subleties that can exist. Thanks @EOF for pointing out the implementation defined behavior in bitfields. 回答1: No, unsigned and unsigned int both refer to the same type. See cppreference

Workaround to get codes to work in MSVC++ 2010 due to “Type name is not allowed”

由 左心房为你撑大大i 提交于 2019-12-08 09:19:32
问题 I am trying to implement a finger detection, which link is given here. Am I am going through the code in MSVC2010, it gives me error as shown in Figure as shown below. Could someone tell me why the following codes gives me error? Is this related to these following questions; 1, 2, 3? Is there a possible workaround? I already included: #include <cstdint> #include <stdint.h> I also tried: unsigned short depth = (unsigned short) (v[2] * 1000.0f); // hand depth unsigned short near = (unsigned

Linux操作系统 进程之间的通信

由 你。 提交于 2019-12-08 05:12:31
进程之间的通信 预备知识： 1、用户态和内核态，当一个进程在执行用户自己的代码时处于用户运行态（用户态）；当一个进程因为系统调用陷入内核代码中执行时处于内核运行态（内核态）。 2、进程之间的通信（Inter Processs Communication- IPC ）实现机制有：管道、消息队列、信号值、信号、共享内存、共享映射文件、套接字等。 3、及时通信：信号（类似中断）；非及时通信：共享内存、邮箱、管道、套接字、 4、常见的信号：终止信号、定时器信号、用户自定义信号等 5、信号： 用户、系统或者进程 发送给 目标进程 的 信息 ，以通知目标进程某个 状态的改变 或 系统异常 。 6、 PCB（progress control block- 进程控制块），系统通过PCB，描述进程和控制进程。在Linux系统下，PCB是 task_struct结构体（进程描述符） 。 　　1、 进程状态 ：记录进程是处于运行状态还是等待状态 　　2、 调度信息 ：进程由哪个函数调度，具体怎样调度等 　　3、进程之间的 通讯状况 　　4、进程之间的 亲属关系 ：在父进程和子进程之间有task_struct类型的指针，将父进程和子进程联系起来 　　5、 时间数据信息 ：每个进程执行所占用CPU的时间 　　6、 进程的标志 　　7、 进程的标识符 ：该进程唯一的标识符用来区别其他进程 　　8、
订阅 unsigned