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  • problem with QDataStream & QDataStream::operator>> ( char *& s )

    - by yan bellavance
    QFile msnLogFile(item->data(Qt::UserRole).toString()); QDataStream logDataStream; if(msnLogFile.exists()){ msnLogFile.open(QIODevice::ReadOnly); logDataStream.setDevice(&msnLogFile); QByteArray logBlock; logDataStream >> logBlock; } This code doesnt work. The QByte that results is empty. Same thing if I use a char* . Oddely enough the same code works in another program. Im tying to find the difference between both. This works if i use int,uint, quint8, etc

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  • C++ overide global operator comma gives error

    - by uray
    the second function gives error C2803 http://msdn.microsoft.com/en-us/library/zy7kx46x%28VS.80%29.aspx : 'operator ,' must have at least one formal parameter of class type. any clue? template<class T,class A = std::allocator<T>> class Sequence : public std::vector<T,A> { public: Sequence<T,A>& operator,(const T& a) { this->push_back(a); return *this; } Sequence<T,A>& operator,(const Sequence<T,A>& a) { for(Sequence<T,A>::size_type i=0 ; i<a.size() ; i++) { this->push_back(a.at(i)); } return *this; } }; //this works! template<typename T> Sequence<T> operator,(const T& a, const T&b) { Sequence<T> seq; seq.push_back(a); seq.push_back(b); return seq; } //this gives error C2803! Sequence<double> operator,(const double& a, const double& b) { Sequence<double> seq; seq.push_back(a); seq.push_back(b); return seq; }

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  • Is there an exponent operator in C#?

    - by Charlie
    For example, does an operator exist to handle this? float Result, Number1, Number2; Number1 = 2; Number2 = 2; Result = Number1 (operator) Number2; In the past the ^ operator has served as an exponential operator in other languages, but in C# it is a bit-wise operator. Do I have to write a loop or include another namespace to handle exponential operations? If so, how do I handle exponential operations using non-integers?

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  • Why is Decimal('0') > 9999.0 True in Python?

    - by parxier
    This is somehow related to my question Why is ''0 True in Python? In Python 2.6.4: >> Decimal('0') > 9999.0 True From the answer to my original question I understand that when comparing objects of different types in Python 2.x the types are ordered by their name. But in this case: >> type(Decimal('0')).__name__ > type(9999.0).__name__ False Why is Decimal('0') > 9999.0 == True then? UPDATE: I usually work on Ubuntu (Linux 2.6.31-20-generic #57-Ubuntu SMP Mon Feb 8 09:05:19 UTC 2010 i686 GNU/Linux, Python 2.6.4 (r264:75706, Dec 7 2009, 18:45:15) [GCC 4.4.1] on linux2). On Windows (WinXP Professional SP3, Python 2.6.4 (r264:75706, Nov 3 2009, 13:23:17) [MSC v.1500 32 bit (Intel)] on win32) my original statement works differently: >> Decimal('0') > 9999.0 False I even more puzzled now. %-(

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  • undefined C/C++ symbol as operator

    - by uray
    I notice that the character/symbol '`' and '@' is not used as an operator in C/C++, does anyone know the reason or historically why its so? if its really not used, is it safe to define those symbols as another operator/statement using #define?

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  • Performance difference in for loop condition?

    - by CSharperWithJava
    Hello all, I have a simple question that I am posing mostly for my curiousity. What are the differences between these two lines of code? (in C++) for(int i = 0; i < N, N > 0; i++) for(int i = 0; i < N && N > 0; i++) The selection of the conditions is completely arbitrary, I'm just interested in the differences between , and &&. I'm not a beginner to coding by any means, but I've never bothered with the comma operator. Are there performance/behavior differences or is it purely aesthetic? One last note, I know there are bigger performance fish to fry than a conditional operator, but I'm just curious. Indulge me. Edit Thanks for your answers. It turns out the code that prompted this question had misused the comma operator in the way I've described. I wondered what the difference was and why it wasn't a && operator, but it was just written incorrectly. I didn't think anything was wrong with it because it worked just fine. Thanks for straightening me out.

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  • What is the safest way to subtract two System.Runtime.InteropServices.ComTypes.FILETIME objects

    - by Anindya Chatterjee
    I wonder what is the safest way to subtract two System.Runtime.InteropServices.ComTypes.FILETIME objects? I used the following code but sometimes it gives me ArithmaticOverflow exception due to the negative number in Low 32-bit values. I am not sure enclosing the body with unchecked will serve the purpose or not. Please give me some suggestion on how to do it safely without getting any runtime exception or CS0675 warning message. private static UInt64 SubtractTimes(FILETIME a, FILETIME b) { UInt64 aInt = ((UInt64)(a.dwHighDateTime << 32)) | (UInt32)a.dwLowDateTime; UInt64 bInt = ((UInt64)(b.dwHighDateTime << 32)) | (UInt32)b.dwLowDateTime; return aInt - bInt; }

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  • using 'new' operator

    - by notLikeCpp
    I have simple task concerning 'new' operator. I need to create array of 10 chars and then input those chars using 'cin'. Should it look like this ? : char c = new char[10]; for(int i=0; i < 10; i++) { cin >> char[i] >> endl; }

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  • Providing less than operator for one element of a pair

    - by Koszalek Opalek
    What would be the most elegant way too fix the following code: #include <vector> #include <map> #include <set> using namespace std; typedef map< int, int > row_t; typedef vector< row_t > board_t; typedef row_t::iterator area_t; bool operator< ( area_t const& a, area_t const& b ) { return( a->first < b->first ); }; int main( int argc, char* argv[] ) { int row_num; area_t it; set< pair< int, area_t > > queue; queue.insert( make_pair( row_num, it ) ); // does not compile }; One way to fix it is moving the definition of less< to namespace std (I know, you are not supposed to do it.) namespace std { bool operator< ( area_t const& a, area_t const& b ) { return( a->first < b->first ); }; }; Another obvious solution is defining less than< for pair< int, area_t but I'd like to avoid that and be able to define the operator only for the one element of the pair where it is not defined.

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  • R: How to pass a list of selection expressions (strings in this case) to the subset function?

    - by John
    Here is some example data: data = data.frame(series = c("1a", "1b", "1e"), reading = c(0.1, 0.4, 0.6)) > data series reading 1 1a 0.1 2 1b 0.4 3 1e 0.6 Which I can pull out selective single rows using subset: > subset (data, series == "1a") series reading 1 1a 0.1 And pull out multiple rows using a logical OR > subset (data, series == "1a" | series == "1e") series reading 1 1a 0.1 3 1e 0.6 But if I have a long list of series expressions, this gets really annoying to input, so I'd prefer to define them in a better way, something like this: series_you_want = c("1a", "1e") (although even this sucks a little) and be able to do something like this, subset (data, series == series_you_want) The above obviously fails, I'm just not sure what the best way to do this is?

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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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  • 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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  • How does the bitwise operator '^' work?

    - by SpawnCxy
    I'm a little confused when I see the output of following code: $x = "a"; $y = "b"; $x ^= $y; $y ^= $x; $x ^= $y; echo $x; //got b echo $y; //got a And I wonder how does the operator ^ work here?Explanations with clarity would be greatly appreciated!

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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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  • 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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  • 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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  • 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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