declaration

This is a beginners question but I haven't found a comprehensive answer. What are the differences (if any) of the following declarations? CHARACTER(5) :: a CHARACTER, DIMENSION (5) :: b CHARACTER(LEN=5) :: c CHARACTER :: d(5) CHARACTER :: e*5 And: are all of these declarations possible with other types, e.g REAL ? Regardless of the type, <type>,dimension(5) :: b and <type> :: b(5) are identical and denote an array of length 5 . <type> can be e.g. character , integer , real , logical , etc. character(5) is a short-hand of character(len=5) and declares a string of length 5 . If the length is

in a c++ class declaration, you can label a group of members as private or public, e.g. private: int x; double y; seems like there's no way to do this in c#. am I wrong? No, you can't not do this in C#. At best, you can use the default visibility for members, which is private, and not use private, but for public, you have to indicate it for all members. You're correct. Although, if you leave visibility keyword off altogether, a member defaults to private. No you're not wrong. If you don't write any modifiers it will be assumed as private. 来源： https://stackoverflow.com/questions/546869/labeling

just wondering the difference between the presence of the last comma in the array, if there is any at all >> [1,2,3] => [1, 2, 3] >> [1,2,3,] => [1, 2, 3] The second array still works, no exception raised Thanks There's no difference. In Ruby, you're free to add a trailing comma to an array. It makes syntax like this: a = [ 1, 2, 3, ] A bit nicer, in some cases (e.g., if you want to add an element, you simply add a 4, line and don't have to worry about checking for a comma on the last line). It isn't a error, just a empty value(ignored by the compiler), but I suggest you to read Understanding

This question already has an answer here: How does the Comma Operator work 9 answers When I use two variables in a for loop with different conditions two conditions like I have used below i<3,j<2 the for loop is always executing till the second condition fails. #include<iostream> #include<conio.h> using namespace std ; int main() { int i,j ; for(i=0,j=0;i<3,j<2;i++,j++) { cout<<"hello" ; } getch() ; return 0 ; } In that code, hello is printed 2 times. Why? If I use i<3,j<10 , "Hello" is printed 10 times. I can't understand why the first condition is being neglected. Is it compiler dependent or

I'm a beginner and I'm doing an exercise following a book. Below is my code and i got this "Missing partial modifier on declaration of type 'Windowsform.Form1'; another partial declaration of this type exists" . What should i do? My code is as follows: using System; using System.Windows.Forms; namespace Windowsform { public class Form1 : Form { private TextBox txtEnter; private Label lblDisplay; private Button btnOk; public Form1() { this.txtEnter = new TextBox(); this.lblDisplay = new Label(); this.btnOk = new Button(); this.Text = "My Hellowin app!"; //txtEnter this.txtEnter.Location = new

This question already has an answer here: How to initialize HashSet values by construction? 23 answers In Java, how do I create a final Set that's populated at construction? I want to do something like the following: static final Set<Integer> NECESSARY_PERMISSIONS = new HashSet<Integer>([1,2,3,6]); but I don't know the proper syntax in Java. Try this idiom: import java.util.Arrays; new HashSet<Integer>(Arrays.asList(1, 2, 3, 6)) Paul Bellora You might consider using Guava's ImmutableSet : static final Set<Integer> NECESSARY_PERMISSIONS = ImmutableSet.<Integer>builder() .add(1) .add(2) .add(3)

I'm starting to learn Generics for Java and I read several tutorials, but I'm a bit confused and not sure how a generic method is declared. When I use a generic type, what is the correct order to define the method? I found this sample, when do I need to use the angle brackets and when not? public class Box<A> { private A a; ... public void setA(A a) { this.a = a; } public <A> List<A> transform(List<A> in) { return null; } public static <A> A getFirstElement(List<A> list) { return null; } public A getA() { return a; } The problem is that your code is using the same character A, but it has