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  • Conditional operator in Mako using Pylons

    - by Antoine Leclair
    In PHP, I often use the conditional operator to add an attribute to an html element if it applies to the element in question. For example: <select name="blah"> <option value="1"<?= $blah == 1 ? ' selected="selected"' : '' ?>> One </option> <option value="2"<?= $blah == 2 ? ' selected="selected"' : '' ?>> Two </option> </select> I'm starting a project with Pylons using Mako for the templating. How can I achieve something similar? Right now, I see two possibilities that are not ideal. Solution 1: <select name="blah"> % if blah == 1: <option value="1" selected="selected">One</option> % else: <option value="1">One</option> % endif % if blah == 2: <option value="2" selected="selected">Two</option> % else: <option value="2">Two</option> % endif </select> Solution 2: <select name="blah"> <option value="1" % if blah == 1: selected="selected" % endif >One</option> <option value="2" % if blah == 2: selected="selected" % endif >Two</option> </select> In this particular case, the value is equal to the variable tested (value="1" = blah == 1), but I use the same pattern in other situations, like <?= isset($variable) ? ' value="$variable" : '' ?>. I am looking for a clean way to achieve this using Mako.

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  • Most readable way to write simple conditional check

    - by JRL
    What would be the most readable/best way to write a multiple conditional check such as shown below? Two possibilities that I could think of (this is Java but the language really doesn't matter here): Option 1: boolean c1 = passwordField.getPassword().length > 0; boolean c2 = !stationIDTextField.getText().trim().isEmpty(); boolean c3 = !userNameTextField.getText().trim().isEmpty(); if (c1 && c2 && c3) { okButton.setEnabled(true); } Option 2: if (passwordField.getPassword().length > 0 && !stationIDTextField.getText().trim().isEmpty() && !userNameTextField.getText().trim().isEmpty() { okButton.setEnabled(true); } What I don't like about option 2 is that the line wraps and then indentation becomes a pain. What I don't like about option 1 is that it creates variables for nothing and requires looking at two places. So what do you think? Any other options?

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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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  • Replace conditional with polymorphism refactoring or similar?

    - by Anders Svensson
    Hi, I have tried to ask a variant of this question before. I got some helpful answers, but still nothing that felt quite right to me. It seems to me this shouldn't really be that hard a nut to crack, but I'm not able to find an elegant simple solution. (Here's my previous post, but please try to look at the problem stated here as procedural code first so as not to be influenced by the earlier explanation which seemed to lead to very complicated solutions: http://stackoverflow.com/questions/2772858/design-pattern-for-cost-calculator-app ) Basically, the problem is to create a calculator for hours needed for projects that can contain a number of services. In this case "writing" and "analysis". The hours are calculated differently for the different services: writing is calculated by multiplying a "per product" hour rate with the number of products, and the more products are included in the project, the lower the hour rate is, but the total number of hours is accumulated progressively (i.e. for a medium-sized project you take both the small range pricing and then add the medium range pricing up to the number of actual products). Whereas for analysis it's much simpler, it is just a bulk rate for each size range. How would you be able to refactor this into an elegant and preferably simple object-oriented version (please note that I would never write it like this in a purely procedural manner, this is just to show the problem in another way succinctly). I have been thinking in terms of factory, strategy and decorator patterns, but can't get any to work well. (I read Head First Design Patterns a while back, and both the decorator and factory patterns described have some similarities to this problem, but I have trouble seeing them as good solutions as stated there. The decorator example seems very complicated for just adding condiments, but maybe it could work better here, I don't know. And the factory pattern example with the pizza factory...well it just seems to create such a ridiculous explosion of classes, at least in their example. I have found good use for factory patterns before, but I can't see how I could use it here without getting a really complicated set of classes) The main goal would be to only have to change in one place (loose coupling etc) if I were to add a new parameter (say another size, like XSMALL, and/or another service, like "Administration"). Here's the procedural code example: public class Conditional { private int _numberOfManuals; private string _serviceType; private const int SMALL = 2; private const int MEDIUM = 8; public int GetHours() { if (_numberOfManuals <= SMALL) { if (_serviceType == "writing") return 30 * _numberOfManuals; if (_serviceType == "analysis") return 10; } else if (_numberOfManuals <= MEDIUM) { if (_serviceType == "writing") return (SMALL * 30) + (20 * _numberOfManuals - SMALL); if (_serviceType == "analysis") return 20; } else //i.e. LARGE { if (_serviceType == "writing") return (SMALL * 30) + (20 * (MEDIUM - SMALL)) + (10 * _numberOfManuals - MEDIUM); if (_serviceType == "analysis") return 30; } return 0; //Just a default fallback for this contrived example } } All replies are appreciated! I hope someone has a really elegant solution to this problem that I actually thought from the beginning would be really simple... Regards, Anders

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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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  • How to Highlight a Row in Excel Using Conditional Formatting

    - by Erez Zukerman
    Conditional formatting is an Excel feature you can use when you want to format cells based on their content. For example, you can have a cell turn red when it contains a number lower than 100. But how do you highlight an entire row? If you’ve never used Conditional Formatting before, you might want to look at Using Conditional Cell Formatting in Excel 2007. It’s one version back, but the interface really hasn’t changed much. But what if you wanted to highlight other cells based on a cell’s value? The screenshot above shows some codenames used for Ubuntu distributions. One of these is made up; when I entered “No” in the “Really” column, the entire row got different background and font colors. To see how this was done, read on.How To Make a Youtube Video Into an Animated GIFHTG Explains: What Are Character Encodings and How Do They Differ?How To Make Disposable Sleeves for Your In-Ear Monitors

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  • Class Assignment Operators

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

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

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

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  • operator for enums

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

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  • Operator overloading outside class

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

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  • Java operator overload

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

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  • Operator overloading C++ outside class

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

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  • null coalescing operator for javascript?

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

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

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

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  • Problem with operator ==

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

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