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  • Haskell Cons Operator (:)

    - by Carson Myers
    I am really new to Haskell (Actually I saw "Real World Haskell" from O'Reilly and thought "hmm, I think I'll learn functional programming" yesterday) and I am wondering: I can use the construct operator to add an item to the beginning of a list: 1 : [2,3] [1,2,3] I tried making an example data type I found in the book and then playing with it: --in a file data BillingInfo = CreditCard Int String String | CashOnDelivery | Invoice Int deriving (Show) --in ghci $ let order_list = [Invoice 2345] $ order_list [Invoice 2345] $ let order_list = CashOnDelivery : order_list $ order_list [CashOnDelivery, CashOnDelivery, CashOnDelivery, CashOnDelivery, CashOnDelivery, CashOnDelivery, CashOnDelivery, CashOnDelivery, CashOnDelivery, CashOnDelivery, CashOnDelivery, CashOnDelivery, CashOnDelivery, CashOnDelivery, ...- etc... it just repeats forever, is this because it uses lazy evaluation? -- EDIT -- okay, so it is being pounded into my head that let order_list = CashOnDelivery:order_list doesn't add CashOnDelivery to the original order_list and then set the result to order_list, but instead is recursive and creates an infinite list, forever adding CashOnDelivery to the beginning of itself. Of course now I remember that Haskell is a functional language and I can't change the value of the original order_list, so what should I do for a simple "tack this on to the end (or beginning, whatever) of this list?" Make a function which takes a list and BillingInfo as arguments, and then return a list? -- EDIT 2 -- well, based on all the answers I'm getting and the lack of being able to pass an object by reference and mutate variables (such as I'm used to)... I think that I have just asked this question prematurely and that I really need to delve further into the functional paradigm before I can expect to really understand the answers to my questions... I guess what i was looking for was how to write a function or something, taking a list and an item, and returning a list under the same name so the function could be called more than once, without changing the name every time (as if it was actually a program which would add actual orders to an order list, and the user wouldn't have to think of a new name for the list each time, but rather append an item to the same list).

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  • Implementing operator< in C++

    - by Vulcan Eager
    I have a class with a few numeric fields such as: class Class1 { int a; int b; int c; public: // constructor and so on... bool operator<(const Class1& other) const; }; I need to use objects of this class as a key in an std::map. I therefore implement operator<. What is the simplest implementation of operator< to use here?

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  • What is Ruby's double-colon (::) all about?

    - by Meltemi
    I'd probably be able to answer this for myself if "::" wasn't so hard to Google. Didn't see anything on SO so thought I'd try my luck. What is this double-colon :: all about? I see it everywhere in Rails: class User < ActiveRecord::Base or… ActionController::Routing::Routes.draw do |map| I found a definition from this guy: The :: is a unary operator that allows: constants, instance methods and class methods defined within a class or module, to be accessed from anywhere outside the class or module. but that just leads to more questions. What good is scope (private, protected) if you can just use :: to expose anything?

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  • C# implicit conversions and == operator

    - by Arnis L.
    Some code for context: class a { } class b { public a a{get;set;} public static implicit operator a(b b) { return b.a; } } a a=null; b b=null; a = b; //compiler: cannot apply operator '==' to operands of type tralala... bool c = a == b; Is it possible to use == operator on different type instances, where one can implicitly convert to another? What did i miss? Edit: If types must be the same calling ==, then why int a=1; double b=1; bool c=a==b; works?

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  • why i^=j^=i^=j isn't equal to *i^=*j^=*i^=*j

    - by klvoek
    In c , when there is variables (assume both as int) i less than j , we can use the equation i^=j^=i^=j to exchange the value of the two variables. For example, let int i = 3, j = 5; after computed i^=j^=i^=j, I got i = 5, j = 3 . What is so amazing to me. But, if i use two int pointers to re-do this , with *i^=*j^=*i^=*j , use the example above what i got will be i = 0 and j = 3. Then, describe it simply: In C 1 int i=3, j=5; i^=j^=i^=j; // after this i = 5, j=3 2 int i = 3, j= 5; int *pi = &i, *pj = &j; *pi^=*pj^=*pi^=*pj; // after this, $pi = 0, *pj = 5 In JavaScript var i=3, j=5; i^=j^=i^=j; // after this, i = 0, j= 3 the result in JavaScript makes this more interesting to me my sample code , on ubuntu server 11.0 & gcc #include <stdio.h> int main(){ int i=7, j=9; int *pi=&i, *pj=&j; i^=j^=i^=j; printf("i=%d j=%d\n", i, j); i=7, j==9; *pi^=*pj^=*pi^=*pj printf("i=%d j=%d\n", *pi, *pj); } however, i had spent hours to test and find out why, but nothing means. So, please help me. Or, just only i made some mistake???

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  • printing using one '\n'

    - by Alex
    I am pretty sure all of you are familiar with the concept of the Big4, and I have several stuffs to do print in each of the constructor, assignment, destructor, and copy constructor. The restriction is this: I CAN'T use more than one newline (e.g., ƒn or std::endl) in any method I can have a method called print, so I am guessing print is where I will put that precious one and only '\n', my problem is that how can the method print which prints different things on each of the element I want to print in each of the Big4? Any idea? Maybe overloading the Big4?

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  • Any way to allow classes implementing IEntity and downcast to have operator == comparisons?

    - by George Mauer
    Basically here's the issue. All entities in my system are identified by their type and their id. new Customer() { Id = 1} == new Customer() {Id = 1}; new Customer() { Id = 1} != new Customer() {Id = 2}; new Customer() { Id = 1} != new Product() {Id = 1}; Pretty standard scenario. Since all Entities have an Id I define an interface for all entities. public interface IEntity { int Id { get; set;} } And to simplify creation of entities I make public abstract class BaseEntity<T> : where T : IEntity { int Id { get; set;} public static bool operator ==(BaseEntity<T> e1, BaseEntity<T> e2) { if (object.ReferenceEquals(null, e1)) return false; return e1.Equals(e2); } public static bool operator !=(BaseEntity<T> e1, BaseEntity<T> e2) { return !(e1 == e2); } } where Customer and Product are something like public class Customer : BaseEntity<Customer>, IEntity {} public class Product : BaseEntity<Product>, IEntity {} I think this is hunky dory. I think all I have to do is override Equals in each entity (if I'm super clever, I can even override it only once in the BaseEntity) and everything with work. So now I'm expanding my test coverage and find that its not quite so simple! First of all , when downcasting to IEntity and using == the BaseEntity< override is not used. So what's the solution? Is there something else I can do? If not, this is seriously annoying. Upadate It would seem that there is something wrong with my tests - or rather with comparing on generics. Check this out [Test] public void when_created_manually_non_generic() { // PASSES! var e1 = new Terminal() {Id = 1}; var e2 = new Terminal() {Id = 1}; Assert.IsTrue(e1 == e2); } [Test] public void when_created_manually_generic() { // FAILS! GenericCompare(new Terminal() { Id = 1 }, new Terminal() { Id = 1 }); } private void GenericCompare<T>(T e1, T e2) where T : class, IEntity { Assert.IsTrue(e1 == e2); } Whats going on here? This is not as big a problem as I was afraid, but is still quite annoying and a completely unintuitive way for the language to behave. Update Update Ah I get it, the generic implicitly downcasts to IEntity for some reason. I stand by this being unintuitive and potentially problematic for my Domain's consumers as they need to remember that anything happening within a generic method or class needs to be compared with Equals()

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  • What does the ^ operator do in Java?

    - by joroj
    What function does the "^" operator serve in Java? When I try this: int a = 5^n; ...it gives me: for n = 5, returns 0 for n = 4, returns 1 for n = 6, returns 3 ...so I guess it doesn't indicate exponentiation. But what is it then?

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  • Who can give me a link for the operator= of vector in MSDN?

    - by 8888q8888
    Who can give me a link for the operator= of vector in MSDN? Why I can only find operator[]? If operator= is just something default, like copy everything in A to B, how this following code works? vector<double> v(100,1); v = vector<double>(200,2); // if operator= is just a trivail version, how to make sure the old v get cleared?

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  • Operator Overloading in C++ as int + obj

    - by Azher
    Hi Guys, I have following class:- class myclass { size_t st; myclass(size_t pst) { st=pst; } operator int() { return (int)st; } int operator+(int intojb) { return int(st) + intobj; } }; this works fine as long as I use it like this:- char* src="This is test string"; int i= myclass(strlen(src)) + 100; but I am unable to do this:- int i= 100+ myclass(strlen(src)); Any idea, how can I achieve this?? Thanks in advance. Regards,

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  • Why is overloading operator&() prohibited for classes stored in STL containers?

    - by sharptooth
    Suddenly in this article ("problem 2") I see a statement that C++ Standard prohibits using STL containers for storing elemants of class if that class has an overloaded operator&(). Having overloaded operator&() can indeed be problematic, but looks like a default "address-of" operator can be used easily through a set of dirty-looking casts that are used in boost::addressof() and are believed to be portable and standard-compilant. Why is having an overloaded operator&() prohibited for classes stored in STL containers while the boost::addressof() workaround exists?

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  • How to change the meaning of pointer access operator

    - by kumar_m_kiran
    Hi All, This may be very obvious question, pardon me if so. I have below code snippet out of my project, #include <stdio.h> class X { public: int i; X() : i(0) {}; }; int main(int argc,char *arv[]) { X *ptr = new X[10]; unsigned index = 5; cout<<ptr[index].i<<endl; return 0; } Question Can I change the meaning of the ptr[index] ? Because I need to return the value of ptr[a[index]] where a is an array for subindexing. I do not want to modify existing source code. Any new function added which can change the behavior is needed. Since the access to index operator is in too many places (536 to be precise) in my code, and has complex formulas inside the index subscript operator, I am not inclined to change the code in many locations. PS : 1. I tried operator overload and came to conclusion that it is not possible. 2. Also p[i] will be transformed into *(p+i). I cannot redefine the basic operator '+'. So just want to reconfirm my understanding and if there are any possible short-cuts to achieve. Else I need fix it by royal method of changing every line of code :) .

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  • Why do I need an intermediate conversion to go from struct to decimal, but not struct to int?

    - by Jesse McGrew
    I have a struct like this, with an explicit conversion to float: struct TwFix32 { public static explicit operator float(TwFix32 x) { ... } } I can convert a TwFix32 to int with a single explicit cast: (int)fix32 But to convert it to decimal, I have to use two casts: (decimal)(float)fix32 There is no implicit conversion from float to either int or decimal. Why does the compiler let me omit the intermediate cast to float when I'm going to int, but not when I'm going to decimal?

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  • How do I write an overload operator where both arguments are interface

    - by Eric Girard
    I'm using interface for most of my stuff. I can't find a way to create an overload operator + that would allow me to perform an addition on any objects implementing the IPoint interface Code interface IPoint { double X { get; set; } double Y { get; set; } } class Point : IPoint { double X { get; set; } double Y { get; set; } //How and where do I create this operator/extension ??? public static IPoint operator + (IPoint a,IPoint b) { return Add(a,b); } public static IPoint Add(IPoint a,IPoint b) { return new Point { X = a.X + b.X, Y = a.Y + b.Y }; } } //Dumb use case : public class Test { IPoint _currentLocation; public Test(IPoint initialLocation) { _currentLocation = intialLocation } public MoveOf(IPoint movement) { _currentLocation = _currentLocation + intialLocation; //Much cleaner/user-friendly than _currentLocation = Point.Add(_currentLocation,intialLocation); } }

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  • Oracle: What does `(+)` do in a WHERE clause?

    - by Jonathan Lonowski
    Found the following in an Oracle-based application that we're migrating (generalized): SELECT Table1.Category1, Table1.Category2, count(*) as Total, count(Tab2.Stat) AS Stat FROM Table1, Table2 WHERE (Table1.PrimaryKey = Table2.ForeignKey(+)) GROUP BY Table1.Category1, Table1.Category2 What does (+) do in a WHERE clause? I've never seen it used like that before.

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  • C++: ptr->hello(); /* VERSUS */ (*ptr).hello();

    - by Joey
    i was learning about c++ pointers... so the "-" operator seemed strange to me... instead of ptr-hello(); one could write (*ptr).hello(); because it also seems to work, so i thought the former is just a more convenient way is that the case or is there any difference?

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  • Php plugin to replace '->' with '.' as the member access operator ? Or even better: alternative synt

    - by Gigi
    Present day usable solution: Note that if you use an ide or an advanced editor, you could make a code template, or record a macro that inserts '->' when you press Ctrl and '.' or something. Netbeans has macros, and I have recorded a macro for this, and I like it a lot :) (just click the red circle toolbar button (start record macro),then type -> into the editor (thats all the macro will do, insert the arrow into the editor), then click the gray square (stop record macro) and assign the 'Ctrl dot' shortcut to it, or whatever shortcut you like) The php plugin: The php plugin, would also have to have a different string concatenation operator than the dot. Maybe a double dot ? Yea... why not. All it has to do is set an activation tag so that it doesnt replace / interpreter '.' as '->' for old scripts and scripts that dont intent do use this. Something like this: <php+ $obj.i = 5 ?> (notice the modified '<?php' tag to '<?php+' ) This way it wouldnt break old code. (and you can just add the '<?php+' code template to your editor and then type 'php tab' (for netbeans) and it would insert '<?php+' ) With the alternative syntax method you could even have old and new syntax cohabitating on the same page like this (I am illustrating this to show the great compatibility of this method, not because you would want to do this): <?php+ $obj.i = 5; ?> <?php $obj->str = 'a' . 'b'; ?> You could change the tag to something more explanatory, in case somebody who doesnt know about the plugin reads the script and thinks its a syntax error <?php-dot.com $obj.i = 5; ?> This is easy because most editors have code templates, so its easy to assign a shortcut to it. And whoever doesnt want the dot replacement, doesnt have to use it. These are NOT ultimate solutions, they are ONLY examples to show that solutions exist, and that arguments against replacing '->' with '.' are only excuses. (Just admit you like the arrow, its ok : ) With this potential method, nobody who doesnt want to use it would have to use it, and it wouldnt break old code. And if other problems (ahem... excuses) arise, they could be fixed too. So who can, and who will do such a thing ?

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  • Operator + for matrices in C++

    - by cibercitizen1
    I suppose the naive implementation of a + operator for matrices (2D for instance) in C++ would be: class Matrix { Matrix operator+ (Matrix other) const { Matrix result; // fill result with *this.data plus other.data return result; } } so we could use it like Matrix a; Matrix b; Matrix c; c = a + b; Right? But if matrices are big this is not efficient as we are doing one not-necessary copy (return result). Therefore, If we wan't to be efficient we have to forget the clean call: c = a + b; Right? What would you suggest / prefer ? Thanks.

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  • Using the AND and NOT Operator in Python

    - by NoahClark
    Here is my custom class that I have that represents a triangle. I'm trying to write code that checks to see if self.a, self.b, and self.c are greater than 0, which would mean that I have Angle, Angle, Angle. Below you will see the code that checks for A and B, however when I use just self.a != 0 then it works fine. I believe I'm not using & correctly. Any ideas? Here is how I am calling it: print myTri.detType() class Triangle: # Angle A To Angle C Connects Side F # Angle C to Angle B Connects Side D # Angle B to Angle A Connects Side E def __init__(self, a, b, c, d, e, f): self.a = a self.b = b self.c = c self.d = d self.e = e self.f = f def detType(self): #Triangle Type AAA if self.a != 0 & self.b != 0: return self.a #If self.a > 10: #return AAA #Triangle Type AAS #elif self.a = 0: #return AAS #Triangle Type ASA #Triangle Type SAS #Triangle Type SSS #else: #return unknown

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  • Calling all the 3 functions while using or operator even after returning true as a result.

    - by Shantanu Gupta
    I am calling three functions in my code where i want to validate some of my fields. When I tries to work with the code given below. It checks only for first value until it gets false result. I want some thing like that if fisrt function returns true then it should also call next function and so on. What can be used instead of Or Operator to do this. if (IsFieldEmpty(ref txtFactoryName, true, "Required") || IsFieldEmpty(ref txtShortName, true, "Required") || IsFieldEmpty(ref cboGodown, true, "Required")) { }

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  • Logical value of an assignment in C

    - by Andy Shulman
    while (curr_data[1] != (unsigned int)NULL && ((curr_ptr = (void*)curr_data[1]) || 1)) Two part question. What will (curr_ptr = (void*)curr_data[1]) evaluate to, logically. TRUE? Also, I know its rather hack-ish, but is the while statement legal C? I would have to go through great contortions to put the assignment elsewhere in the code, so I'd be really nice if I could leave it there, but if it's so egregious that it makes everyone's eyeballs burst into flames, I'll change it.

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