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Search found 1908 results on 77 pages for 'relational operators'.

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  • C++ [] array operator with multiple arguments?

    - by genesys
    Can I define in C++ an array operator that takes multiple arguments? I tried it like this: const T& operator[](const int i, const int j, const int k) const{ return m_cells[k*m_resSqr+j*m_res+i]; } T& operator[](const int i, const int j, const int k){ return m_cells[k*m_resSqr+j*m_res+i]; } But I'm getting this error: error C2804 binary operator '[' has too many parameters

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  • Please explain this php expression "!$variable"

    - by DogBot
    What does an exclamaton mark in front of a variable mean? And how is it being used in this piece of code? EDIT: From the answers so far I suspect that I also should mention that this code is in a function where one of the parameters is $mytype ....would this be a way of checking if $mytype was passed? - Thanks to all of the responders so far. $myclass = null; if ($mytype == null && ($PAGE->pagetype <> 'site-index' && $PAGE->pagetype <>'admin-index')) { return $myclass; } elseif ($mytype == null && ($PAGE->pagetype == 'site-index' || $PAGE->pagetype =='admin-index')) { $myclass = ' active_tree_node'; return $myclass; } elseif (!$mytype == null && ($PAGE->pagetype == 'site-index' || $PAGE->pagetype =='admin-index')) { return $myclass; }`

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  • Simple question about operator ||

    - by Tristan
    HEllo, i try to do that in FlashBuilder (FlexProject) protected function btn_detail_view_clickHandler(event:MouseEvent):void { CurrentState="Statistiques" || "PartMarche"; } But it's not working, i guess this is not the right syntax but what's the right syntax ? Thanks

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  • Why isn't our c# graphics code working any more?

    - by Jared
    Here's the situation: We have some generic graphics code that we use for one of our projects. After doing some clean-up of the code, it seems like something isn't working anymore (The graphics output looks completely wrong). I ran a diff against the last version of the code that gave the correct output, and it looks like we changed one of our functions as follows: static public Rectangle FitRectangleOld(Rectangle rect, Size targetSize) { if (rect.Width <= 0 || rect.Height <= 0) { rect.Width = targetSize.Width; rect.Height = targetSize.Height; } else if (targetSize.Width * rect.Height > rect.Width * targetSize.Height) { rect.Width = rect.Width * targetSize.Height / rect.Height; rect.Height = targetSize.Height; } else { rect.Height = rect.Height * targetSize.Width / rect.Width; rect.Width = targetSize.Width; } return rect; } to static public Rectangle FitRectangle(Rectangle rect, Size targetSize) { if (rect.Width <= 0 || rect.Height <= 0) { rect.Width = targetSize.Width; rect.Height = targetSize.Height; } else if (targetSize.Width * rect.Height > rect.Width * targetSize.Height) { rect.Width *= targetSize.Height / rect.Height; rect.Height = targetSize.Height; } else { rect.Height *= targetSize.Width / rect.Width; rect.Width = targetSize.Width; } return rect; } All of our unit tests are all passing, and nothing in the code has changed except for some syntactic shortcuts. But like I said, the output is wrong. We'll probably just revert back to the old code, but I'm curious if anyone has any idea what's going on here. Thanks.

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  • Using operator+ without leaking memory?

    - by xokmzxoo
    So the code in question is this: const String String::operator+ (const String& rhs) { String tmp; tmp.Set(this->mString); tmp.Append(rhs.mString); return tmp; } This of course places the String on the stack and it gets removed and returns garbage. And placing it on the heap would leak memory. So how should I do this?

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  • while(0=0) evaluates to false

    - by paque
    b=10; while(a=b) { b--; if(b==-10)break; } B goes from 10 to -10. In my world, the while-statement, a=b, should always be true (since the assigment always "goes well"). That is not the case. When the loop stops, b will have a value of 0. In my world, it should pass 0 and go all the way to -10, when the if-statement kicks in. Have I misunderstood something major? (Code tested in IE8 and Adobe Acrobat)

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  • Understanding Incrementing

    - by Chad
    For example this: var a = 123; var b = a++; now a contains 124 and b contains 123 I understand that b is taking the value of a and then a is being incremented. However, I don't understand why this is so. The principal reason for why the creators of JavaScript would want this. Is this really more useful than doing it the PHP way? What is the advantage to this other than confusing newbies?

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  • Binary operator overloading on a templated class (C++)

    - by GRB
    Hi all, I was recently trying to gauge my operator overloading/template abilities and as a small test, created the Container class below. While this code compiles fine and works correctly under MSVC 2008 (displays 11), both MinGW/GCC and Comeau choke on the operator+ overload. As I trust them more than MSVC, I'm trying to figure out what I'm doing wrong. Here is the code: #include <iostream> using namespace std; template <typename T> class Container { friend Container<T> operator+ <> (Container<T>& lhs, Container<T>& rhs); public: void setobj(T ob); T getobj(); private: T obj; }; template <typename T> void Container<T>::setobj(T ob) { obj = ob; } template <typename T> T Container<T>::getobj() { return obj; } template <typename T> Container<T> operator+ <> (Container<T>& lhs, Container<T>& rhs) { Container<T> temp; temp.obj = lhs.obj + rhs.obj; return temp; } int main() { Container<int> a, b; a.setobj(5); b.setobj(6); Container<int> c = a + b; cout << c.getobj() << endl; return 0; } This is the error Comeau gives: Comeau C/C++ 4.3.10.1 (Oct 6 2008 11:28:09) for ONLINE_EVALUATION_BETA2 Copyright 1988-2008 Comeau Computing. All rights reserved. MODE:strict errors C++ C++0x_extensions "ComeauTest.c", line 27: error: an explicit template argument list is not allowed on this declaration Container<T> operator+ <> (Container<T>& lhs, Container<T>& rhs) ^ 1 error detected in the compilation of "ComeauTest.c". I'm having a hard time trying to get Comeau/MingGW to play ball, so that's where I turn to you guys. It's been a long time since my brain has melted this much under the weight of C++ syntax, so I feel a little embarrassed ;). Thanks in advance. EDIT: Eliminated an (irrelevant) lvalue error listed in initial Comeau dump.

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  • Logical xor operator in c++?

    - by RAC
    Is there such a thing? First time I encountered a practical need for it, but I don't see one listed in stroustrup. I intend to write: // Detect when exactly one of A,B is equal to five. return (A==5) ^^ (B==5); But there is no ^^ operator. Can I use bitwise ^ here and get the right answer (regardless of machine representation of true and false)? I never mix & and &&, or | and ||, so I hesitate to do that with ^ and ^^. I'd be more comfortable writing my own "bool XOR(bool,bool)" function instead.

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  • Suppress error with @ operator in PHP

    - by Mez
    In your opinion, is it ever valid to use the @ operator to suppress an error/warning in PHP whereas you may be handling the error? If so, in what circumstances would you use this? Code examples are welcome. Edit: Note to repliers. I'm not looking to turn error reporting off, but, for example, common practice is to use @fopen($file); and then check afterwards... but you can get rid of the @ by doing if (file_exists($file)) { fopen($file); } else { die('File not found'); } or similar. I guess the question is - is there anywhere that @ HAS to be used to supress an error, that CANNOT be handled in any other manner?

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  • Is return an operator or a function?

    - by eSKay
    This is too basic I think, but how do both of these work? return true; // 1 and return (true); // 2 Similar: sizeof, exit My guess: If return was a function, 1 would be erroneous. So, return should be a unary operator that can also take in brackets... pretty much like unary minus: -5 and -(5), both are okay. Is that what it is - a unary operator?

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  • Implicit conversion while using += operator?

    - by bdhar
    Conside the following code: int main() { signed char a = 10; a += a; // Line 5 a = a + a; return 0; } I am getting this warning at Line 5: d:\codes\operator cast\operator cast\test.cpp(5) : warning C4244: '+=' : conversion from 'int' to 'signed char', possible loss of data Does this mean that += operator makes an implicit cast of the right hand operator to int? P.S: I am using Visual studio 2005

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  • difference between -> and . for member selection operator.

    - by TimothyTech
    in this book i have I'm learning pointers, and i just got done with the chapter about OOP (spits on ground) anyways its telling me i can use a member selection operator like this ( - ). it sayd that is is like the "." except points to objects rather than member objects. whats the difference, it looks like it is used the same way...

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  • Does this language feature already exist?

    - by Pindatjuh
    I'm currently developing a new language for programming in a continuous environment (compare it to electrical engineering), and I've got some ideas on a certain language construction. Let me explain the feature by explanation and then by definition: x = a U b; Where x is a variable and a and b are other variables (or static values). This works like a union between a and b; no duplicates and no specific order. with(x) { // regular 'with' usage; using the global interpretation of "x" x = 5; // will replace the original definition of "x = a U b;" } with(x = a) { // this code block is executed when the "x" variable // has the "a" variable assigned. All references in // this code-block to "x" are references to "a". So saying: x = 5; // would only change the variable "a". If the variable "a" // later on changes, x still equals to 5, in this fashion: // 'x = a U b U 5;' // '[currentscope] = 5;' // thus, 'a = 5;' } with(x = b) { // same but with "b" } with(x != a) { // here the "x" variable refers to any variable // but "a"; thus saying x = 5; // is equal to the rewriting of // 'x = a U b U 5;' // 'b = 5;' (since it was the scope of this block) } with(x = (a U b)) { // guaranteed that "x" is 'a U b'; interacting with "x" // will interact with both "a" and "b". x = 5; // makes both "a" and "b" equal to 5; also the "x" variable // is updated to contain: // 'x = a U b U 5;' // '[currentscope] = 5;' // 'a U b = 5;' // and thus: 'a = 5; b = 5;'. } // etc. In the above, all code-blocks are executed, but the "scope" changes in each block how x is interpreted. In the first block, x is guaranteed to be a: thus interacting with x inside that block will interact on a. The second and the third code-block are only equal in this situation (because not a: then there only remains b). The last block guarantees that x is at least a or b. Further more; U is not the "bitwise or operator", but I've called it the "and/or"-operator. Its definition is: "U" = "and" U "or" (On my blog, http://cplang.wordpress.com/2009/12/19/binop-and-or/, there is more (mathematical) background information on this operator. It's loosely based on sets. Using different syntax, changed it in this question.) Update: more examples. print = "Hello world!" U "How are you?"; // this will print // both values, but the // order doesn't matter. // 'userkey' is a variable containing a key. with(userkey = "a") { print = userkey; // will only print "a". } with(userkey = ("shift" U "a")) { // pressed both "shift" and the "a" key. print = userkey; // will "print" shift and "a", even // if the user also pressed "ctrl": // the interpretation of "userkey" is changed, // such that it only contains the matched cases. } with((userkey = "shift") U (userkey = "a")) { // same as if-statement above this one, showing the distributivity. } x = 5 U 6 U 7; y = x + x; // will be: // y = (5 U 6 U 7) + (5 U 6 U 7) // = 10 U 11 U 12 U 13 U 14 somewantedkey = "ctrl" U "alt" U "space" with(userkey = somewantedkey) { // must match all elements of "somewantedkey" // (distributed the Boolean equals operated) // thus only executed when all the defined keys are pressed } with(somewantedkey = userkey) { // matches only one of the provided "somewantedkey" // thus when only "space" is pressed, this block is executed. } Update2: more examples and some more context. with(x = (a U b)) { // this } // can be written as with((x = a) U (x = b)) { // this: changing the variable like x = 5; // will be rewritten as: // a = 5 and b = 5 } Some background information: I'm building a language which is "time-independent", like Java is "platform-independant". Everything stated in the language is "as is", and is continuously actively executed. This means; the programmer does not know in which order (unless explicitly stated using constructions) elements are, nor when statements are executed. The language is completely separated from the "time"-concept, i.e. it's continuously executed: with(a < 5) { a++; } // this is a loop-structure; // how and when it's executed isn't known however. with(a) { // everytime the "a" variable changes, this code-block is executed. b = 4; with(b < 3) { // runs only three times. } with(b > 0) { b = b - 1; // runs four times } } Update 3: After pondering on the type of this language feature; it closely resemblances Netbeans Platform's Lookup, where each "with"-statement a synchronized agent is, working on it's specific "filter" of objects. Instead of type-based, this is variable-based (fundamentally quite the same; just a different way of identifiying objects). I greatly thank all of you for providing me with very insightful information and links/hints to great topics I can research. Thanks. I do not know if this construction already exists, so that's my question: does this language feature already exist?

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  • Operator as and generic classes

    - by abatishchev
    I'm writing .NET On-the-Fly compiler for CLR scripting and want execution method make generic acceptable: object Execute() { return type.InvokeMember(..); } T Execute<T>() { return Execute() as T; /* doesn't work: The type parameter 'T' cannot be used with the 'as' operator because it does not have a class type constraint nor a 'class' constraint */ // also neither typeof(T) not T.GetType(), so on are possible return (T) Execute(); // ok } But I think operator as will be very useful: if result type isn't T method will return null, instead of an exception! Is it possible to do?

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  • STL: how to overload operator= for <vector> ?

    - by MBes
    There's simple example: #include <vector> int main() { vector<int> veci; vector<double> vecd; for(int i = 0;i<10;++i){ veci.push_back(i); vecd.push_back(i); } vecd = veci; // <- THE PROBLEM } The thing I need to know is how to overload operator = so that I could make assignment like this: vector<double> = vector<int>; I've just tried a lot of ways, but always compiler has been returning errors... Is there any option to make this code work without changing it? I can write some additional lines, but can't edit or delete the existing ones. Ty.

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  • PHP alias @ function

    - by SyaZ
    I'm new to PHP and I'm confused seeing some examples calling a function with a @ prefix like @mysql_ping(). What is it for? Googling / searching is not much of a help since @ gets discarded and 'alias' is not good enough keyword.

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  • C++ enforce conditions on inherited classes

    - by user231536
    I would like to define an abstract base class X and enforce the following: a) every concrete class Y that inherits from X define a constructor Y(int x) b) it should be possible to test whether two Y objects are equal. For a, one not very good solution is to put a pure virtual fromInt method in X which concrete class will have to define. But I cannot enforce construction. For b), I cannot seem to use a pure virtual method in X bool operator == (const X& other) const =0; because in overridden classes this remains undefined. It is not enough to define bool operator == (const Y& other) const { //stuff} because the types don't match. How do I solve these problems?

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  • C++ operator[] syntax.

    - by Lanissum
    Just a quick syntax question. I'm writing a map class (for school). If I define the following operator overload: template<typename Key, typename Val> class Map {... Val* operator[](Key k); What happens when a user writes: Map<int,int> myMap; map[10] = 3; Doing something like that will only overwrite a temporary copy of the [null] pointer at Key k. Is it even possible to do: map[10] = 3; printf("%i\n", map[10]); with the same operator overload?

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