小编liv*_*hak的帖子

我是一个C++新手.我在这里尝试了我的第一个程序.我的眼睛这个程序是正确的.

#include <iostream>

using namespace std;

class mystruct
{
    private:
        int m_a;
        float m_b;

    public:
        mystruct(int x, float y)
        {
                m_a = x;
                m_b = y;
        }

};


int main()
{

        mystruct m = mystruct(5,3.14);

        cout << "my structure " << m << endl;

        return 0;
}

但是我得到了很多错误.不知道为什么？

cout.cpp: In function ‘int main()’:
cout.cpp:26:29: error: no match for ‘operator<<’ in ‘std::operator<< [with _Traits = std::char_traits<char>]((* & std::cout), ((const char*)"my structure ")) << m’
cout.cpp:26:29: note: candidates are:
/usr/include/c++/4.6/ostream:110:7: note: std::basic_ostream<_CharT, …

我试图了解各种sysconf宏.我编写了一个程序如下.

int main()
{
    fprintf(stdout, "No. of clock ticks per sec : %ld\n",sysconf(_SC_CLK_TCK));
    return 0;
}

我总是得到100的结果.我在一个时钟频率为2.93GHz的CPU上运行它.数字100的确切含义是什么.

我有一个unsigned char指针,其中包含一个结构.现在我想要执行以下操作

unsigned char buffer[24];

//code to fill the buffer with the relevant information.

int len = ntohs((record_t*)buffer->len);

其中record_t结构包含一个名为len的字段.我无法这样做并且收到错误.

error: request for member ‘len’ in something not a structure or union.

然后我尝试了:

int len = ntohs(((record_t*)buffer)->len);

这样才能使操作员优先.那给了我warning: dereferencing type-punned pointer will break strict-aliasing rules.

然后我宣布

record_t *rec = null;

rec = (record_t*)

我在这做错了什么？

我想要一个排序的字典.但之间的项目的顺序mydict和orddict似乎并没有改变.

from collections import OrderedDict

mydict = {'a':1,'b':2,'c':3,'d':4}

orddict = OrderedDict(mydict)

print(mydict,orddict)

# print items in mydict:
print('mydict')
for k,v in mydict.items():
    print(k,v)

print('ordereddict')
# print items in ordered dictionary
for k,v in orddict.items():
    print(k,v)


# print the dictionary keys
# for key in mydict.keys():
#     print(key)


#  print the dictionary values
# for value in mydict.values():
#     print(value)

我试图在运行12.04的Ubuntu机器上使用mysql创建一个远程数据库.

它有一个root用户,启用了远程登录,没有密码.我已经启动了服务器.

输出

sudo netstat -tap | grep mysql

节目

tcp        0      0 localhost:mysql         *:*                     LISTEN      13246/mysqld

我创建了一个名为nwtopology的数据库(如上所述root还没有密码.)

 create database nwtopology
 grant all privileges on *.* to root@192.168.129.221
 FLUSH PRIVILEGES;

从也运行Ubuntu 12.04的客户端机器我使用python脚本使用sqlalchemy连接到远程mysql数据库.

from pox.core import core
import pox.openflow.libopenflow_01 as of
import re
import datetime
import time
from sqlalchemy import create_engine, ForeignKey
from sqlalchemy import Column, Date, Integer, String
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import relationship, backref
from sqlalchemy import create_engine
from sqlalchemy.orm import sessionmaker
from sqlalchemy.sql.expression import exists

log …

我从这个链接(https://gist.github.com/jiewmeng/3787223)获得了这个程序.我一直在网上搜索,以便更好地理解处理器缓存(L1和L2).我想成为能够编写一个程序,让我能够猜测我的新笔记本电脑上L1和L2缓存的大小.(仅用于学习目的.我知道我可以检查规格.)

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

#define KB 1024
#define MB 1024 * 1024

int main() {
    unsigned int steps = 256 * 1024 * 1024;
    static int arr[4 * 1024 * 1024];
    int lengthMod;
    unsigned int i;
    double timeTaken;
    clock_t start;
    int sizes[] = {
        1 * KB, 4 * KB, 8 * KB, 16 * KB, 32 * KB, 64 * KB, 128 * KB, 256 * KB,
        512 * KB, 1 * MB, 1.5 …

我遇到了以下迷宫定义代码:

typedef struct mazeNode {
    int hasCheese;
    int tag;
    struct mazeNode *left;
    struct mazeNode *right;
} maze_t;

maze_t maze = {
    .tag = 1,
    .left = &(maze_t) {
        .left = &(maze_t) {
            .left = &(maze_t) {},
            .right = &(maze_t) {}
        },
        .right = &(maze_t) {
            .right = &(maze_t) {}
        }
    },
    .right = &(maze_t) {
        .tag = 8,
        .left = &(maze_t) {},
        .right = &(maze_t) {
            .tag = 10,
            .left = &(maze_t) {
                .tag = 11,
                .left = …

我试图运行一个java程序,我得到以下运行时错误.错误如下所示.

Exception in thread "main" java.lang.NoSuchFieldError: DEF_CONTENT_CHARSET
    at org.apache.http.impl.client.DefaultHttpClient.setDefaultHttpParams(DefaultHttpClient.java:175)
    at org.apache.http.impl.client.DefaultHttpClient.createHttpParams(DefaultHttpClient.java:158)
    at org.apache.http.impl.client.AbstractHttpClient.getParams(AbstractHttpClient.java:448)
    at org.apache.http.impl.client.AbstractHttpClient.createClientConnectionManager(AbstractHttpClient.java:309)
    at org.apache.http.impl.client.AbstractHttpClient.getConnectionManager(AbstractHttpClient.java:466)
    at org.apache.http.impl.client.AbstractHttpClient.createHttpContext(AbstractHttpClient.java:286)
    at org.apache.http.impl.client.AbstractHttpClient.execute(AbstractHttpClient.java:851)
    at org.apache.http.impl.client.AbstractHttpClient.execute(AbstractHttpClient.java:805)
    at org.apache.http.impl.client.AbstractHttpClient.execute(AbstractHttpClient.java:784)
    at net.floodlightcontroller.core.internal.PacketStreamerClient.registerForPackets(PacketStreamerClient.java:90)
    at net.floodlightcontroller.core.internal.PacketStreamerClient.main(PacketStreamerClient.java:51)

现在我添加到类路径的文件如下.

export CLASSPATH=$(JARS=(./lib/*.jar); IFS=:; echo "${JARS[*]}")
export CLASSPATH=$CLASSPATH:~/.m2/repository/org/apache/httpcomponents/httpclient/4.0.1/httpclient-4.0.1.jar
export CLASSPATH=$CLASSPATH:~/.m2/repository/org/apache/httpcomponents/httpcore/4.0.1/httpcore-4.0.1.jar
export CLASSPATH=$CLASSPATH:~/.m2/repository/commons-logging/commons-logging/1.1.1/commons-logging-1.1.1.jar
export CLASSPAHT=$CLASSPATH:~/ms_thesis/ONOS/httpcore-4.1.jar
#export CLASSPATH=$CLASSPATH:~/ms_thesis/ONOS/lib/httpclient-4.2.jar
export CLASSPATH=$CLASSPATH:~/google-gson-2.2.4/gson-2.2.4.jar

"main"java.lang.NoSuchFieldError:DEF_CONTENT_CHARSET的原因是什么

我下载了,http-core-4.1-alpha因为这是org/apache/http/params/SyncBasicHttpParams class来自findjar.com 的jar .那个版本的http-core是不可协商的.我如何找到与该版本的http-core兼容的httpclient版本？

我有以下示例程序

#include <stdio.h>                              
#include <stdlib.h>

#pragma pack(push)
#pragma pack(1)
typedef struct{
  char a;
  int b;
  char c;
}st_a;
#pragma pack(pop)

typedef struct{
  char a;
  int b;
  char c;
}st_b;


int main()
{
  printf("size of struct a %zd \n",sizeof(st_a));
  printf("size of struct b %zd \n",sizeof(st_b));

  return 0;
}

上述计划的输出是

size of struct a 6 
size of struct b 12 

现在,如果我更改结构声明如下:

#pragma pack(1)
typedef struct{
  char a;
  int b;
  char c;
}st_a;
#pragma unpack()

该计划的输出是

size of struct a 6 
size of …

我正在编写一些代码,我需要在for循环中使用两个变量.以下代码似乎没问题吗？

它确实给了我预期的结果.

for (loop_1 = offset,loop_2 = (offset + 2); loop_1 >= (offset - 190),loop_2 <= (190 + offset + 2); loop_1--,loop_2++)
{
    if (  (*(uint8_t*)(in_payload + loop_1) == get_a1_byte(bitslip)) &&
         ((*(uint8_t*)(in_payload + loop_2) == get_a2_byte(bitslip)))
       )
    {
          a1_count++;
    }
}

但我得到一个编译器警告说:

file.c:499:73:警告:逗号表达式的左侧操作数无效

这是什么意思？