operator-keyword

Just spent a frustrating couple of hours debugging this code: LinkedHashMap<String, Integer> rsrqs = new LinkedHashMap<String, Integer>(); Integer boxedPci = 52; Integer boxedRsrq = boxedPci != null ? rsrqs.get(boxedPci.toString()) : -1; The above produces a NullPointerException. The below code doesn't: LinkedHashMap<String, Integer> rsrqs = new LinkedHashMap<String, Integer>(); Integer boxedPci = 52; Integer boxedRsrq = boxedPci != null ? rsrqs.get(boxedPci.toString()) : Integer.valueOf(-1); The only difference is wrapping the -1 with Integer.valueOf(). I'm sure I'm going to smack my forehead

When it comes to constructors, adding the keyword explicit prevents an enthusiastic compiler from creating an object when it was not the programmer’s first intention. Is such mechanism available for casting operators too? struct Foo { operator std::string() const; }; Here, for instance, I would like to be able to cast Foo into a std::string , but I don’t want such cast to happen implicitly. Nawaz Yes and No. It depends on which version of C++, you're using. C++98 and C++03 do not support explicit type conversion operators But C++11 does. Example, struct A { //implicit conversion to int

In Scala, I can make a caseclass, case class Foo(x:Int) , and then put it in a list like so: List(Foo(42)) Now, nothing strange here. The following is strange to me. The operator :: is a function on a list, right? With any function with one argument in Scala, I can call it with infix notation. An example is 1 + 2 is a function (+) on the object Int . The class Foo I just defined does not have the :: operator, so how is the following possible? Foo(40) :: List(Foo(2)) In Scala 2.8 RC1, I get the following output from the interactive prompt: scala> case class Foo(x:Int) defined class Foo scala>

I have difficulties in understanding the sequence of calls in the code below. I was expecting to see the output below A1B2 While I can see that the output I get is BA12 I thought that the call std::cout<< b->fooA() << b->fooB() << std::endl was equivalent to call std::cout.operator<<( b->fooA() ).operator<< ( b->fooB() ) but I can see that this is not the case. Can you help me understanding better how this does it work and the relationship with the global operator<< ? Is this last ever called in this sequence? Regards AFAG #include <iostream> struct cbase{ int fooA(){ std::cout<<"A"; return 1;

I have class CMatrix, where is "double pointer" to array of values. class CMatrix { public: int rows, cols; int **arr; }; I simply need to access the values of matrix by typing: CMatrix x; x[0][0] = 23; I know how to do that using: x(0,0) = 23; But I really need to do that the other way. Can anyone help me with that? Please? Thank you guys for help at the end I did it this way... class CMatrix { public: int rows, cols; int **arr; public: int const* operator[]( int const y ) const { return &arr[0][y]; } int* operator[]( int const y ) { return &arr[0][y]; } .... Thank you for your help I really

I have following class:- class myclass { size_t st; myclass(size_t pst) { st=pst; } operator int() { return (int)st; } int operator+(int intojb) { return int(st) + intobj; } }; this works fine as long as I use it like this:- char* src="This is test string"; int i= myclass(strlen(src)) + 100; but I am unable to do this:- int i= 100+ myclass(strlen(src)); Any idea, how can I achieve this?? Implement the operator overloading outside of the class: class Num { public: Num(int i) { this->i = i; } int i; }; int operator+(int i, const Num& n) { return i + n.i; } You have to implement the operator as a