Search Results

Search found 2995 results on 120 pages for 'logical operators'.

Page 19/120 | < Previous Page | 15 16 17 18 19 20 21 22 23 24 25 26  | Next Page >

  • 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

    Read the article

  • 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.

    Read the article

  • 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?

    Read the article

  • (C++) What's the difference between these overloaded operator functions?

    - by cv3000
    What is the difference between these two ways of overloading the != operator below. Which is consider better? Class Test { ...// private: int iTest public: BOOL operator==(const &Test test) const; BOOL operator!=(const &Test test) const; } BOOL operator==(const &Test test) const { return (iTest == test.iTest); } //overload function 1 BOOL Test::operator!=(const &Test test) const { return !operator==(test); } //overload function 2 BOOL Test::operator!=(const &Test test) const { return (iTest != test.iTest); } I've just recently seen function 1's syntax for calling a sibling operator function and wonder if writing it that way provides any benefits.

    Read the article

  • Is there any way to maximize PHP_INT_MAX?

    - by Tom
    I can't find this row in php.ini, so is there any way to do that? So ok, I understood, the answer is not, then how should I calculate the result of this task if I want it do with php? The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ?

    Read the article

  • What does the C# operator => mean?

    - by Mr. Mark
    Answers to a recent post (Any chances to imitate times() Ruby method in C#?) use the = operator in the usage examples. What does this operator do? I can't locate it in my C# book, and it is hard to search for symbols like this online. (I couldn't find it.)

    Read the article

  • C# XOR on two byte variables will not compile without a cast

    - by Ash
    Why does the following raise a compile time error: 'Cannot implicitly convert type 'int' to 'byte': byte a = 25; byte b = 60; byte c = a ^ b; This would make sense if I were using an arithmentic operator because the result of a + b could be larger than can be stored in a single byte. However applying this to the XOR operator is pointless. XOR here it a bitwise operation that can never overflow a byte. using a cast around both operands works: byte c = (byte)(a ^ b);

    Read the article

  • C++ vector<T>::iterator operator +

    - by Tom
    Hi, Im holding an iterator that points to an element of a vector, and I would like to compare it to the next element of the vector. Here is what I have Class Point{ public: float x,y; } //Somewhere in my code I do this vector<Point> points = line.getPoints(); foo (points.begin(),points.end()); where foo is: void foo (Vector<Point>::iterator begin,Vector<Point>::iterator end) { std::Vector<Point>::iterator current = begin; for(;current!=end-1;++current) { std::Vector<Point>::iterator next = current + 1; //Compare between current and next. } } I thought that this would work, but current + 1 is not giving me the next element of the vector. I though operator+ was the way to go, but doesnt seem so. Is there a workaround on this? THanks

    Read the article

  • 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.

    Read the article

  • 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.

    Read the article

  • What is operator<< <> in C++?

    - by Austin Hyde
    I have seen this in a few places, and to confirm I wasn't crazy, I looked for other examples. Apparently this can come in other flavors as well, eg operator+ <>. However, nothing I have seen anywhere mentions what it is, so I thought I'd ask. It's not the easiest thing to google operator<< <>( :-)

    Read the article

  • 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; }`

    Read the article

  • 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

    Read the article

  • 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?

    Read the article

  • 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?

    Read the article

  • 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?

    Read the article

  • 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

    Read the article

  • 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)

    Read the article

  • 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?

    Read the article

  • 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?

    Read the article

  • 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

    Read the article

< Previous Page | 15 16 17 18 19 20 21 22 23 24 25 26  | Next Page >