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  • == Operator and operands

    - by rahul
    I want to check whether a value is equal to 1. Is there any difference in the following lines of code Evaluated value == 1 1 == evaluated value in terms of the compiler execution

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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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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 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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  • 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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  • 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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  • 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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