assignment-operator

I have a class that requires a non-default copy constructor and assignment operator (it contains lists of pointers). Is there any general way to reduce the code duplication between the copy constructor and the assignment operator? There's no "general way" for writing custom copy constructors and assignment operators that works in all cases. But there's an idiom called "copy-&-swap": class myclass { ... public: myclass(myclass const&); void swap(myclass & with); myclass& operator=(myclass copy) { this->swap(copy); return *this; } ... }; It's useful in many (but not all) situations. Sometimes

This question already has an answer here: problems with char array = char array 2 answers Well here is my first post. I've been trying to do this choice choosing thing and I want the user to choose only numbers instead of typing them down (easier) but when I want the numbers to equal a string, it says "array type char[30] is not assignable". Even if at the back I put semi-colon or not. #include <stdio.h> int main() { int choice1; char word[30]; printf("You have three choice.\n"); printf("[1] Jump [2] Run [3] Dance\n"); scanf("%d",&choice1); if (choice1 == 1) { word = "Jump" //Error #1 } else

If T is a class type with the default signature for assignment operator, then we can write: T const &ref = ( T{} = something ); which creates a dangling reference. However, with the signature: T &operator=(T t) & the above code with dangling reference will fail to compile. This would prevent some situations where we return an lvalue that designates a temporary object -- undesirable situations because they can lead to dangling references. Is there any reason not to do this; would we be disabling any valid use cases for the assignment operators? I think the same comments can apply to the

I know in some languages the following: a += b is more efficient than: a = a + b because it removes the need for creating a temporary variable. Is this the case in C? Is it more efficient to use += (and, therefore also -= *= etc) So here's a definitive answer... $ cat junk1.c #include <stdio.h> int main() { long a, s = 0; for (a = 0; a < 1000000000; a++) { s = s + a * a; } printf("Final sum: %ld\n", s); } michael@isolde:~/junk$ cat junk2.c #include <stdio.h> int main() { long a, s = 0; for (a = 0; a < 1000000000; a++) { s += a * a; } printf("Final sum: %ld\n", s); } michael@isolde:~/junk$ for

Are both these PHP statements doing the same thing?: $o =& $thing; $o = &$thing; judda Yes, they are both the exact same thing. They just take the reference of the object and reference it within the variable $o . Please note, thing should be variables. They're not the same thing, syntactically speaking. The operator is the atomic =& and this actually matters. For instance you can't use the =& operator in a ternary expression. Neither of the following are valid syntax: $f = isset($field[0]) ? &$field[0] : &$field; $f =& isset($field[0]) ? $field[0] : $field; So instead you would use this: isset

I'm studying the C++ programming language and I'm reading the chapter about assignment operator ( = ). In C++ initalization and assignment are operation so similar that we can use the same notation. But my question is : when i initialize a variable am I doing it with the assignment operator ? When i assign to a variable, am I doing it with the assignment operator ? I think that the only difference is between initialization and assignment because when we initalize a variable we are giving it a new value with the assignmnet operator, when we assign to a variable we are replacing the old value of