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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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  • How operator oveloading works

    - by Rasmi Ranjan Nayak
    I have below code class rectangle { ..... .....//Some code int operator+(rectangle r1) { return(r1.length+length); } }; In main fun. int main() { rectangle r1(10,20); rectangle r2(40,60); rectangle r3(30,60); int len = r1+r3; } Here if we will see in operator+(), we are doing r1.length + length. How the compiler comes to know that the 2nd length in return statement belong to object r3 not to r1 or r2? I think answer may be in main() we have writeen int len = r1+r3; If that is the case then why do we need to write in operator+(....) { r1.lenth + lenth; //Why not length + length? } Why not length + length? Bcause compiler already knows from main() that the first length belong to object r1 and 2nd to object r3.

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  • perl - universal operator overload

    - by Todd Freed
    I have an idea for perl, and I'm trying to figure out the best way to implement it. The idea is to have new versions of every operator which consider the undefined value as the identity of that operation. For example: $a = undef + 5; # undef treated as 0, so $a = 5 $a = undef . "foo"; # undef treated as '', so $a = foo $a = undef && 1; # undef treated as false, $a = true and so forth. ideally, this would be in the language as a pragma, or something. use operators::awesome; However, I would be satisfied if I could implement this special logic myself, and then invoke it where needed: use My::Operators; The problem is that if I say "use overload" inside My::Operators only affects objects blessed into My::Operators. So the question is: is there a way (with "use overoad" or otherwise) to do a "universal operator overload" - which would be called for all operations, not just operations on blessed scalars. If not - who thinks this would be a great idea !? It would save me a TON of this kind of code if($object && $object{value} && $object{value} == 15) replace with if($object{value} == 15) ## the special "is-equal-to" operator

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  • Recursion problem overloading an operator

    - by Tronfi
    I have this: typedef string domanin_name; And then, I try to overload the operator< in this way: bool operator<(const domain_name & left, const domain_name & right){ int pos_label_left = left.find_last_of('.'); int pos_label_right = right.find_last_of('.'); string label_left = left.substr(pos_label_left); string label_right = right.substr(pos_label_right); int last_pos_label_left=0, last_pos_label_right=0; while(pos_label_left!=string::npos && pos_label_right!=string::npos){ if(label_left<label_right) return true; else if(label_left>label_right) return false; else{ last_pos_label_left = pos_label_left; last_pos_label_right = pos_label_right; pos_label_left = left.find_last_of('.', last_pos_label_left); pos_label_right = right.find_last_of('.', last_pos_label_left); label_left = left.substr(pos_label_left, last_pos_label_left); label_right = right.substr(pos_label_right, last_pos_label_right); } } } I know it's a strange way to overload the operator <, but I have to do it this way. It should do what I want. That's not the point. The problem is that it enter in an infinite loop right in this line: if(label_left<label_right) return true; It seems like it's trying to use this overloading function itself to do the comparision, but label_left is a string, not a domain name! Any suggestion?

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  • overload == (and != , of course) operator, can I bypass == to determine whether the object is null

    - by LLS
    Hello, when I try to overload operator == and != in C#, and override Equal as recommended, I found I have no way to distinguish a normal object and null. For example, I defined a class Complex. public static bool operator ==(Complex lhs, Complex rhs) { return lhs.Equals(rhs); } public static bool operator !=(Complex lhs, Complex rhs) { return !lhs.Equals(rhs); } public override bool Equals(object obj) { if (obj is Complex) { return (((Complex)obj).Real == this.Real && ((Complex)obj).Imaginary == this.Imaginary); } else { return false; } } But when I want to use if (temp == null) When temp is really null, some exception happens. And I can't use == to determine whether the lhs is null, which will cause infinite loop. What should I do in this situation. One way I can think of is to us some thing like Class.Equal(object, object) (if it exists) to bypass the == when I do the check. What is the normal way to solve the problem?

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  • Operator Overloading << in c++

    - by thlgood
    I'm a fresh man in C++. I write this simple program to practice Overlaoding. This is my code: #include <iostream> #include <string> using namespace std; class sex_t { private: char __sex__; public: sex_t(char sex_v = 'M'):__sex__(sex_v) { if (sex_v != 'M' && sex_v != 'F') { cerr << "Sex type error!" << sex_v << endl; __sex__ = 'M'; } } const ostream& operator << (const ostream& stream) { if (__sex__ == 'M') cout << "Male"; else cout << "Female"; return stream; } }; int main(int argc, char *argv[]) { sex_t me('M'); cout << me << endl; return 0; } When I compiler it, It print a lots of error message: The error message was in a mess. It's too hard for me to found useful message sex.cpp: ???‘int main(int, char**)’?: sex.cpp:32:10: ??: ‘operator<<’?‘std::cout << me’????? sex.cpp:32:10: ??: ???: /usr/include/c++/4.6/ostream:110:7: ??: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(std::basic_ostre

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  • Use new hosting which is already used buy a domain, for a new domain

    - by Yasser
    I am new to web hosting n stuffs, I am having a domain called www.yassershaikh.com from domainz.in, from where I had taken a LINUX hosting. I am running a wordpress blog there. Now I have taken a another domain from goDaddy called www.codera.org, now here I want to use Windows hosting, but that I am not planning to buy for atleast 6 months. So I was thinking if it's possible to use the hosting(LINUX) which I am using for my first site. Is this possible ? Please guide me on this please. (Also for now I am just redirecting it to one page on my blog.)

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  • re-partition new drive and use new partition as 'home'

    - by vector
    Linux noob here. I installed Ubuntu 12.04 on a brand new drive (dual boot with windows on another drive) and re-partitioned it afterwards (with gparted off of live cd) like so (sudo fdisk -l) : Device Boot Start End Blocks Id System /dev/sdb1 * 2048 63735807 31866880 83 Linux /dev/sdb2 1448509438 1465147391 8318977 5 Extended Partition 2 does not start on physical sector boundary. /dev/sdb3 63735808 1448507391 692385792 83 Linux /dev/sdb5 1448509440 1465147391 8318976 82 Linux swap / Solaris I'd like to use sdb3 as default home for all work and fun related program installs and files, but I haven't even gotten as far as changing permissions on it. Any help will be most appreciated.

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  • Class Assignment Operators

    - by Maxpm
    I made the following operator overloading test: #include <iostream> #include <string> using namespace std; class TestClass { string ClassName; public: TestClass(string Name) { ClassName = Name; cout << ClassName << " constructed." << endl; } ~TestClass() { cout << ClassName << " destructed." << endl; } void operator=(TestClass Other) { cout << ClassName << " in operator=" << endl; cout << "The address of the other class is " << &Other << "." << endl; } }; int main() { TestClass FirstInstance("FirstInstance"); TestClass SecondInstance("SecondInstance"); FirstInstance = SecondInstance; SecondInstance = FirstInstance; return 0; } The assignment operator behaves as-expected, outputting the address of the other class. Now, how would I actually assign something from the other class? For example, something like this: void operator=(TestClass Other) { ClassName = Other.ClassName; }

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  • Why overload true and false instead of defining bool operator?

    - by Joe Enos
    I've been reading about overloading true and false in C#, and I think I understand the basic difference between this and defining a bool operator. The example I see around is something like: public static bool operator true(Foo foo) { return (foo.PropA > 0); } public static bool operator false(Foo foo) { return (foo.PropA <= 0); } To me, this is the same as saying: public static implicit operator bool(Foo foo) { return (foo.PropA > 0); } The difference, as far as I can tell, is that by defining true and false separately, you can have an object that is both true and false, or neither true nor false: public static bool operator true(Foo foo) { return true; } public static bool operator false(Foo foo) { return true; } //or public static bool operator true(Foo foo) { return false; } public static bool operator false(Foo foo) { return false; } I'm sure there's a reason this is allowed, but I just can't think of what it is. To me, if you want an object to be able to be converted to true or false, a single bool operator makes the most sense. Can anyone give me a scenario where it makes sense to do it the other way? Thanks

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  • null-coalescing operator or conditional operator

    - by rkrauter
    Which coding style do you prefer: object o = new object(); //string s1 = o ?? "Tom"; // Cannot implicitly convert type 'object' to 'string' CS0266 string s3 = Convert.ToString(o ?? "Tom"); string s2 = (o != null) ? o.ToString() : "Tom"; s2 or s3? Is it possible to make it shorter? s1 does not obviously work.

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  • C++ operator new, object versions, and the allocation sizes

    - by mizubasho
    Hi. I have a question about different versions of an object, their sizes, and allocation. The platform is Solaris 8 (and higher). Let's say we have programs A, B, and C that all link to a shared library D. Some class is defined in the library D, let's call it 'classD', and assume the size is 100 bytes. Now, we want to add a few members to classD for the next version of program A, without affecting existing binaries B or C. The new size will be, say, 120 bytes. We want program A to use the new definition of classD (120 bytes), while programs B and C continue to use the old definition of classD (100 bytes). A, B, and C all use the operator "new" to create instances of D. The question is, when does the operator "new" know the amount of memory to allocate? Compile time or run time? One thing I am afraid of is, programs B and C expect classD to be and alloate 100 bytes whereas the new shared library D requires 120 bytes for classD, and this inconsistency may cause memory corruption in programs B and C if I link them with the new library D. In other words, the area for extra 20 bytes that the new classD require may be allocated to some other variables by program B and C. Is this assumption correct? Thanks for your help.

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  • operator for enums

    - by Veer
    Hi all, Just out of curiosity, asking this Like the expression one below a = (condition) ? x : y; // two outputs why can't we have an operator for enums? say, myValue = f ??? fnApple() : fnMango() : fnOrange(); // no. of outputs specified in the enum definition instead of switch statements (eventhough refractoring is possible) enum Fruit { apple, mango, orange }; Fruit f = Fruit.apple; Or is it some kind of useless operator?

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  • Operator overloading outside class

    - by bobobobo
    There are two ways to overload operators for a C++ class: Inside class class Vector2 { public: float x, y ; Vector2 operator+( const Vector2 & other ) { Vector2 ans ; ans.x = x + other.x ; ans.y = y + other.y ; return ans ; } } ; Outside class class Vector2 { public: float x, y ; } ; Vector2 operator+( const Vector2& v1, const Vector2& v2 ) { Vector2 ans ; ans.x = v1.x + v2.x ; ans.y = v1.y + v2.y ; return ans ; } (Apparently in C# you can only use the "outside class" method.) In C++, which way is more correct? Which is preferable?

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  • Java operator overload

    - by rengolin
    Coming from C++ to Java, the obvious unanswered question is why not operator overload. On the web some go about: "it's clearly obfuscated and complicate maintenance" but no one really elaborates that further (I completely disagree, actually). Other people pointed out that some objects do have an overload (like String operator +) but that is not extended to other objects nor is extensible to the programmer's decision. I've heard that they're considering extending the favour to BigInt and similar, but why not open that for our decisions? How exactly if complicates maintenance and where on earth does this obfuscate code? Isn't : Complex a, b, c; a = b + c; much simpler than: Complex a, b, c; a.equals( b.add(c) ); ??? Or is it just habit?

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  • null coalescing operator for javascript?

    - by Daniel Schaffer
    I assumed this question has already been asked here, but I couldn't find any, so here it goes: Is there a null coalescing operator in Javascript? For example, in C#, I can do this: String someString = null; var whatIWant = someString ?? "Cookies!"; The best approximation I can figure out for Javascript is using the conditional operator: var someString = null; var whatIWant = someString ? someString : 'Cookies!'; Which is sorta icky IMHO. Can I do better?

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  • Operator overloading C++ outside class

    - by bobobobo
    Well, so there are 2 ways to overload operators for a C++ class INSIDE CLASS class Vector2 { public: float x, y ; Vector2 operator+( const Vector2 & other ) { Vector2 ans ; ans.x = x + other.x ; ans.y = y + other.y ; return ans ; } } ; OUTSIDE CLASS class Vector2 { public: float x, y ; } ; Vector2 operator+( const Vector2& v1, const Vector2& v2 ) { Vector2 ans ; ans.x = v1.x + v2.x ; ans.y = v1.y + v2.y ; return ans ; } In C# apparently you can only use the OUTSIDE class method The question is, in C++, which is "morer-correcter?" Which is preferable? When is one way better than another?

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  • no match for operator= using a std::vector

    - by Max
    I've got a class declared like this: class Level { private: std::vector<mapObject::MapObject> features; (...) }; and in one of its member functions I try to iterate through that vector like this: vector<mapObject::MapObject::iterator it; for(it=features.begin(); it<features.end(); it++) { /* loop code */ } This seems straightforward to me, but g++ gives me this error: src/Level.cpp:402: error: no match for ‘operator=’ in ‘it = ((const yarl::level::Level*)this)-yarl::level::Level::features.std::vector<_Tp, _Alloc::begin [with _Tp = yarl::mapObject::MapObject, _Alloc = std::allocator<yarl::mapObject::MapObject>]()’ /usr/include/c++/4.4/bits/stl_iterator.h:669: note: candidates are: __gnu_cxx::__normal_iterator<yarl::mapObject::MapObject*,std::vector & __gnu_cxx::__normal_iterator<yarl::mapObject::MapObject*,std::vector >::operator=(const __gnu_cxx::__normal_iterator<yarl::mapObject::MapObject*, ``std::vector<yarl::mapObject::MapObject, std::allocator<yarl::mapObject::MapObject> > >&) Anyone know why this is happening?

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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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  • C++ overloading operator comma for variadic arguments

    - by uray
    is it possible to construct variadic arguments for function by overloading operator comma of the argument? i want to see an example how to do so.., maybe something like this: template <typename T> class ArgList { public: ArgList(const T& a); ArgList<T>& operator,(const T& a,const T& b); } //declaration void myFunction(ArgList<int> list); //in use: myFunction(1,2,3,4); //or maybe: myFunction(ArgList<int>(1),2,3,4);

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  • C# String Operator Overloading

    - by ScottSEA
    G'Day Mates - What is the right way (excluding the argument of whether it is advisable) to overload the string operators <, , <= and = ? I've tried it five ways to Sunday and I get various errors - my best shot was declaring a partial class and overloading from there, but it won't work for some reason. namespace System { public partial class String { public static Boolean operator <(String a, String b) { return a.CompareTo(b) < 0; } public static Boolean operator >(String a, String b) { return a.CompareTo(b) > 0; } } }

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  • An operator == whose parameters are non-const references

    - by Eduardo León
    I this post, I've seen this: class MonitorObjectString: public MonitorObject { // some other declarations friend inline bool operator==(/*const*/ MonitorObjectString& lhs, /*const*/ MonitorObjectString& rhs) { return lhs.fVal==rhs.fVal; } } Before we can continue, THIS IS VERY IMPORTANT: I am not questioning anyone's ability to code. I am just wondering why someone would need non-const references in a comparison. The poster of that question did not write that code. This was just in case. This is important too: I added both /*const*/s and reformatted the code. Now, we get back to the topic: I can't think of a sane use of the equality operator that lets you modify its by-ref arguments. Do you?

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  • Dynamic Operator Overloading on dict classes in Python

    - by Ishpeck
    I have a class that dynamically overloads basic arithmetic operators like so... import operator class IshyNum: def __init__(self, n): self.num=n self.buildArith() def arithmetic(self, other, o): return o(self.num, other) def buildArith(self): map(lambda o: setattr(self, "__%s__"%o,lambda f: self.arithmetic(f, getattr(operator, o))), ["add", "sub", "mul", "div"]) if __name__=="__main__": number=IshyNum(5) print number+5 print number/2 print number*3 print number-3 But if I change the class to inherit from the dictionary (class IshyNum(dict):) it doesn't work. I need to explicitly def __add__(self, other) or whatever in order for this to work. Why?

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  • Scala :: operator, how it works?

    - by Felix
    Hello Guys, in Scala, I can make a caseclass case class Foo(x:Int) and then put it in a list like so: List(Foo(42)) Now, nothing strange here. The following is strange to me. The operator :: is a function on a list, right? With any function with 1 argument in Scala, I can call it with infix notation. An example is 1 + 2 is a function (+) on the object Int. The class Foo I just defined does not have the :: operator, so how is the following possible: Foo(40) :: List(Foo(2)) ? In scala 2.8 rc1, I get the following output from the interactive prompt: scala> case class Foo(x:Int) defined class Foo scala> Foo(40) :: List(Foo(2)) res2: List[Foo] = List(Foo(40), Foo(2)) scala> I can go on and use it, but if someone can explain it I will be glad :)

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