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  • Why do not C++11's move constructor/assignment operator act as expected

    - by xmllmx
    #include <iostream> using namespace std; struct A { A() { cout << "A()" << endl; } ~A() { cout << "~A()" << endl; } A(A&&) { cout << "A(A&&)" << endl; } A& operator =(A&&) { cout << "A& operator =(A&&)" << endl; return *this; } }; struct B { // According to the C++11, the move ctor/assignment operator // should be implicitly declared and defined. The move ctor // /assignment operator should implicitly call class A's move // ctor/assignment operator to move member a. A a; }; B f() { B b; // The compiler knows b is a temporary object, so implicitly // defined move ctor/assignment operator of class B should be // called here. Which will cause A's move ctor is called. return b; } int main() { f(); return 0; } My expected output should be: A() A(A&&) ~A() ~A() However, the actual output is: (The C++ compiler is: Visual Studio 2012) A() ~A() ~A() Is this a bug of VC++? or just my misunderstanding?

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  • C++ operator lookup rules / Koenig lookup

    - by John Bartholomew
    While writing a test suite, I needed to provide an implementation of operator<<(std::ostream&... for Boost unit test to use. This worked: namespace theseus { namespace core { std::ostream& operator<<(std::ostream& ss, const PixelRGB& p) { return (ss << "PixelRGB(" << (int)p.r << "," << (int)p.g << "," << (int)p.b << ")"); } }} This didn't: std::ostream& operator<<(std::ostream& ss, const theseus::core::PixelRGB& p) { return (ss << "PixelRGB(" << (int)p.r << "," << (int)p.g << "," << (int)p.b << ")"); } Apparently, the second wasn't included in the candidate matches when g++ tried to resolve the use of the operator. Why (what rule causes this)? The code calling operator<< is deep within the Boost unit test framework, but here's the test code: BOOST_AUTO_TEST_SUITE(core_image) BOOST_AUTO_TEST_CASE(test_output) { using namespace theseus::core; BOOST_TEST_MESSAGE(PixelRGB(5,5,5)); // only compiles with operator<< definition inside theseus::core std::cout << PixelRGB(5,5,5) << "\n"; // works with either definition BOOST_CHECK(true); // prevent no-assertion error } BOOST_AUTO_TEST_SUITE_END() For reference, I'm using g++ 4.4 (though for the moment I'm assuming this behaviour is standards-conformant).

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  • On C++ global operator new: why it can be replaced

    - by Jimmy
    I wrote a small program in VS2005 to test whether C++ global operator new can be overloaded. It can. #include "stdafx.h" #include "iostream" #include "iomanip" #include "string" #include "new" using namespace std; class C { public: C() { cout<<"CTOR"<<endl; } }; void * operator new(size_t size) { cout<<"my overload of global plain old new"<<endl; // try to allocate size bytes void *p = malloc(size); return (p); } int main() { C* pc1 = new C; cin.get(); return 0; } In the above, my definition of operator new is called. If I remove that function from the code, then operator new in C:\Program Files (x86)\Microsoft Visual Studio 8\VC\crt\src\new.cpp gets called. All is good. However, in my opinion, my implementations of operator new does NOT overload the new in new.cpp, it CONFLICTS with it and violates the one-definition rule. Why doesn't the compiler complain about it? Or does the standard say since operator new is so special, one-definition rule does not apply here? Thanks.

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  • Operator+ for a subtype of a template class.

    - by baol
    I have a template class that defines a subtype. I'm trying to define the binary operator+ as a template function, but the compiler cannot resolve the template version of the operator+. #include <iostream> template<typename other_type> struct c { c(other_type v) : cs(v) {} struct subtype { subtype(other_type v) : val(v) {} other_type val; } cs; }; template<typename other_type> typename c<other_type>::subtype operator+(const typename c<other_type>::subtype& left, const typename c<other_type>::subtype& right) { return typename c<other_type>::subtype(left.val + right.val); } // This one works // c<int>::subtype operator+(const c<int>::subtype& left, // const c<int>::subtype& right) // { return c<int>::subtype(left.val + right.val); } int main() { c<int> c1 = 1; c<int> c2 = 2; c<int>::subtype cs3 = c1.cs + c2.cs; std::cerr << cs3.val << std::endl; } I think the reason is because the compiler (g++4.3) cannot guess the template type so it's searching for operator+<int> instead of operator+. What's the reason for that? What elegant solution can you suggest?

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  • Operator+ for a subtype of a template classe.

    - by baol
    I have a template class that defines a subtype. I'm trying to define the binary operator+ as a template function, but the compiler cannot resolve the template version of the operator+. #include <iostream> template<typename other_type> struct c { c(other_type v) : cs(v) {} struct subtype { subtype(other_type v) : val(v) {} other_type val; } cs; }; template<typename other_type> typename c<other_type>::subtype operator+(const typename c<other_type>::subtype& left, const typename c<other_type>::subtype& right) { return typename c<other_type>::subtype(left.val + right.val); } // This one works // c<a>::subtype operator+(const c<a>::subtype& left, // const c<a>::subtype& right) // { return c<a>::subtype(left.val + right.val); } int main() { c<int> c1 = 1; c<int> c2 = 2; c<int>::subtype cs3 = c1.cs + c2.cs; std::cerr << cs3.val << std::endl; } I think the reason is because the compiler (g++4.3) cannot guess the template type so it's searching for operator+<int> instead of operator+. What's the reason for that? What elegant solution can you suggest?

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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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  • Calling all the 3 functions while using or operator even after returning true as a result.

    - by Shantanu Gupta
    I am calling three functions in my code where i want to validate some of my fields. When I tries to work with the code given below. It checks only for first value until it gets false result. I want some thing like that if fisrt function returns true then it should also call next function and so on. What can be used instead of Or Operator to do this. if (IsFieldEmpty(ref txtFactoryName, true, "Required") || IsFieldEmpty(ref txtShortName, true, "Required") || IsFieldEmpty(ref cboGodown, true, "Required")) { }

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  • Custom string class (C++)

    - by Sanctus2099
    Hey guys. I'm trying to write my own C++ String class for educational and need purposes. The first thing is that I don't know that much about operators and that's why I want to learn them. I started writing my class but when I run it it blocks the program but does not do any crash. Take a look at the following code please before reading further: class CString { private: char* cstr; public: CString(); CString(char* str); CString(CString& str); ~CString(); operator char*(); operator const char*(); CString operator+(const CString& q)const; CString operator=(const CString& q); }; First of all I'm not so sure I declared everything right. I tried googleing about it but all the tutorials about overloading explain the basic ideea which is very simple but lack to explain how and when each thing is called. For instance in my = operator the program calls CString(CString& str); but I have no ideea why. I have also attached the cpp file below: CString::CString() { cstr=0; } CString::CString(char *str) { cstr=new char[strlen(str)]; strcpy(cstr,str); } CString::CString(CString& q) { if(this==&q) return; cstr = new char[strlen(q.cstr)+1]; strcpy(cstr,q.cstr); } CString::~CString() { if(cstr) delete[] cstr; } CString::operator char*() { return cstr; } CString::operator const char* () { return cstr; } CString CString::operator +(const CString &q) const { CString s; s.cstr = new char[strlen(cstr)+strlen(q.cstr)+1]; strcpy(s.cstr,cstr); strcat(s.cstr,q.cstr); return s; } CString CString::operator =(const CString &q) { if(this!=&q) { if(cstr) delete[] cstr; cstr = new char[strlen(q.cstr)+1]; strcpy(cstr,q.cstr); } return *this; } For testing I used a code just as simple as this CString a = CString("Hello") + CString(" World"); printf(a); I tried debugging it but at a point I get lost. First it calls the constructor 2 times for "hello" and for " world". Then it get's in the + operator which is fine. Then it calls the constructor for the empty string. After that it get's into "CString(CString& str)" and now I'm lost. Why is this happening? After this I noticed my string containing "Hello World" is in the destructor (a few times in a row). Again I'm very puzzeled. After converting again from char* to Cstring and back and forth it stops. It never get's into the = operator but neither does it go further. printf(a) is never reached. I use VisualStudio 2010 for this but it's basically just standard c++ code and thus I don't think it should make that much of a difference

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  • Why isn't my operator overloading working properly?

    - by Mithrax
    I have the following Polynomial class I'm working on: #include <iostream> using namespace std; class Polynomial { //define private member functions private: int coef[100]; // array of coefficients // coef[0] would hold all coefficients of x^0 // coef[1] would hold all x^1 // coef[n] = x^n ... int deg; // degree of polynomial (0 for the zero polynomial) //define public member functions public: Polynomial::Polynomial() //default constructor { for ( int i = 0; i < 100; i++ ) { coef[i] = 0; } } void set ( int a , int b ) //setter function { //coef = new Polynomial[b+1]; coef[b] = a; deg = degree(); } int degree() { int d = 0; for ( int i = 0; i < 100; i++ ) if ( coef[i] != 0 ) d = i; return d; } void print() { for ( int i = 99; i >= 0; i-- ) { if ( coef[i] != 0 ) { cout << coef[i] << "x^" << i << " "; } } } // use Horner's method to compute and return the polynomial evaluated at x int evaluate ( int x ) { int p = 0; for ( int i = deg; i >= 0; i-- ) p = coef[i] + ( x * p ); return p; } // differentiate this polynomial and return it Polynomial differentiate() { if ( deg == 0 ) { Polynomial t; t.set ( 0, 0 ); return t; } Polynomial deriv;// = new Polynomial ( 0, deg - 1 ); deriv.deg = deg - 1; for ( int i = 0; i < deg; i++ ) deriv.coef[i] = ( i + 1 ) * coef[i + 1]; return deriv; } Polynomial Polynomial::operator + ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) c.coef[i] += a.coef[i]; for ( int i = 0; i <= b.deg; i++ ) c.coef[i] += b.coef[i]; c.deg = c.degree(); return c; } Polynomial Polynomial::operator += ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) c.coef[i] += a.coef[i]; for ( int i = 0; i <= b.deg; i++ ) c.coef[i] += b.coef[i]; c.deg = c.degree(); for ( int i = 0; i < 100; i++) a.coef[i] = c.coef[i]; a.deg = a.degree(); return a; } Polynomial Polynomial::operator -= ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) c.coef[i] += a.coef[i]; for ( int i = 0; i <= b.deg; i++ ) c.coef[i] -= b.coef[i]; c.deg = c.degree(); for ( int i = 0; i < 100; i++) a.coef[i] = c.coef[i]; a.deg = a.degree(); return a; } Polynomial Polynomial::operator *= ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) for ( int j = 0; j <= b.deg; j++ ) c.coef[i+j] += ( a.coef[i] * b.coef[j] ); c.deg = c.degree(); for ( int i = 0; i < 100; i++) a.coef[i] = c.coef[i]; a.deg = a.degree(); return a; } Polynomial Polynomial::operator - ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) c.coef[i] += a.coef[i]; for ( int i = 0; i <= b.deg; i++ ) c.coef[i] -= b.coef[i]; c.deg = c.degree(); return c; } Polynomial Polynomial::operator * ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) for ( int j = 0; j <= b.deg; j++ ) c.coef[i+j] += ( a.coef[i] * b.coef[j] ); c.deg = c.degree(); return c; } }; int main() { Polynomial a, b, c, d; a.set ( 7, 4 ); //7x^4 a.set ( 1, 2 ); //x^2 b.set ( 6, 3 ); //6x^3 b.set ( -3, 2 ); //-3x^2 c = a - b; // (7x^4 + x^2) - (6x^3 - 3x^2) a -= b; c.print(); cout << "\n"; a.print(); cout << "\n"; c = a * b; // (7x^4 + x^2) * (6x^3 - 3x^2) c.print(); cout << "\n"; d = c.differentiate().differentiate(); d.print(); cout << "\n"; cout << c.evaluate ( 2 ); //substitue x with 2 cin.get(); } Now, I have the "-" operator overloaded and it works fine: Polynomial Polynomial::operator - ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) c.coef[i] += a.coef[i]; for ( int i = 0; i <= b.deg; i++ ) c.coef[i] -= b.coef[i]; c.deg = c.degree(); return c; } However, I'm having difficulty with my "-=" operator: Polynomial Polynomial::operator -= ( Polynomial b ) { Polynomial a = *this; //a is the poly on the L.H.S Polynomial c; for ( int i = 0; i <= a.deg; i++ ) c.coef[i] += a.coef[i]; for ( int i = 0; i <= b.deg; i++ ) c.coef[i] -= b.coef[i]; c.deg = c.degree(); // overwrite value of 'a' with the newly computed 'c' before returning 'a' for ( int i = 0; i < 100; i++) a.coef[i] = c.coef[i]; a.deg = a.degree(); return a; } I just slightly modified my "-" operator method to overwrite the value in 'a' and return 'a', and just use the 'c' polynomial as a temp. I've put in some debug print statement and I confirm that at the time of computation, both: c = a - b; and a -= b; are computed to the same value. However, when I go to print them, their results are different: Polynomial a, b; a.set ( 7, 4 ); //7x^4 a.set ( 1, 2 ); //x^2 b.set ( 6, 3 ); //6x^3 b.set ( -3, 2 ); //-3x^2 c = a - b; // (7x^4 + x^2) - (6x^3 - 3x^2) a -= b; c.print(); cout << "\n"; a.print(); cout << "\n"; Result: 7x^4 -6x^3 4x^2 7x^4 1x^2 Why is my c = a - b and a -= b giving me different results when I go to print them?

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  • Conditional Operator Example

    - by mbcrump
    If you haven’t taken the time to learn conditional operators, then now is the time. I’ve added a quick and dirty example for those on the forums.   Code Snippet using System; using System.Net.Mail; using System.Net; using System.Globalization; using System.Windows.Forms;   class Demo {     //Please use conditional statements in your code. See example below.       public static void Main()     {         int dollars = 10;           //Bad Coder Bad !!! Don't do this         if (dollars == 1)         {             Console.WriteLine("Please deposit {0} dollar.", dollars);         }         else         {             Console.WriteLine("Please deposit {0} dollars.", dollars);         }             //Good Coder Good !!! Do this         Console.WriteLine("Please deposit {0} dollar{1}.", dollars, dollars == 1 ? ' ' : 's');         //                                                          expression   ? true : false           Console.ReadLine();          } }

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  • Showplan Operator of the Week – BookMark/Key Lookup

    Fabiano continues in his mission to describe the major Showplan Operators used by SQL Server's Query Optimiser. This week he meets a star, the Key Lookup, a stalwart performer, but most famous for its role in ill-performing queries where an index does not 'cover' the data required to execute the query. If you understand why, and in what circumstances, key lookups are slow, it helps greatly with optimising query performance.

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  • LINQ and the use of Repeat and Range operator

    - by vik20000in
    LINQ is also very useful when it comes to generation of range or repetition of data.  We can generate a range of data with the help of the range method.     var numbers =         from n in Enumerable.Range(100, 50)         select new {Number = n, OddEven = n % 2 == 1 ? "odd" : "even"}; The above query will generate 50 records where the record will start from 100 till 149. The query also determines if the number is odd or even. But if we want to generate the same number for multiple times then we can use the Repeat method.     var numbers = Enumerable.Repeat(7, 10); The above query will produce a list with 10 items having the value 7. Vikram

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  • Operator of the Week - Spools, Eager Spool

    For the fifth part of Fabiano's mission to describe the major Showplan Operators used by SQL Server's Query Optimiser, he introduces the spool operators and particularly the Eager Spool, explains blocking and non-blocking and then describes how the Halloween Problem is avoided.

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  • Showplan Operator of the week - Assert

    As part of his mission to explain the Query Optimiser in practical terms, Fabiano attempts the feat of describing, one week at a time, all the major Showplan Operators used by SQL Server's Query Optimiser to build the Query Plan. He starts with Assert

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  • Ruby: if statement using regexp and boolean operator [migrated]

    - by bev
    I'm learning Ruby and have failed to make a compound 'if' statement work. Here's my code (hopefully self explanatory) commentline = Regexp.new('^;;') blankline = Regexp.new('^(\s*)$') if (line !~ commentline || line !~ blankline) puts line end the variable 'line' is gotten from reading the following file: ;; alias filename backupDir Prog_i Prog_i.rb ./store Prog_ii Prog_ii.rb ./store This fails and I'm not sure why. Basically I want the comment lines and blank lines to be ignored during the processing of the lines in the file. Thanks for your help.

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  • Showplan Operator of the Week – BookMark/Key Lookup

    Fabiano continues in his mission to describe the major Showplan Operators used by SQL Server's Query Optimiser. This week he meets a star, the Key Lookup, a stalwart performer, but most famous for its role in ill-performing queries where an index does not 'cover' the data required to execute the query. If you understand why, and in what circumstances, key lookups are slow, it helps greatly with optimising query performance.

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  • How do I overload an operator for an enumeration in C#?

    - by ChrisHDog
    I have an enumerated type that I would like to define the , <, =, and <= operators for. I know that these operators are implictly created on the basis of the enumerated type (as per the documentation) but I would like to explictly define these operators (for clarity, for control, to know how to do it, etc...) I was hoping I could do something like: public enum SizeType { Small = 0, Medium = 1, Large = 2, ExtraLarge = 3 } public SizeType operator >(SizeType x, SizeType y) { } But this doesn't seem to work ("unexpected toke") ... is this possible? It seems like it should be since there are implictly defined operators. Any suggestions?

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  • How does delete deal with pointer constness?

    - by aJ
    I was reading this question Deleting a const pointer and wanted to know more about delete behavior. Now, as per my understanding: delete expression works in two steps: invoke destructor then releases the memory (often with a call to free()) by calling operator delete. operator delete accepts a void*. As part of a test program I overloaded operator delete and found that operator delete doesn't accept const pointer. Since operator delete does not accept const pointer and delete internally calls operator delete, how does Deleting a const pointer work ? Does delete uses const_cast internally?

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  • Why would the assignment operator ever do something different than its matching constructor?

    - by Neil G
    I was reading some boost code, and came across this: inline sparse_vector &assign_temporary(sparse_vector &v) { swap(v); return *this; } template<class AE> inline sparse_vector &operator=(const sparse_vector<AE> &ae) { self_type temporary(ae); return assign_temporary(temporary); } It seems to be mapping all of the constructors to assignment operators. Great. But why did C++ ever opt to make them do different things? All I can think of is scoped_ptr?

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  • Why does javascript's "in" operator return true when testing if 0 exists in an array that doesn't co

    - by Mariano Peterson
    For example, this returns true, and makes sense: var x = [1,2]; 1 in x; // true This returns false, and makes sense: var x = [1,2]; 3 in x; // false However this returns true, and I don't understand why: var x = [1,2]; 0 in x; You can quickly test it by putting this in your browser's address bar: javascript:var x=[1,2]; alert(0 in x); Why does the "in" operator in Javascript return true when testing if "0" exists in array, even when the array doesn't appear to contain "0"?

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