compare

I have a basic programming background and have been self sufficient for many years but this problem I can't seem to solve. I have a program in VBA and I need to compare two strings. I have tried using the following methods to compare my strings below but to no avail: //Assume Cells(1, 1).Value = "Cat" Dim A As String, B As String A="Cat" B=Cell(1, 1).Value If A=B Then... If A Like B Then... If StrCmp(A=B, 1)=0 Then... I've even tried inputting the Strings straight into the code to see if it would work: If "Cat" = "Cat" Then... If "Cat" Like "Cat" Then... If StrCmp("Cat" = "Cat", 1) Then... VBA

This question already has answers here : Why does the `is` operator behave differently in a script vs the REPL? (2 answers) Closed 8 months ago . I want to preface this by saying that I know the difference between == and is one is for references and the other is for objects. I also know that python caches the integers in the range (-5, 256) at startup so they should work when comparing them with is . However I have seen a strange behaviour. >>> 2**7 is 2**7 True >>> 2**10 is 2**10 False This is to be expected, 2**7 is 128 and 2**10 is 1024 , one is in the interval (-5, 256) and the other is

What is the difference between CDialog and CDialogEx? I decided to use CDialog because I can't assign m_pMainWnd to Dlg if Dlg is derived from CDialogEx. BOOL CPreparationApp::InitInstance() { MyClass2 Dlg; //derived from CDialog m_pMainWnd = &Dlg; Dlg.DoModal(); return TRUE; } What kind of problems I might fave by not using CDialogEx like form wizard was offering? How to assign m_pMainWnd variable derived from CDialogEx? CDialogEX is derived from CDialog, so, setting m_pMainWnd to a CDialogEx derived object should not be a problem. CDialogEx provides the abillity to set the background color

Assuming I have to (small-to-medium) arrays: tokens = ["aaa", "ccc", "xxx", "bbb", "ccc", "yyy", "zzz"] template = ["aaa", "bbb", "ccc"] How can I determine whether tokens contains all entries of template , in that same order? (Note that in the example above, the first "ccc" should be ignored, resulting in a match due to the last "ccc".) Cleanest, I think, to do this via recursion: class Array def align(other) if pos = index(other.first) other.size == 1 || slice(pos..-1).align(other.drop(1)) end end end so: [1,2,3,4,3,2,1].align([1,2,3]) => true [1,2,3,4,3,2,1].align([1,4,1]) => true [1,2,3,4

I have a simple data file. Each line in the file has four elements. Some lines are filled with no blank entries. Other lines have a first entry and the remaining three are blank, or rather "filled" with a space. It is a tab delimited file. Example of the input file: . . . 30 13387412 34.80391242 sSN_FIRST 30 13387412 34.80391242 sSN5_40 30.1 30.2 30.3 30.4 31 14740248 65.60590089 s32138223_44 31 14740248 65.60590089 s321382_LAST . . . To reiterate, the "blanks" in my file actually contain a single space, if this matters. My overall goal is to "fill in" the second and third column (the fourth

Code: public void placeO(int xpos, int ypos) { for(int i=0; i<3;i++) for(int j = 0;j<3;j++) { // The line below does not work. what can I use to replace this? if(position[i][j]==' ') { position[i][j]='0'; } } } Change it to: if(position[i][j] == 0) Each char can be compared with an int. The default value is '\u0000' i.e. 0 for a char array element. And that's exactly what you meant by empty cell , I assume. To test this you can run this. class Test { public static void main(String[] args) { char[][] x = new char[3][3]; for (int i=0; i<3; i++){ for (int j=0; j<3; j++){ if (x[i][j] == 0){ System

本文将介绍 Java 8 新增的 Lambda 表达式，包括 Lambda 表达式的常见用法以及方法引用的用法，并对 Lambda 表达式的原理进行分析，最后对 Lambda 表达式的优缺点进行一个总结。 1. 概述 Java 8 引入的 Lambda 表达式的主要作用就是简化部分匿名内部类的写法。 能够使用 Lambda 表达式的一个重要依据是必须有相应的函数接口。所谓函数接口，是指内部有且仅有一个抽象方法的接口。 Lambda 表达式的另一个依据是类型推断机制。在上下文信息足够的情况下，编译器可以推断出参数表的类型，而不需要显式指名。 2. 常见用法 2.1 无参函数的简写 无参函数就是没有参数的函数，例如 Runnable 接口的 run() 方法，其定义如下： @FunctionalInterface public interface Runnable { public abstract void run(); } 在 Java 7 及之前版本，我们一般可以这样使用： new Thread(new Runnable() { @Override public void run() { System.out.println("Hello"); System.out.println("Jimmy"); } }).start(); 从 Java 8 开始

一、技术总结 这个题目是属于set容器的内容，使用可以减少很多代码量 开始试过在同一个for循环中定义两个auto，结果编译通不过，有时候构思很重要，就比如这一题，开始我是一个一个去加，而代码中是，先同意其中一个容器中的所有数量，然后再遍历另一个容器，查找是否在该容器中有相同值，如果没有再在之前的所有数量上加一，这样就统计出来了两个set容器中值的并集。 同时，如何查找两个容器中的相同元素也是这个需要考查的点，这里使用find的函数，find函数是返回找到值的迭代器，如果没有找到就返回该容器的end()处迭代器，通过遍历其中一个set容器，然后调用find函数代码如下： int Nt = q[compare[i][0]].size(), Nc = 0;//Nc表示共同元素的个数，Nt表示总共元素，即并集。 for(auto it = q[compare[i][1]].begin(); it != q[compare[i][1]].end(); it++){ if(q[compare[i][0]].find(*it) != q[compare[i][0]].end()) Nc++; else Nt++; } 二、参考代码 #include<iostream> #include<set> using namespace std; const int N = 51; const int

Is making system call to "md5sum file1" and "md5sum file2" and compare two return values enough in this case? Well, that will tell you whether they're definitely different or probably the same. It's possible for two files to have the same hash but not actually have the same data... just very unlikely. In your situation, what is the impact if you get a false positive (i.e. if you think they're the same, but they're not)? MD5 is probably good enough not to worry about collisions if they would only occur accidentally ... but if you've got security (or money) at stake and someone could plant a