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  • [C++] Can all/any struct assignment operator be Overloaded? (and specifically struct tm = sql::Resu

    - by Luke Mcneice
    Hi all, Generally, i was wondering if there was any exceptions of types that cant have thier assignment operator overloaded. Specifically, I'm wanting to overload the assignment operator of a tm struct, (time.h) so i can assign a sql::ResultSet to it. I have already have the conversion logic: sscanf(sqlresult->getString("StoredAt").c_str(),"%d-%d-%d %d:%d:%d",&TempTimeStruct->tm_year,&TempTimeStruct->tm_mon,&TempTimeStruct->tm_mday,&TempTimeStruct->tm_hour,&TempTimeStruct->tm_min,&TempTimeStruct->tm_sec); //populating the struct I tried the overload with this: tm& tm::operator=(sql::ResultSet & results) { //CODE return *this; } however VS08 reports: error C2511: 'tm &tm::operator =(sql::ResultSet &)' : overloaded member function not found in 'tm'

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  • Using php's magic function inside another function does not work

    - by Sirber
    I want to use magic function __set() and __get() for storing SQL data inside a php5 class and I get some strange issue using them inside a function: Works: if (!isset($this->sPrimaryKey) || !isset($this->sTable)) return false; $id = $this->{$this->sPrimaryKey}; if (empty($id)) return false; echo 'yaay!'; Does not work: if (!isset($this->sPrimaryKey) || !isset($this->sTable)) return false; if (empty($this->{$this->sPrimaryKey})) return false; echo 'yaay!'; would this be a php bug?

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  • How `is_base_of` works?

    - by Alexey Malistov
    Why the following code works? typedef char (&yes)[1]; typedef char (&no)[2]; template <typename B, typename D> struct Host { operator B*() const; operator D*(); }; template <typename B, typename D> struct is_base_of { template <typename T> static yes check(D*, T); static no check(B*, int); static const bool value = sizeof(check(Host<B,D>(), int())) == sizeof(yes); }; //Test sample class Base {}; class Derived : private Base {}; //Exspression is true. int test[is_base_of<Base,Derived>::value && !is_base_of<Derived,Base>::value]; Note that B is private base. Note that operator B*() is const. How does this work? Why this works? Why static yes check(D*, T); is better than static yes check(B*, int); ?

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  • How is method group overload resolution different to method call overload resolution?

    - by thecoop
    The following code doesn't compile (error CS0123: No overload for 'System.Convert.ToString(object)' matches delegate 'System.Converter<T,string>'): class A<T> { void Method(T obj) { Converter<T, string> toString = Convert.ToString; } } however, this does: class A<T> { void Method(T obj) { Converter<T, string> toString = o => Convert.ToString(o); } } intellisense gives o as a T, and the Convert.ToString call as using Convert.ToString(object). In c# 3.5, delegates can be created from co/contra-variant methods, so the ToString(object) method can be used as a Converter<T, string>, as T is always guarenteed to be an object. So, the first example (method group overload resolution) should be finding the only applicable method string Convert.ToString(object o), the same as the method call overload resolution. Why is the method group & method call overload resolution producing different results?

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  • Why does operator<< not work with something returned by operator-?

    - by Felix
    Here's a small test program I wrote: #include <iostream> using namespace std; class A { public: int val; A(int _val=0):val(_val) { } A operator+(A &a) { return A(val + a.val); } A operator-(A &a) { return A(val - a.val); } friend ostream& operator<<(ostream &, A &); }; ostream& operator<<(ostream &out, A &a) { out<<a.val; return out; } int main() { A a(3), b(4), c = b - a; cout<<c<<endl; // this works cout<<(b-a)<<endl; // this doesn't return 0; } I can't seem to get why the line marked "this works" works and the one marked "this doesn't" doesn't. When I try to compile the program with the cout<<(b-a); line, here's what I get: [felix@the-machine C]$ g++ test.cpp test.cpp: In function ‘int main()’: test.cpp:26:13: error: no match for ‘operator<<’ in ‘std::cout << b.A::operator-(((A&)(& a)))’ /usr/lib/gcc/i686-pc-linux-gnu/4.5.0/../../../../include/c++/4.5.0/ostream:108:7: note: candidates are: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(std::basic_ostream<_CharT, _Traits>::__ostream_type& (*)(std::basic_ostream<_CharT, _Traits>::__ostream_type&)) [with _CharT = char, _Traits = std::char_traits<char>, std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>] /usr/lib/gcc/i686-pc-linux-gnu/4.5.0/../../../../include/c++/4.5.0/ostream:117:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(std::basic_ostream<_CharT, _Traits>::__ios_type& (*)(std::basic_ostream<_CharT, _Traits>::__ios_type&)) [with _CharT = char, _Traits = std::char_traits<char>, std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>, std::basic_ostream<_CharT, _Traits>::__ios_type = std::basic_ios<char>] /usr/lib/gcc/i686-pc-linux-gnu/4.5.0/../../../../include/c++/4.5.0/ostream:127:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(std::ios_base& (*)(std::ios_base&)) [with _CharT = char, _Traits = std::char_traits<char>, std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>] /usr/lib/gcc/i686-pc-linux-gnu/4.5.0/../../../../include/c++/4.5.0/ostream:165:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(long int) [with _CharT = char, _Traits = std::char_traits<char>, std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>] /usr/lib/gcc/i686-pc-linux-gnu/4.5.0/../../../../include/c++/4.5.0/ostream:169:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(long unsigned int) [with _CharT = char, _Traits = std::char_traits<char>, std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>] /usr/lib/gcc/i686-pc-linux-gnu/4.5.0/../../../../include/c++/4.5.0/ostream:173:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(bool) [with _CharT = char, _Traits = std::char_traits<char>, std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>] /usr/lib/gcc/i686-pc-linux-gnu/4.5.0/../../../../include/c++/4.5.0/bits/ostream.tcc:91:5: note: std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(short int) [with _CharT = char, _Traits = std::char_traits<char>] /usr/lib/gcc/i686-pc-linux-gnu/4.5.0/../../../../include/c++/4.5.0/ostream:180:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(short unsigned int) [with _CharT = char, _Traits = std::char_traits<char>, std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>] /usr/lib/gcc/i686-pc-linux-gnu/4.5.0/../../../../include/c++/4.5.0/bits/ostream.tcc:105:5: note: std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(int) [with _CharT = char, _Traits = std::char_traits<char>] /usr/lib/gcc/i686-pc-linux-gnu/4.5.0/../../../../include/c++/4.5.0/ostream:191:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(unsigned int) [with _CharT = char, _Traits = std::char_traits<char>, std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>] /usr/lib/gcc/i686-pc-linux-gnu/4.5.0/../../../../include/c++/4.5.0/ostream:200:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(long long int) [with _CharT = char, _Traits = std::char_traits<char>, std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>] /usr/lib/gcc/i686-pc-linux-gnu/4.5.0/../../../../include/c++/4.5.0/ostream:204:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(long long unsigned int) [with _CharT = char, _Traits = std::char_traits<char>, std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>] /usr/lib/gcc/i686-pc-linux-gnu/4.5.0/../../../../include/c++/4.5.0/ostream:209:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(double) [with _CharT = char, _Traits = std::char_traits<char>, std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>] /usr/lib/gcc/i686-pc-linux-gnu/4.5.0/../../../../include/c++/4.5.0/ostream:213:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(float) [with _CharT = char, _Traits = std::char_traits<char>, std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>] /usr/lib/gcc/i686-pc-linux-gnu/4.5.0/../../../../include/c++/4.5.0/ostream:221:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(long double) [with _CharT = char, _Traits = std::char_traits<char>, std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>] /usr/lib/gcc/i686-pc-linux-gnu/4.5.0/../../../../include/c++/4.5.0/ostream:225:7: note: std::basic_ostream<_CharT, _Traits>::__ostream_type& std::basic_ostream<_CharT, _Traits>::operator<<(const void*) [with _CharT = char, _Traits = std::char_traits<char>, std::basic_ostream<_CharT, _Traits>::__ostream_type = std::basic_ostream<char>] /usr/lib/gcc/i686-pc-linux-gnu/4.5.0/../../../../include/c++/4.5.0/bits/ostream.tcc:119:5: note: std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT, _Traits>::operator<<(std::basic_ostream<_CharT, _Traits>::__streambuf_type*) [with _CharT = char, _Traits = std::char_traits<char>, std::basic_ostream<_CharT, _Traits>::__streambuf_type = std::basic_streambuf<char>] test.cpp:18:11: note: std::ostream& operator<<(std::ostream&, A&) [felix@the-machine C]$ Quite nasty.

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  • C++ Unary - Operator Overload Won't Compile

    - by Brian Hooper
    I am attempting to create an overloaded unary - operator but can't get the code to compile. A cut-down version of the code is as follows:- class frag { public: frag myfunc (frag oper1, frag oper2); frag myfunc2 (frag oper1, frag oper2); friend frag operator + (frag &oper1, frag &oper2); frag operator - () { frag f; f.element = -element; return f; } private: int element; }; frag myfunc (frag oper1, frag oper2) { return oper1 + -oper2; } frag myfunc2 (frag oper1, frag oper2) { return oper1 + oper2; } frag operator+ (frag &oper1, frag &oper2) { frag innerfrag; innerfrag.element = oper1.element + oper2.element; return innerfrag; } The compiler reports... /home/brian/Desktop/frag.hpp: In function ‘frag myfunc(frag, frag)’: /home/brian/Desktop/frag.hpp:41: error: no match for ‘operator+’ in ‘oper1 + oper2.frag::operator-()’ /home/brian/Desktop/frag.hpp:16: note: candidates are: frag operator+(frag&, frag&) Could anyone suggest what I need to be doing here? Thanks.

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  • Static Variables in Overloaded Functions

    - by BSchlinker
    I have a function which does the following: When the function is called and passed a true bool value, it sets a static bool value to true When the function is called and passed a string, if the static bool value is set to true, it will do something with that string Here is my concern -- will a static variable remain the same between two overloaded functions? If not, I can simply create a separate function designed to keep track of the bool value, but I try to keep things simple.

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  • Should you declare methods using overloads or optional parameters in C# 4.0?

    - by Greg Beech
    I was watching Anders' talk about C# 4.0 and sneak preview of C# 5.0, and it got me thinking about when optional parameters are available in C# what is going to be the recommended way to declare methods that do not need all parameters specified? For example something like the FileStream class has about fifteen different constructors which can be divided into logical 'families' e.g. the ones below from a string, the ones from an IntPtr and the ones from a SafeFileHandle. FileStream(string,FileMode); FileStream(string,FileMode,FileAccess); FileStream(string,FileMode,FileAccess,FileShare); FileStream(string,FileMode,FileAccess,FileShare,int); FileStream(string,FileMode,FileAccess,FileShare,int,bool); It seems to me that this type of pattern could be simplified by having three constructors instead, and using optional parameters for the ones that can be defaulted, which would make the different families of constructors more distinct [note: I know this change will not be made in the BCL, I'm talking hypothetically for this type of situation]. What do you think? From C# 4.0 will it make more sense to make closely related groups of constructors and methods a single method with optional parameters, or is there a good reason to stick with the traditional many-overload mechanism?

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  • Templated << friend not working when in interrelationship with other templated union types

    - by Dwight
    While working on my basic vector library, I've been trying to use a nice syntax for swizzle-based printing. The problem occurs when attempting to print a swizzle of a different dimension than the vector in question. In GCC 4.0, I originally had the friend << overloaded functions (with a body, even though it duplicated code) for every dimension in each vector, which caused the code to work, even if the non-native dimension code never actually was called. This failed in GCC 4.2. I recently realized (silly me) that only the function declaration was needed, not the body of the code, so I did that. Now I get the same warning on both GCC 4.0 and 4.2: LINE 50 warning: friend declaration 'std::ostream& operator<<(std::ostream&, const VECTOR3<TYPE>&)' declares a non-template function Plus the five identical warnings more for the other function declarations. The below example code shows off exactly what's going on and has all code necessary to reproduce the problem. #include <iostream> // cout, endl #include <sstream> // ostream, ostringstream, string using std::cout; using std::endl; using std::string; using std::ostream; // Predefines template <typename TYPE> union VECTOR2; template <typename TYPE> union VECTOR3; template <typename TYPE> union VECTOR4; typedef VECTOR2<float> vec2; typedef VECTOR3<float> vec3; typedef VECTOR4<float> vec4; template <typename TYPE> union VECTOR2 { private: struct { TYPE x, y; } v; struct s1 { protected: TYPE x, y; }; struct s2 { protected: TYPE x, y; }; struct s3 { protected: TYPE x, y; }; struct s4 { protected: TYPE x, y; }; struct X : s1 { operator TYPE() const { return s1::x; } }; struct XX : s2 { operator VECTOR2<TYPE>() const { return VECTOR2<TYPE>(s2::x, s2::x); } }; struct XXX : s3 { operator VECTOR3<TYPE>() const { return VECTOR3<TYPE>(s3::x, s3::x, s3::x); } }; struct XXXX : s4 { operator VECTOR4<TYPE>() const { return VECTOR4<TYPE>(s4::x, s4::x, s4::x, s4::x); } }; public: VECTOR2() {} VECTOR2(const TYPE& x, const TYPE& y) { v.x = x; v.y = y; } X x; XX xx; XXX xxx; XXXX xxxx; // Overload for cout friend ostream& operator<<(ostream& os, const VECTOR2<TYPE>& toString) { os << "(" << toString.v.x << ", " << toString.v.y << ")"; return os; } friend ostream& operator<<(ostream& os, const VECTOR3<TYPE>& toString); friend ostream& operator<<(ostream& os, const VECTOR4<TYPE>& toString); }; template <typename TYPE> union VECTOR3 { private: struct { TYPE x, y, z; } v; struct s1 { protected: TYPE x, y, z; }; struct s2 { protected: TYPE x, y, z; }; struct s3 { protected: TYPE x, y, z; }; struct s4 { protected: TYPE x, y, z; }; struct X : s1 { operator TYPE() const { return s1::x; } }; struct XX : s2 { operator VECTOR2<TYPE>() const { return VECTOR2<TYPE>(s2::x, s2::x); } }; struct XXX : s3 { operator VECTOR3<TYPE>() const { return VECTOR3<TYPE>(s3::x, s3::x, s3::x); } }; struct XXXX : s4 { operator VECTOR4<TYPE>() const { return VECTOR4<TYPE>(s4::x, s4::x, s4::x, s4::x); } }; public: VECTOR3() {} VECTOR3(const TYPE& x, const TYPE& y, const TYPE& z) { v.x = x; v.y = y; v.z = z; } X x; XX xx; XXX xxx; XXXX xxxx; // Overload for cout friend ostream& operator<<(ostream& os, const VECTOR3<TYPE>& toString) { os << "(" << toString.v.x << ", " << toString.v.y << ", " << toString.v.z << ")"; return os; } friend ostream& operator<<(ostream& os, const VECTOR2<TYPE>& toString); friend ostream& operator<<(ostream& os, const VECTOR4<TYPE>& toString); }; template <typename TYPE> union VECTOR4 { private: struct { TYPE x, y, z, w; } v; struct s1 { protected: TYPE x, y, z, w; }; struct s2 { protected: TYPE x, y, z, w; }; struct s3 { protected: TYPE x, y, z, w; }; struct s4 { protected: TYPE x, y, z, w; }; struct X : s1 { operator TYPE() const { return s1::x; } }; struct XX : s2 { operator VECTOR2<TYPE>() const { return VECTOR2<TYPE>(s2::x, s2::x); } }; struct XXX : s3 { operator VECTOR3<TYPE>() const { return VECTOR3<TYPE>(s3::x, s3::x, s3::x); } }; struct XXXX : s4 { operator VECTOR4<TYPE>() const { return VECTOR4<TYPE>(s4::x, s4::x, s4::x, s4::x); } }; public: VECTOR4() {} VECTOR4(const TYPE& x, const TYPE& y, const TYPE& z, const TYPE& w) { v.x = x; v.y = y; v.z = z; v.w = w; } X x; XX xx; XXX xxx; XXXX xxxx; // Overload for cout friend ostream& operator<<(ostream& os, const VECTOR4& toString) { os << "(" << toString.v.x << ", " << toString.v.y << ", " << toString.v.z << ", " << toString.v.w << ")"; return os; } friend ostream& operator<<(ostream& os, const VECTOR2<TYPE>& toString); friend ostream& operator<<(ostream& os, const VECTOR3<TYPE>& toString); }; // Test code int main (int argc, char * const argv[]) { vec2 my2dVector(1, 2); cout << my2dVector.x << endl; cout << my2dVector.xx << endl; cout << my2dVector.xxx << endl; cout << my2dVector.xxxx << endl; vec3 my3dVector(3, 4, 5); cout << my3dVector.x << endl; cout << my3dVector.xx << endl; cout << my3dVector.xxx << endl; cout << my3dVector.xxxx << endl; vec4 my4dVector(6, 7, 8, 9); cout << my4dVector.x << endl; cout << my4dVector.xx << endl; cout << my4dVector.xxx << endl; cout << my4dVector.xxxx << endl; return 0; } The code WORKS and produces the correct output, but I prefer warning free code whenever possible. I followed the advice the compiler gave me (summarized here and described by forums and StackOverflow as the answer to this warning) and added the two things that supposedly tells the compiler what's going on. That is, I added the function definitions as non-friends after the predefinitions of the templated unions: template <typename TYPE> ostream& operator<<(ostream& os, const VECTOR2<TYPE>& toString); template <typename TYPE> ostream& operator<<(ostream& os, const VECTOR3<TYPE>& toString); template <typename TYPE> ostream& operator<<(ostream& os, const VECTOR4<TYPE>& toString); And, to each friend function that causes the issue, I added the <> after the function name, such as for VECTOR2's case: friend ostream& operator<< <> (ostream& os, const VECTOR3<TYPE>& toString); friend ostream& operator<< <> (ostream& os, const VECTOR4<TYPE>& toString); However, doing so leads to errors, such as: LINE 139: error: no match for 'operator<<' in 'std::cout << my2dVector.VECTOR2<float>::xxx' What's going on? Is it something related to how these templated union class-like structures are interrelated, or is it due to the unions themselves? Update After rethinking the issues involved and listening to the various suggestions of Potatoswatter, I found the final solution. Unlike just about every single cout overload example on the internet, I don't need access to the private member information, but can use the public interface to do what I wish. So, I make a non-friend overload functions that are inline for the swizzle parts that call the real friend overload functions. This bypasses the issues the compiler has with templated friend functions. I've added to the latest version of my project. It now works on both versions of GCC I tried with no warnings. The code in question looks like this: template <typename SWIZZLE> inline typename EnableIf< Is2D< typename SWIZZLE::PARENT >, ostream >::type& operator<<(ostream& os, const SWIZZLE& printVector) { os << (typename SWIZZLE::PARENT(printVector)); return os; } template <typename SWIZZLE> inline typename EnableIf< Is3D< typename SWIZZLE::PARENT >, ostream >::type& operator<<(ostream& os, const SWIZZLE& printVector) { os << (typename SWIZZLE::PARENT(printVector)); return os; } template <typename SWIZZLE> inline typename EnableIf< Is4D< typename SWIZZLE::PARENT >, ostream >::type& operator<<(ostream& os, const SWIZZLE& printVector) { os << (typename SWIZZLE::PARENT(printVector)); return os; }

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  • Can you overload controller methods in ASP.Net MVC?

    - by Eric Brown
    Im curious to see if you can overload controller methods in ASP.Net MVC. Whenever I try, I get the error below. The two methods accept different arguements. Is this something that cannot be done? The current request for action 'MyMethod' on controller type 'MyController' is ambiguous between the following action methods:

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  • templates and casting operators

    - by Jonathan Swinney
    This code compiles in CodeGear 2009 and Visual Studio 2010 but not gcc. Why? class Foo { public: operator int() const; template <typename T> T get() const { return this->operator T(); } }; Foo::operator int() const { return 5; } The error message is: test.cpp: In member function `T Foo::get() const': test.cpp:6: error: 'const class Foo' has no member named 'operator T'

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  • Properly removing an Integer from a List<Integer>

    - by Yuval A
    Here's a nice pitfall I just encountered. Consider a list of integers: List<Integer> list = new ArrayList<Integer>(); list.add(5); list.add(6); list.add(7); list.add(1); Any educated guess on what happens when you execute list.remove(1)? What about list.remove(new Integer(1))? This can cause some nasty bugs. What is the proper way to differentiate between remove(int index), which removes an element from given index and remove(Object o), which removes an element by reference, when dealing with lists of integers? The main point to consider here is the one @Nikita mentioned - exact parameter matching takes precedence over auto-boxing.

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  • Trying to reduce the speed overhead of an almost-but-not-quite-int number class

    - by Fumiyo Eda
    I have implemented a C++ class which behaves very similarly to the standard int type. The difference is that it has an additional concept of "epsilon" which represents some tiny value that is much less than 1, but greater than 0. One way to think of it is as a very wide fixed point number with 32 MSBs (the integer parts), 32 LSBs (the epsilon parts) and a huge sea of zeros in between. The following class works, but introduces a ~2x speed penalty in the overall program. (The program includes code that has nothing to do with this class, so the actual speed penalty of this class is probably much greater than 2x.) I can't paste the code that is using this class, but I can say the following: +, -, +=, <, > and >= are the only heavily used operators. Use of setEpsilon() and getInt() is extremely rare. * is also rare, and does not even need to consider the epsilon values at all. Here is the class: #include <limits> struct int32Uepsilon { typedef int32Uepsilon Self; int32Uepsilon () { _value = 0; _eps = 0; } int32Uepsilon (const int &i) { _value = i; _eps = 0; } void setEpsilon() { _eps = 1; } Self operator+(const Self &rhs) const { Self result = *this; result._value += rhs._value; result._eps += rhs._eps; return result; } Self operator-(const Self &rhs) const { Self result = *this; result._value -= rhs._value; result._eps -= rhs._eps; return result; } Self operator-( ) const { Self result = *this; result._value = -result._value; result._eps = -result._eps; return result; } Self operator*(const Self &rhs) const { return this->getInt() * rhs.getInt(); } // XXX: discards epsilon bool operator<(const Self &rhs) const { return (_value < rhs._value) || (_value == rhs._value && _eps < rhs._eps); } bool operator>(const Self &rhs) const { return (_value > rhs._value) || (_value == rhs._value && _eps > rhs._eps); } bool operator>=(const Self &rhs) const { return (_value >= rhs._value) || (_value == rhs._value && _eps >= rhs._eps); } Self &operator+=(const Self &rhs) { this->_value += rhs._value; this->_eps += rhs._eps; return *this; } Self &operator-=(const Self &rhs) { this->_value -= rhs._value; this->_eps -= rhs._eps; return *this; } int getInt() const { return(_value); } private: int _value; int _eps; }; namespace std { template<> struct numeric_limits<int32Uepsilon> { static const bool is_signed = true; static int max() { return 2147483647; } } }; The code above works, but it is quite slow. Does anyone have any ideas on how to improve performance? There are a few hints/details I can give that might be helpful: 32 bits are definitely insufficient to hold both _value and _eps. In practice, up to 24 ~ 28 bits of _value are used and up to 20 bits of _eps are used. I could not measure a significant performance difference between using int32_t and int64_t, so memory overhead itself is probably not the problem here. Saturating addition/subtraction on _eps would be cool, but isn't really necessary. Note that the signs of _value and _eps are not necessarily the same! This broke my first attempt at speeding this class up. Inline assembly is no problem, so long as it works with GCC on a Core i7 system running Linux!

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  • uint8_t and unsigned char linking error

    - by mnn
    I'm using template function: template<typename T> void func(const T& value) { obj->func(value); } where obj is object of class: void my_object::func(int64_t value) { ... } void my_object::func(uint64_t value) { ... } void my_object::func(uint32_t value) { ... } void my_object::func(uint16_t value) { ... } void my_object::func(uint8_t value) { ... } The problem is with uint8_t overload of my_object::func() override. Linker complains about unresolved external symbols to overloads, which should have unsigned char parameter. Should I replace uint8_t overload with unsigned char overload? Edit: Just now noticed, that linker complains about uint64_t and int64_t too. I compile on Windows using MSVC++ 2008 Express.

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  • Few Basic Questions in Overriding

    - by Dahlia
    I have few problems with my basic and would be thankful if someone can clear this. What does it mean when I say base *b = new derived; Why would one go for this? We very well separately can create objects for class base and class derived and then call the functions accordingly. I know that this base *b = new derived; is called as Object Slicing but why and when would one go for this? I know why it is not advisable to convert the base class object to derived class object (because base class is not aware of the derived class members and methods). I even read in other StackOverflow threads that if this is gonna be the case then we have to change/re-visit our design. I understand all that, however, I am just curious, Is there any way to do this? class base { public: void f(){cout << "In Base";} }; class derived:public base { public: void f(){cout << "In Derived";} }; int _tmain(int argc, _TCHAR* argv[]) { base b1, b2; derived d1, d2; b2 = d1; d2 = reinterpret_cast<derived*>(b1); //gives error C2440 b1.f(); // Prints In Base d1.f(); // Prints In Derived b2.f(); // Prints In Base d1.base::f(); //Prints In Base d2.f(); getch(); return 0; } In case of my above example, is there any way I could call the base class f() using derived class object? I used d1.base()::f() I just want to know if there any way without using scope resolution operator? Thanks a lot for your time in helping me out!

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  • Are there supposed to be more restrictions on operator->* overloads?

    - by Potatoswatter
    I was perusing section 13.5 after refuting the notion that built-in operators do not participate in overload resolution, and noticed that there is no section on operator->*. It is just a generic binary operator. Its brethren, operator->, operator*, and operator[], are all required to be non-static member functions. This precludes definition of a free function overload to an operator commonly used to obtain a reference from an object. But the uncommon operator->* is left out. In particular, operator[] has many similarities. It is binary (they missed a golden opportunity to make it n-ary), and it accepts some kind of container on the left and some kind of locator on the right. Its special-rules section, 13.5.5, doesn't seem to have any actual effect except to outlaw free functions. (And that restriction even precludes support for commutativity!) So, for example, this is perfectly legal (in C++0x, remove obvious stuff to translate to C++03): #include <utility> #include <iostream> #include <type_traits> using namespace std; template< class F, class S > typename common_type< F,S >::type operator->*( pair<F,S> const &l, bool r ) { return r? l.second : l.first; } template< class T > T & operator->*( pair<T,T> &l, bool r ) { return r? l.second : l.first; } template< class T > T & operator->*( bool l, pair<T,T> &r ) { return l? r.second : r.first; } int main() { auto x = make_pair( 1, 2.3 ); cerr << x->*false << " " << x->*4 << endl; auto y = make_pair( 5, 6 ); y->*(0) = 7; y->*0->*y = 8; // evaluates to 7->*y = y.second cerr << y.first << " " << y.second << endl; } I can certainly imagine myself giving into temp[la]tation. For example, scaled indexes for vector: v->*matrix_width[5] = x; Did the standards committee forget to prevent this, was it considered too ugly to bother, or are there real-world use cases?

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  • How does Java pick which method to call?

    - by Gaurav
    Given the following code: public class Test { public void method(Object o){ System.out.println("object"); } public void method(String s) { System.out.println("String"); } public void method() { System.out.println("blank"); } /** * @param args */ public static void main(String[] args) { // TODO Auto-generated method stub Test test=new Test(); test.method(null); } } Java prints "String". Why is this the case?

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  • Extend and Overload MS and Point Types

    - by dr d b karron
    Do I have make my own Point and Vector types to overload them ? Why does this not work ? namespace System . windows { public partial struct Point : IFormattable { public static Point operator * ( Point P , double D ) { Point Po = new Point ( ); return Po; } } } namespace SilverlightApplication36 { public partial class MainPage : UserControl { public static void ShrinkingRectangle ( WriteableBitmap wBM , int x1 , int y1 , int x2 , int y2 , Color C ) { wBM . DrawRectangle ( x1 , y1 , x2 , y2 , Colors . Red ); Point Center = Mean ( x1 , y1 , x2 , y2 ); wBM . SetPixel ( Center , Colors.Blue , 3 ); Point P1 = new Point ( x1 , y1 ); Point P2 = new Point ( x1 , y2 ); Point P3 = new Point ( x1 , y2 ); Point P4 = new Point ( x2 , y1 ); const int Steps = 10; for ( int i = 0 ; i < Steps ; i++ ) { double iF = (double)(i+1) / (double)Steps; double jF = ( 1.0 - iF ); Point P11 = **P1 * jF;** } }

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  • polymorphism and interfaces

    - by mixm
    if i have two classes x and y, both extend class w. and x implementing interface z. if i have methods doSomething(w object) and doSomething(x object), what would happen if i call doSomething(x)? edit: im implementing this on java, more specifically on android. im asking this because some classes which implement a specific interface mostly does the same thing when doSomething() is called. but there are special cases which i would like to single out.

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  • Java Best Practice for type resolution at runtime.

    - by Brian
    I'm trying to define a class (or set of classes which implement the same interface) that will behave as a loosely typed object (like JavaScript). They can hold any sort of data and operations on them depend on the underlying type. I have it working in three different ways but none seem ideal. These test versions only allow strings and integers and the only operation is add. Adding integers results in the sum of the integer values, adding strings concatenates the strings and adding an integer to a string converts the integer to a string and concatenates it with the string. The final version will have more types (Doubles, Arrays, JavaScript-like objects where new properties can be added dynamically) and more operations. Way 1: public interface DynObject1 { @Override public String toString(); public DynObject1 add(DynObject1 d); public DynObject1 addTo(DynInteger1 d); public DynObject1 addTo(DynString1 d); } public class DynInteger1 implements DynObject1 { private int value; public DynInteger1(int v) { value = v; } @Override public String toString() { return Integer.toString(value); } public DynObject1 add(DynObject1 d) { return d.addTo(this); } public DynObject1 addTo(DynInteger1 d) { return new DynInteger1(d.value + value); } public DynObject1 addTo(DynString1 d) { return new DynString1(d.toString()+Integer.toString(value)); } } ...and similar for DynString1 Way 2: public interface DynObject2 { @Override public String toString(); public DynObject2 add(DynObject2 d); } public class DynInteger2 implements DynObject2 { private int value; public DynInteger2(int v) { value = v; } @Override public String toString() { return Integer.toString(value); } public DynObject2 add(DynObject2 d) { Class c = d.getClass(); if(c==DynInteger2.class) { return new DynInteger2(value + ((DynInteger2)d).value); } else { return new DynString2(toString() + d.toString()); } } } ...and similar for DynString2 Way 3: public class DynObject3 { private enum ObjectType { Integer, String }; Object value; ObjectType type; public DynObject3(Integer v) { value = v; type = ObjectType.Integer; } public DynObject3(String v) { value = v; type = ObjectType.String; } @Override public String toString() { return value.toString(); } public DynObject3 add(DynObject3 d) { if(type==ObjectType.Integer && d.type==ObjectType.Integer) { return new DynObject3(Integer.valueOf(((Integer)value).intValue()+((Integer)value).intValue())); } else { return new DynObject3(value.toString()+d.value.toString()); } } } With the if-else logic I could use value.getClass()==Integer.class instead of storing the type but with more types I'd change this to use a switch statement and Java doesn't allow switch to use Classes. Anyway... My question is what is the best way to go about something thike this?

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  • How to make an ambiguous call distinct in C++?

    - by jcyang
    void outputString(const string &ss) { cout << "outputString(const string& ) " + ss << endl; } void outputString(const string ss) { cout << "outputString(const string ) " + ss << endl; } int main(void) { //! outputString("ambigiousmethod"); const string constStr = "ambigiousmethod2"; //! outputString(constStr); } ///:~ How to make distinct call? EDIT: This piece of code could be compiled with g++ and msvc. thanks.

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  • Visual C++ doesn't operator<< overload

    - by PierreBdR
    I have a vector class that I want to be able to input/output from a QTextStream object. The forward declaration of my vector class is: namespace util { template <size_t dim, typename T> class Vector; } I define the operator<< as: namespace util { template <size_t dim, typename T> QTextStream& operator<<(QTextStream& out, const util::Vector<dim,T>& vec) { ... } template <size_t dim, typename T> QTextStream& operator>>(QTextStream& in,util::Vector<dim,T>& vec) { .. } } However, if I ty to use these operators, Visual C++ returns this error: error C2678: binary '<<' : no operator found which takes a left-hand operand of type 'QTextStream' (or there is no acceptable conversion) A few things I tried: Originaly, the methods were defined as friends of the template, and it is working fine this way with g++. The methods have been moved outside the namespace util I changed the definition of the templates to fit what I found on various Visual C++ websites. The original friend declaration is: friend QTextStream& operator>>(QTextStream& ss, Vector& in) { ... } The "Visual C++ adapted" version is: friend QTextStream& operator>> <dim,T>(QTextStream& ss, Vector<dim,T>& in); with the function pre-declared before the class and implemented after. I checked the file is correctly included using: #pragma message ("Including vector header") And everything seems fine. Doesn anyone has any idea what might be wrong?

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  • Using __str__ representation for printing objects in containers in Python

    - by BobDobbs
    I've noticed that when an instance with an overloaded str method is passed to the print() function as an argument, it prints as intended. However, when passing a container that contains one of those instances to print(), it uses the repr method instead. That is to say, print(x) displays the correct string representation of x, and print(x, y) works correctly, but print([x]) or print((x, y)) prints the repr representation instead. First off, why does this happen? Secondly, is there a way to correct that behavior of print() in this circumstance?

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