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  • Why can operator-> be overloaded manually?

    - by FredOverflow
    Wouldn't it make sense if p->m was just syntactic sugar for (*p).m? Essentially, every operator-> that I have ever written could have been implemented as follows: Foo::Foo* operator->() { return &**this; } Is there any case where I would want p->m to mean something else than (*p).m?

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  • How to push_back without operator=() for const members?

    - by WilliamKF
    How to push_back() to a C++ std::vector without using operator=() for which the default definition violates having const members? struct Item { Item(int value) : _value(value) { } const int _value; } vector<Item> items; items.push_back(Item(3)); I'd like to keep the _value const since it should not change after the object is constructed, so the question is how do I initialize my vector with elements without invoking operator=()?

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  • Android: Cannot cast from View to MapView

    - by Gaz
    I'm trying to run the Android MapView example, and am getting a 'Cannot cast from View to MapView' error in Eclipse. My layout is as follows <?xml version="1.0" encoding="utf-8"?> <RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android" android:id="@+id/mainlayout" android:orientation="vertical" android:layout_width="fill_parent" android:layout_height="fill_parent" > <com.google.android.maps.MapView android:id="@+id/mapview" android:layout_width="fill_parent" android:layout_height="fill_parent" android:clickable="true" android:apiKey="0jwi0saLYCPGfO-t7glg5bQoBz7jVKWCcgyQWQA" /> <LinearLayout android:id="@+id/zoomview" android:layout_width="wrap_content" android:layout_height="wrap_content" android:layout_alignBottom="@id/mapview" android:layout_centerHorizontal="true" /> </RelativeLayout> and the Activity code is package org.gaz.mapapp; import android.os.Bundle; import android.view.View; import android.widget.LinearLayout; import android.widget.ZoomControls; import com.google.android.maps.*; public class MapView extends MapActivity { LinearLayout linearLayout; MapView mapView; ZoomControls mZoom; public void onCreate(Bundle savedInstance) { linearLayout = (LinearLayout) findViewById(R.id.mainlayout); mapView = (MapView) findViewById(R.id.mapview); mZoom = (ZoomControls) mapView.getZoomControls(); } @Override protected boolean isRouteDisplayed() { // TODO Auto-generated method stub return false; } } The error is caused by the line mapView = (MapView) findViewById(R.id.mapview); Can anybody offer an suggestions as to a fix? Cheers, Gaz.

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  • C++ reinterpret cast ?

    - by Ian
    I would like to cast one object of the class PointsList to another object Points3DList (and vice versa) where: template <class T> class PointsList { protected: std::vector <Point <T> *> points; //Only illustration, not possible with templaes }; and template <class T> class Points3DList { protected: std::vector <Point3D<T> *> points; //Only illustration, not possible with templaes }; Between Point and Point3D there is no relationship (inheritance nor composition)... template <class T> class Point { protected: T x; T y; }; template <class T> class Point3D { protected: T x; T y; T z; }; What do you think about conversion Points3DList <T> *pl3D = new Points3DList <T> (); ... PointsList <T> *pl = reinterpret_cast < PointList <T> * > ( pl3D ); where pl3D represents pointer to Points3DList object.. Can reinterpret_cast be used in this case or it is better to create a conversion function? Data model in this case can not be changed...

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  • Polynomial division overloading operator (solved)

    - by Vlad
    Ok. here's the operations i successfully code so far thank's to your help: Adittion: polinom operator+(const polinom& P) const { polinom Result; constIter i = poly.begin(), j = P.poly.begin(); while (i != poly.end() && j != P.poly.end()) { //logic while both iterators are valid if (i->pow > j->pow) { //if the current term's degree of the first polynomial is bigger Result.insert(i->coef, i->pow); i++; } else if (j->pow > i->pow) { // if the other polynomial's term degree is bigger Result.insert(j->coef, j->pow); j++; } else { // if both are equal Result.insert(i->coef + j->coef, i->pow); i++; j++; } } //handle the remaining items in each list //note: at least one will be equal to end(), but that loop will simply be skipped while (i != poly.end()) { Result.insert(i->coef, i->pow); ++i; } while (j != P.poly.end()) { Result.insert(j->coef, j->pow); ++j; } return Result; } Subtraction: polinom operator-(const polinom& P) const //fixed prototype re. const-correctness { polinom Result; constIter i = poly.begin(), j = P.poly.begin(); while (i != poly.end() && j != P.poly.end()) { //logic while both iterators are valid if (i->pow > j->pow) { //if the current term's degree of the first polynomial is bigger Result.insert(-(i->coef), i->pow); i++; } else if (j->pow > i->pow) { // if the other polynomial's term degree is bigger Result.insert(-(j->coef), j->pow); j++; } else { // if both are equal Result.insert(i->coef - j->coef, i->pow); i++; j++; } } //handle the remaining items in each list //note: at least one will be equal to end(), but that loop will simply be skipped while (i != poly.end()) { Result.insert(i->coef, i->pow); ++i; } while (j != P.poly.end()) { Result.insert(j->coef, j->pow); ++j; } return Result; } Multiplication: polinom operator*(const polinom& P) const { polinom Result; constIter i, j, lastItem = Result.poly.end(); Iter it1, it2, first, last; int nr_matches; for (i = poly.begin() ; i != poly.end(); i++) { for (j = P.poly.begin(); j != P.poly.end(); j++) Result.insert(i->coef * j->coef, i->pow + j->pow); } Result.poly.sort(SortDescending()); lastItem--; while (true) { nr_matches = 0; for (it1 = Result.poly.begin(); it1 != lastItem; it1++) { first = it1; last = it1; first++; for (it2 = first; it2 != Result.poly.end(); it2++) { if (it2->pow == it1->pow) { it1->coef += it2->coef; nr_matches++; } } nr_matches++; do { last++; nr_matches--; } while (nr_matches != 0); Result.poly.erase(first, last); } if (nr_matches == 0) break; } return Result; } Division(Edited): polinom operator/(const polinom& P) const { polinom Result, temp2; polinom temp = *this; Iter i = temp.poly.begin(); constIter j = P.poly.begin(); int resultSize = 0; if (temp.poly.size() < 2) { if (i->pow >= j->pow) { Result.insert(i->coef / j->coef, i->pow - j->pow); temp = temp - Result * P; } else { Result.insert(0, 0); } } else { while (true) { if (i->pow >= j->pow) { Result.insert(i->coef / j->coef, i->pow - j->pow); if (Result.poly.size() < 2) temp2 = Result; else { temp2 = Result; resultSize = Result.poly.size(); for (int k = 1 ; k != resultSize; k++) temp2.poly.pop_front(); } temp = temp - temp2 * P; } else break; } } return Result; } }; The first three are working correctly but division doesn't as it seems the program is in a infinite loop. Final Update After listening to Dave, I finally made it by overloading both / and & to return the quotient and the remainder so thanks a lot everyone for your help and especially you Dave for your great idea! P.S. If anyone wants for me to post these 2 overloaded operator please ask it by commenting on my post (and maybe give a vote up for everyone involved).

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  • Polynomial division overloading operator

    - by Vlad
    Ok. here's the operations i successfully code so far thank's to your help: Adittion: polinom operator+(const polinom& P) const { polinom Result; constIter i = poly.begin(), j = P.poly.begin(); while (i != poly.end() && j != P.poly.end()) { //logic while both iterators are valid if (i->pow > j->pow) { //if the current term's degree of the first polynomial is bigger Result.insert(i->coef, i->pow); i++; } else if (j->pow > i->pow) { // if the other polynomial's term degree is bigger Result.insert(j->coef, j->pow); j++; } else { // if both are equal Result.insert(i->coef + j->coef, i->pow); i++; j++; } } //handle the remaining items in each list //note: at least one will be equal to end(), but that loop will simply be skipped while (i != poly.end()) { Result.insert(i->coef, i->pow); ++i; } while (j != P.poly.end()) { Result.insert(j->coef, j->pow); ++j; } return Result; } Subtraction: polinom operator-(const polinom& P) const //fixed prototype re. const-correctness { polinom Result; constIter i = poly.begin(), j = P.poly.begin(); while (i != poly.end() && j != P.poly.end()) { //logic while both iterators are valid if (i->pow > j->pow) { //if the current term's degree of the first polynomial is bigger Result.insert(-(i->coef), i->pow); i++; } else if (j->pow > i->pow) { // if the other polynomial's term degree is bigger Result.insert(-(j->coef), j->pow); j++; } else { // if both are equal Result.insert(i->coef - j->coef, i->pow); i++; j++; } } //handle the remaining items in each list //note: at least one will be equal to end(), but that loop will simply be skipped while (i != poly.end()) { Result.insert(i->coef, i->pow); ++i; } while (j != P.poly.end()) { Result.insert(j->coef, j->pow); ++j; } return Result; } Multiplication: polinom operator*(const polinom& P) const { polinom Result; constIter i, j, lastItem = Result.poly.end(); Iter it1, it2, first, last; int nr_matches; for (i = poly.begin() ; i != poly.end(); i++) { for (j = P.poly.begin(); j != P.poly.end(); j++) Result.insert(i->coef * j->coef, i->pow + j->pow); } Result.poly.sort(SortDescending()); lastItem--; while (true) { nr_matches = 0; for (it1 = Result.poly.begin(); it1 != lastItem; it1++) { first = it1; last = it1; first++; for (it2 = first; it2 != Result.poly.end(); it2++) { if (it2->pow == it1->pow) { it1->coef += it2->coef; nr_matches++; } } nr_matches++; do { last++; nr_matches--; } while (nr_matches != 0); Result.poly.erase(first, last); } if (nr_matches == 0) break; } return Result; } Division(Edited): polinom operator/(const polinom& P) { polinom Result, temp; Iter i = poly.begin(); constIter j = P.poly.begin(); if (poly.size() < 2) { if (i->pow >= j->pow) { Result.insert(i->coef, i->pow - j->pow); *this = *this - Result; } } else { while (true) { if (i->pow >= j->pow) { Result.insert(i->coef, i->pow - j->pow); temp = Result * P; *this = *this - temp; } else break; } } return Result; } The first three are working correctly but division doesn't as it seems the program is in a infinite loop. Update Because no one seems to understand how i thought the algorithm, i'll explain: If the dividend contains only one term, we simply insert the quotient in Result, then we multiply it with the divisor ans subtract it from the first polynomial which stores the remainder. If the polynomial we do this until the second polynomial( P in this case) becomes bigger. I think this algorithm is called long division, isn't it? So based on these, can anyone help me with overloading the / operator correctly for my class? Thanks!

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  • 3x3 Sobel operator and gradient features

    - by pithyless
    Reading a paper, I'm having difficulty understanding the algorithm described: Given a black and white digital image of a handwriting sample, cut out a single character to analyze. Since this can be any size, the algorithm needs to take this into account (if it will be easier, we can assume the size is 2^n x 2^m). Now, the description states given this image we will convert it to a 512-bit feature (a 512-bit hash) as follows: (192 bits) computes the gradient of the image by convolving it with a 3x3 Sobel operator. The direction of the gradient at every edge is quantized to 12 directions. (192 bits) The structural feature generator takes the gradient map and looks in a neighborhood for certain combinations of gradient values. (used to compute 8 distinct features that represent lines and corners in the image) (128 bits) Concavity generator uses an 8-point star operator to find coarse concavities in 4 directions, holes, and lagrge-scale strokes. The image feature maps are normalized with a 4x4 grid. I'm for now struggling with how to take an arbitrary image, split into 16 sections, and using a 3x3 Sobel operator to come up with 12 bits for each section. (But if you have some insight into the other parts, feel free to comment :)

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  • operator<< overload,

    - by mr.low
    //using namespace std; using std::ifstream; using std::ofstream; using std::cout; class Dog { friend ostream& operator<< (ostream&, const Dog&); public: char* name; char* breed; char* gender; Dog(); ~Dog(); }; im trying to overload the << operator. I'm also trying to practice good coding. But my code wont compile unless i uncomment the using namespace std. i keep getting this error and i dont know. im using g++ compiler. Dog.h:20: error: ISO C++ forbids declaration of ‘ostream’ with no type Dog.h:20: error: ‘ostream’ is neither function nor member function; cannot be declared friend. if i add line using std::cout; then i get this error. Dog.h:21: error: ISO C++ forbids declaration of ‘ostream’ with no type. Can somebody tell me the correct way to overload the << operator with out using namespace std;

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  • How to overload operator<< for qDebug

    - by iyo
    Hi, I'm trying to create more useful debug messages for my class where store data. My code is looking something like this #include <QAbstractTableModel> #include <QDebug> /** * Model for storing data. */ class DataModel : public QAbstractTableModel { // for debugging purposes friend QDebug & operator<< (const QDebug &d, DataModel model); //other stuff }; /** * Overloading operator for debugging purposes */ QDebug & operator<< (QDebug &d, DataModel model) { d << "Hello world!"; return d; } I expect qDebug() << model will print "Hello world!". However, there is alway something like "QAbstractTableModel(0x1c7e520)" on the output. Do you have any idea what's wrong?

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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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  • SQL SERVER – Validating If Date is Last Day of the Year, Month or Day

    - by Pinal Dave
    Here is one more question I recently received in an email- “Pinal, is there any ready made function which will display if the given date is the last day or the year, month or day? For example, if a date is the last day of the Year and last day of the month, I want to display Last Day of the Year and if a date is the last date of the month and last day of the week, I want to display the last day of the week. “ Well, very interesting question and there is no such function available for the same. However, here is the function I have written earlier for my personal use where I almost accomplish same task. -- Example of Year DECLARE @Day DATETIME SET @Day = '2014-12-31' SELECT CASE WHEN CAST(@Day AS DATE) = CAST(DATEADD(ms,-3,DATEADD(yy,0,DATEADD(yy,DATEDIFF(yy,0,@Day)+1,0))) AS DATE) THEN 'LastDayofYear' WHEN CAST(@Day AS DATE) = CAST(DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,@Day)+1,0)) AS DATE) THEN 'LastDayofMonth' WHEN CAST(@Day AS DATE) = CAST(DATEADD(s,-1,DATEADD(wk, DATEDIFF(wk,0,@Day),0)) AS DATE) THEN 'LastDayofWeek' ELSE 'Day' END GO -- Example of Month DECLARE @Day DATETIME SET @Day = '2014-06-30' SELECT CASE WHEN CAST(@Day AS DATE) = CAST(DATEADD(ms,-3,DATEADD(yy,0,DATEADD(yy,DATEDIFF(yy,0,@Day)+1,0))) AS DATE) THEN 'LastDayofYear' WHEN CAST(@Day AS DATE) = CAST(DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,@Day)+1,0)) AS DATE) THEN 'LastDayofMonth' WHEN CAST(@Day AS DATE) = CAST(DATEADD(s,-1,DATEADD(wk, DATEDIFF(wk,0,@Day),0)) AS DATE) THEN 'LastDayofWeek' ELSE 'Day' END GO -- Example of Month DECLARE @Day DATETIME SET @Day = '2014-05-04' SELECT CASE WHEN CAST(@Day AS DATE) = CAST(DATEADD(ms,-3,DATEADD(yy,0,DATEADD(yy,DATEDIFF(yy,0,@Day)+1,0))) AS DATE) THEN 'LastDayofYear' WHEN CAST(@Day AS DATE) = CAST(DATEADD(s,-1,DATEADD(mm, DATEDIFF(m,0,@Day)+1,0)) AS DATE) THEN 'LastDayofMonth' WHEN CAST(@Day AS DATE) = CAST(DATEADD(s,-1,DATEADD(wk, DATEDIFF(wk,0,@Day),0)) AS DATE) THEN 'LastDayofWeek' ELSE 'Day' END GO Let me know if you know any other smarter trick and we can post it here with due credit. Reference: Pinal Dave (http://blog.SQLAuthority.com)Filed under: PostADay, SQL, SQL Authority, SQL DateTime, SQL Query, SQL Server, SQL Tips and Tricks, T SQL

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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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  • SQL SERVER – Saturday Fun Puzzle with SQL Server DATETIME2 and CAST

    - by pinaldave
    Note: I have used SQL Server 2012 for this small fun experiment. Here is what we are going to do. We will run the script one at time instead of running them all together and try to guess the answer. I am confident that many will get it correct but if you do not get correct, you learn something new. Let us create database and sample table. CREATE DATABASE DB2012 GO USE DB2012 GO CREATE TABLE TableDT (DT1 VARCHAR(100), DT2 DATETIME2, DT1C AS DT1, DT2C AS DT2); INSERT INTO TableDT (DT1, DT2) SELECT GETDATE(), GETDATE() GO There are four columns in the table. The first column DT1 is regular VARCHAR and second DT2 is DATETIME2. Both of the column are been populated with the same data as I have used the function GETDATE(). Now let us do the SELECT statement and get the result from both the columns. Before running the query please guess the answer and write it down on the paper or notepad. Question 1: Guess the resultset SELECT DT1, DT2 FROM TableDT GO Now once again run the select statement on the same table but this time retrieve the computed columns only. Once again I suggest you write down the result on the notepad. Question 2: Guess the resultset SELECT DT1C, DT2C FROM TableDT GO Now here is the best part. Let us use the CAST function over the computed columns. Here I do want you to stop and guess the answer for sure. If you have not done it so far, stop do it, believe me you will like it. Question 3: Guess the resultset SELECT CAST(DT1C AS DATETIME2) CDT1C, CAST(DT2C AS DATETIME2) CDT1C FROM TableDT GO Now let us inspect all the answers together and see how many of you got it correct. Answer 1: Answer 2: Answer 3:  If you have not tried to run the script so far, you can execute all the three of the above script together over here and see the result together. SELECT CAST(DT1C AS DATETIME2) CDT1C, CAST(DT2C AS DATETIME2) CDT1C FROM TableDT GO Here is the Saturday Fun question to you – why do we get same result from both of the expressions in Question 3, where as in question 2 both the expression have different answer. I will publish the valid answer with explanation in future blog posts. Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: PostADay, SQL, SQL Authority, SQL DateTime, SQL Puzzle, SQL Query, SQL Server, SQL Tips and Tricks, T SQL, Technology

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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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  • Play or Lift: which one is more explicit?

    - by Andrea
    I am going to investigate web development with Scala, and the choice is between learning Lift or Play: probably I will not have enough time to try both, at least at first. Now, many comparisons between the two are available on the internet, but I would like to know how do they compare with respect to being explicit and involving less magic. Let me explain what I mean by example. I have used, to various degrees, CakePHP, symfony2, Django and Grails. I feel a very clear distinction between Django and symfony2, which are very explicit about what you are doing, and Grails and CakePHP, which try to do their best to guess what you are trying to achieve and often feel "magical". Let me give some examples comparing Django and Grails. In Django, views are functions that take a request as input and return a response. You can instantiate explicitly an instance of HttpResponse and populate its body with a string, or you can use shortcut functions to leverage the template system. In any case the return value from your view always has the same type. In contrast, the render method from Grails is highly polymorphic. You can throw a context at it and it will try to render a template which is found by convention using that context. Or you can pass it a pair of a template path and a context and that will work too. Or a string. Or XML. Grails tries hard to make sense of whatever you return from your controller. In the Django ORM, each model class has a static attribute representing the manager for that class. That manager exposes a fluent interface to build querysets. In Grails, you can have a similar functionality by composing detached criteria. Still, the most common way to query objects seems to be the use of runtime-generated methods like FindUserByEmailNotNull or FindPostByDateGreaterThan. I will not go further, but my point is that in Django-like frameworks you have control over the whole flow of the request/response process, while in Grails-like ones I feel I only have to feel the blanks and the framework will manage the rest of the flow for me. This is not to criticize Grails or CakePHP; which type you prefer is mainly a matter of preference. In fact, I happen to like some aspects of Grails, but I feel more comfortable with a framework which does less for me. Back to the point of the question: which one among Play and Lift is more explicit about what you do and which one tries to simplify more what you have to do with a layer of "magic"?

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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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