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  • Allow member to be const while still supporting operator= on the class

    - by LeopardSkinPillBoxHat
    I have several members in my class which are const and can therefore only be initialised via the initialiser list like so: class MyItemT { public: MyItemT(const MyPacketT& aMyPacket, const MyInfoT& aMyInfo) : mMyPacket(aMyPacket), mMyInfo(aMyInfo) { } private: const MyPacketT mMyPacket; const MyInfoT mMyInfo; }; My class can be used in some of our internally defined container classes (e.g. vectors), and these containers require that operator= is defined in the class. Of course, my operator= needs to do something like this: MyItemT& MyItemT::operator=(const MyItemT& other) { mMyPacket = other.mPacket; mMyInfo = other.mMyInfo; return *this; } which of course doesn't work because mMyPacket and mMyInfo are const members. Other than making these members non-const (which I don't want to do), any ideas about how I could fix this?

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  • F# passing an operator with arguments to a function

    - by dan
    Can you pass in an operation like "divide by 2" or "subtract 1" using just a partially applied operator, where "add 1" looks like this: List.map ((+) 1) [1..5];; //equals [2..6] // instead of having to write: List.map (fun x-> x+1) [1..5] What's happening is 1 is being applied to (+) as it's first argument, and the list item is being applied as the second argument. For addition and multiplication, this argument ordering doesn't matter. Suppose I want to subtract 1 from every element (this will probably be a common beginners mistake): List.map ((-) 1) [1..5];; //equals [0 .. -4], the opposite of what we wanted 1 is applied to the (-) as its first argument, so instead of (list_item - 1), I get (1 - list_item). I can rewrite it as adding negative one instead of subtracting positive one: List.map ((+) -1) [1..5];; List.map (fun x -> x-1) [1..5];; // this works too I'm looking for a more expressive way to write it, something like ((-) _ 1), where _ denotes a placeholder, like in the Arc language. This would cause 1 to be the second argument to -, so in List.map, it would evaluate to list_item - 1. So if you wanted to map divide by 2 to the list, you could write: List.map ((/) _ 2) [2;4;6] //not real syntax, but would equal [1;2;3] List.map (fun x -> x/2) [2;4;6] //real syntax equivalent of the above Can this be done or do I have to use (fun x -> x/2)? It seems that the closest we can get to the placeholder syntax is to use a lambda with a named argument.

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  • What does !! (double exclamation point) mean?

    - by molecules
    In the code below, from a blog post by Alias, I noticed the use of the double exclamation point !!. I was wondering what it meant and where I could go in the future to find explanations for Perl syntax like this. (Yes, I already searched for '!!' at perlsyn). package Foo; use vars qw{$DEBUG}; BEGIN { $DEBUG = 0 unless defined $DEBUG; } use constant DEBUG => !! $DEBUG; sub foo { debug('In sub foo') if DEBUG; ... } UPDATE Thanks for all of your answers. Here is something else I just found that is related The List Squash Operator x!!

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  • C#: Shift left assignment operator behavior

    - by Austin Salonen
    I'm running code that sometimes yields this: UInt32 current; int left, right; ... //sometimes left == right and no shift occurs current <<= (32 + left - right); //this works current <<= (32 - right); current <<= left; It appears for any value = 32, only the value % 32 is shifted. Is there some "optimization" occurring in the framework?

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  • What does the operator "<<" mean in C#?

    - by Kurru
    I was doing some basic audio programming in C# using the NAudio package and I came across the following expression and I have no idea what it means, as i've never seen the << operator being used before. So what does << mean? Please give a quick explaination of this expression. short sample = (short)((buffer[index + 1] << 8) | buffer[index + 0]);

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  • assign operator to variable in python?

    - by abhilashm86
    Usual method of applying mathematics to variables is a * b Is it able to calculate and manipulate two operands like this? a = input('enter a value') b = input('enter a value') op = raw_input('enter a operand') then how do i connect op and two variables a and b?? i know i can compare op to +, -, %, $ and then assign and compute.... but can i do something like a op b , how to tell compiler that op is an operator?? any tweaks possible?

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  • sOperator 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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  • Dynamically allocated structure and casting.

    - by Simone Margaritelli
    Let's say I have a first structure like this: typedef struct { int ivalue; char cvalue; } Foo; And a second one: typedef struct { int ivalue; char cvalue; unsigned char some_data_block[0xFF]; } Bar; Now let's say I do the following: Foo *pfoo; Bar *pbar; pbar = new Bar; pfoo = (Foo *)pbar; delete pfoo; Now, when I call the delete operator, how much memory does it free? sizeof(int) + sizeof(char) Or sizeof(int) + sizeof(char) + sizeof(char) * 0xFF ? And if it's the first case due to the casting, is there any way to prevent this memory leak from happening? Note: please don't answer "use C++ polymorphism" or similar, I am using this method for a reason.

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  • Is Perl's flip-flop operator bugged? It has global state, how can I reset it?

    - by Evan Carroll
    I'm dismayed. Ok, so this was probably the most fun perl bug I've ever found. Even today I'm learning new stuff about perl. Essentially, the flip-flop operator .. which returns false until the left-hand-side returns true, and then true until the right-hand-side returns false keep global state (or that is what I assume.) My question is can I reset it, (perhaps this would be a good addition to perl4-esque hardly ever used reset())? Or, is there no way to use this operator safely? I also don't see this (the global context bit) documented anywhere in perldoc perlop is this a mistake? Code use feature ':5.10'; use strict; use warnings; sub search { my $arr = shift; grep { !( /start/ .. /never_exist/ ) } @$arr; } my @foo = qw/foo bar start baz end quz quz/; my @bar = qw/foo bar start baz end quz quz/; say 'first shot - foo'; say for search \@foo; say 'second shot - bar'; say for search \@bar; Spoiler $ perl test.pl first shot foo bar second shot

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  • SQL with Regular Expressions vs Indexes with Logical Merging Functions

    - by geeko
    Hello Lads, I am trying to develop a complex textual search engine. I have thousands of textual pages from many books. I need to search pages that contain specified complex logical criterias. These criterias can contain virtually any compination of the following: A: Full words. B: Word roots (semilar to stems; i.e. all words with certain key letters). C: Word templates (in some languages are filled in certain templates to form various part of speech such as adjactives, past/present verbs...). D: Logical connectives: AND/OR/XOR/NOT/IF/IFF and parentheses to state priorities. Now, would it be faster to have the pages' full text in database (not indexed) and search though them all using SQL and Regular Expressions ? Or would it be better to construct indexes of word/root/template-page-location tuples. Hence, we can boost searching for individual words/roots/templates. However, it gets tricky as we interdouce logical connectives into our query. I thought of doing the following steps in such cases: 1: Seperately search for each individual words/roots/templates in the specified query. 2: On priority bases, we merge two result lists (from step 1) at a time depedning on the logical connective For example, if we are searching for "he AND (is OR was)": 1: We shall search for "he", "is" and "was" seperately and get result lists for each word. 2: Merge the result lists of "is" and "was" using the merging function OR-MERGE 3: Merge the merged result list from the OR-MERGE function with the one of "he" using the merging function AND-MERGE The result of step 3 is then returned as the result of the specified query. What do you think gurues ? Which is faster ? Any better ideas ? Thank you all in advance.

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  • What does the symbol :=: mean

    - by Dan Maguire
    I've found the symbol :=: in some Clarion code and I can't seem to figure out exactly what it does. The code was written by a previous developer many years ago, so I can't ask him. I also have not been able to find any results for "colon equals colon" in Google. Here is an example of the code, where bufSlcdpaDtl is a file object: lCCRecord Like(bufSlcdpaDtl),Pre(lCCRecord) ! ...other stuff... lCCRecord :=: bufSlcdpaDtl I'm wondering if it's something similar to ::= in Python or possibly the assignment operator :=.

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  • Operator Overloading in C

    - by Leif Andersen
    In C++, I can change the operator on a specific class by doing something like this: MyClass::operator==/*Or some other operator such as =, >, etc.*/(Const MyClass rhs) { /* Do Stuff*/; } But with there being no classes (built in by default) in C. So, how could I do operator overloading for just general functions? For example, if I remember correctly, importing stdlib.h gives you the - operator, which is just syntactic sugar for (*strcut_name).struct_element. So how can I do this in C? Thank you.

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  • Problem with operator ==

    - by CPPDev
    I am facing some problem with use of operator == in the following c++ program. #include < iostream> using namespace std; class A { public: A(char *b) { a = b; } A(A &c) { a = c.a; } bool operator ==(A &other) { return strcmp(a, other.a); } private: char *a; }; int main() { A obj("test"); A obj1("test1"); if(obj1 == A("test1")) { cout<<"This is true"<<endl; } } What's wrong with if(obj1 == A("test1")) line ?? Any help is appreciated.

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  • Why does virtual assignment behave differently than other virtual functions of the same signature?

    - by David Rodríguez - dribeas
    While playing with implementing a virtual assignment operator I have ended with a funny behavior. It is not a compiler glitch, since g++ 4.1, 4.3 and VS 2005 share the same behavior. Basically, the virtual operator= behaves differently than any other virtual function with respect to the code that is actually being executed. struct Base { virtual Base& f( Base const & ) { std::cout << "Base::f(Base const &)" << std::endl; return *this; } virtual Base& operator=( Base const & ) { std::cout << "Base::operator=(Base const &)" << std::endl; return *this; } }; struct Derived : public Base { virtual Base& f( Base const & ) { std::cout << "Derived::f(Base const &)" << std::endl; return *this; } virtual Base& operator=( Base const & ) { std::cout << "Derived::operator=( Base const & )" << std::endl; return *this; } }; int main() { Derived a, b; a.f( b ); // [0] outputs: Derived::f(Base const &) (expected result) a = b; // [1] outputs: Base::operator=(Base const &) Base & ba = a; Base & bb = b; ba = bb; // [2] outputs: Derived::operator=(Base const &) Derived & da = a; Derived & db = b; da = db; // [3] outputs: Base::operator=(Base const &) ba = da; // [4] outputs: Derived::operator=(Base const &) da = ba; // [5] outputs: Derived::operator=(Base const &) } The effect is that the virtual operator= has a different behavior than any other virtual function with the same signature ([0] compared to [1]), by calling the Base version of the operator when called through real Derived objects ([1]) or Derived references ([3]) while it does perform as a regular virtual function when called through Base references ([2]), or when either the lvalue or rvalue are Base references and the other a Derived reference ([4],[5]). Is there any sensible explanation to this odd behavior?

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