标签: operator-keyword

(注意:我正在编写这个项目仅供学习;有关冗余的评论是......呃,多余.;)

我正在尝试实现一个随机访问迭代器,但是我发现关于这个主题的文献很少,所以我将通过试验和错误结合维基多数运算符重载原型列表.到目前为止它运作良好,但我遇到了障碍.

代码如

exscape::string::iterator i = string_instance.begin();
std::cout << *i << std::endl;

工作,并打印字符串的第一个字符.但是,*(i + 1)不起作用,*(1 + i)也不起作用.我的完整实现显然有点太多了,但这是它的要点:

namespace exscape {
    class string {
        friend class iterator;
    ...
    public:
        class iterator : public std::iterator<std::random_access_iterator_tag, char> {
            ...
            char &operator*(void) {
                return *p; // After some bounds checking
            }
            char *operator->(void) {
                return p;
            }

            char &operator[](const int offset) {
                return *(p + offset); // After some bounds checking
            }

            iterator &operator+=(const int offset) {
                p += offset;
                return *this;
            }

            const …

我想在C#中重载运算符'++',但是当我编写下面的代码时,VS 2012会给我一个错误消息.

    public LogItem operator ++()
    {
        ++ visitTimes;
    }

错误是重载的一元运算符++需要一个参数

我在这里是LogItem 类的定义:

     public class LogItem
     {
        /**
         * Constructor
         */
        public LogItem(string ip)
        {
            ipAddress = ip;
            visitTimes = 0;
        }

        /**
         * Operator Overload Function
         */
        public LogItem operator ++() 
        {
            ++ visitTimes;
        }

        public string ipAddress { get; private set; }
        public string location { get; set; }
        public int visitTimes { get; private set; }
     }

我该怎么做才能重载运算符'++'？

编写一个简单的代码并遇到问题我不知道如何处理.我试着通过搜索来调查它,但我找不到任何帮助,每个人的答案都有点高于我的头脑.请有人像小孩子一样解释这个,哈哈.谢谢.

#include <iostream>
#include <string>

using namespace std;

int main()
{
    string invCode = "";
    string lastTwoChars = "";

    cout << "Use this program to determine color of furniture.";
    cout << "Enter five-character inventory code: ";
    cin >> invCode;

    if (invCode.length() == 5)
    {
        lastTwoChars = invCode.substr(3,2);
         if (lastTwoChars == 41)
         { 
              cout << "Red";
              }
         if (lastTwoChars == 25)
         { 
              cout << "Black";
              }
         if (lastTwoChars == 30)
         { 
              cout << "Green";
              }
    }
    else
         cout << "Invalid inventory …

看来我们不能在C#泛型类中轻松调用类型转换运算符.这是代码.为什么？

T006最终归档我们的目标.

using System;
using System.Collections.Generic;
using System.Text;
using System.Diagnostics;
using System.Reflection;
using System.Linq;

namespace ConsoleApplication1
{
    class vec<T, T2> : List<T> where T : class
    {
        public vec(IEnumerable<T2> other)
        {
            //Converter<T2, T> cvt = (v) => (T)v; // T004 failed, try defined function dynamicly, cannot compile, too.

            // T006 pass, client happy, we not happy, but anyway passed, performance may not happy.
            var conversionOperator = typeof(T).GetMethods(BindingFlags.Static | BindingFlags.Public)
                                    .Where(m => m.Name == "op_Explicit" || m.Name == "op_Implicit") …

我正在尝试编写一个类似于这个人的程序,该程序靠近页面顶部的Learn Python the Hard Way程序.

http://learnpythonthehardway.org/book/ex16.html

这是我的下面版本.但它告诉我"%r"最后使用它为什么这样做？我认为这就是你在括号中要做的事情.

# -- coding: utf-8 --

from sys import argv

script, filename = argv

print "Would you like file %r to be overwritten?" % filename
print "Press RETURN if you do, and CTRL-C otherwise."

raw_input('> ')

print "Opening the file ..."
target = open(filename, 'w')
target.truncate()

print "Now type three lines to replace the contents of %r" % filename

line1 = raw_input("line 1: ")
line2 = raw_input("line 2: ")
line3 …

你好我有两个号码

int a= 6;
int b = 20;

当我在做orie |操作时,它没有给出正确的结果.

int result = a | b ;

result = 22但正确的答案是26.

数组yn并且zn等于数字,但有一个奇怪的区别:yn += 7正如预期的那样,行不会改变tn数组,但第二行最后一行会zn += 7改变tn数组!

这是代码:

import numpy as np
def f(x): return (1*x)
def g(x): return (x)
nn = 5
tn = np.zeros(nn)
yn = np.zeros(nn)
zn = np.zeros(nn)

tn = np.linspace(0,1,nn)
yn = f(tn)
zn = g(tn)

print('tn at step1 =',tn)
yn += 7  #as expected, this line does not change tn.
print('tn at step2 =',tn)
zn += 7  #why this line adds 7 to tn …

大家好，我正在尝试重载ifstream和ofstream，但没有成功。

头文件：

#include <iostream>
#include <fstream>

using namespace std;

class Complex
{
    private:
        double real;
        double imaginary;
    public:
        //constructors
        Complex();
        Complex(double newreal, double newimaginary);
        ~Complex();
        //setter
        void setReal(double newreal);
        void setImaginary(double newimaginary);
        //getter
        double getReal();
        double getImaginary();
        //print
        void print();
        //methods
        Complex conjugate();
        Complex add(Complex c2);
        Complex subtraction(Complex c2);
        Complex division(Complex c2);
        Complex multiplication(Complex c2);

friend ifstream& operator >> (ifstream& in, Complex &c1)
{

    in >> c1;

    return in;
}

};

测试文件：

#include <iostream>
#include <fstream>
#include <string>
#include "Complex2.h" …

我不明白为什么一元运算符不起作用.我相信我错过了一些概念.请帮忙.

#include <iostream>
using namespace std;

class Distance {
   private:
      int feet;             // 0 to infinite
      int inches;           // 0 to 12
   public:
      // required constructors
      Distance(){
         feet = 0;
         inches = 0;
      }
      Distance(int f, int i){
         feet = f;
         inches = i;
      }
      // method to display distance
      void displayDistance() {
         cout << "F: " << feet << " I:" << inches <<endl;
      }
      // overloaded minus (-) operator
      Distance operator- () {
         return Distance(-feet, -inches);
      }
}; …

在VB.Net中是否有相当于C#的管道运算符(|)？

我从这里有一些代码如何为我的应用程序为所有用户创建的文件授予完全权限？

它在C#中,我想将其转换为VB.Net.我到目前为止(VS说有一个错误:| InheritanceFlags.ContainerInherit):

Sub ZugriffsrechteEinstellen()

    Dim dInfo As New DirectoryInfo(strPfadSpracheINI)
    Dim dSecurity As New DirectorySecurity

    dSecurity = dInfo.GetAccessControl()
    dSecurity.AddAccessRule(New FileSystemAccessRule(New SecurityIdentifier(WellKnownSidType.WorldSid, Nothing), FileSystemRights.FullControl, InheritanceFlags.ObjectInherit | InheritanceFlags.ContainerInherit, PropagationFlags.NoPropagateInherit, AccessControlType.Allow))

    dInfo.SetAccessControl(dSecurity)

End Sub