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In VBA, an assignment of variables of type Object is a shallow copy, an assignment for all other types (numeric types, String, user defined types, arrays) is a deep copy. So the keyword Set for an assignment signals a shallow copy and the (optional) keyword Let signals a deep copy. There is no built-in method for deep copies of Objects in VBA.
The rule of three (also known as the law of the big three or the big three) is a rule of thumb in C++ (prior to C++11) that claims that if a class defines any of the following then it should probably explicitly define all three: [1] destructor; copy constructor; copy assignment operator; These three functions are special member functions. If ...
C++ objects in general behave like primitive types, so to copy a C++ object one could use the '=' (assignment) operator. There is a default assignment operator provided for all classes, but its effect may be altered through the use of operator overloading. There are dangers when using this technique (see slicing).
Array's destructor deletes the data array of the original, therefore, when it deleted copy's data, because they share the same pointer, it also deleted first's data. Line (2) now accesses invalid data and writes to it. This produces a segmentation fault. If we write our own copy constructor that performs a deep copy then this problem goes away.
In the C++ programming language, the assignment operator, =, is the operator used for assignment.Like most other operators in C++, it can be overloaded.. The copy assignment operator, often just called the "assignment operator", is a special case of assignment operator where the source (right-hand side) and destination (left-hand side) are of the same class type.
SOURCE: Integrated Postsecondary Education Data System, University of Connecticut (2014, 2013, 2012, 2011, 2010).Read our methodology here.. HuffPost and The Chronicle examined 201 public D-I schools from 2010-2014.
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Another loophole [11] applies both to C and C++. Specifically, the languages dictate that member pointers and references are "shallow" with respect to the const-ness of their owners – that is, a containing object that is const has all const members except that member pointees (and referees) are still mutable. To illustrate, consider this C++ ...