I am new to C++ and I am trying to understand some code. What does it mean to have a * in front of the datatype ? and why is the class Name in front of the method name CAStar::LinkChild
void CAStar::LinkChild(_asNode *node, _asNode *temp)
{
}
views:
174answers:
3
+3
A:
The * means that it's a pointer. You will also find that _asNode *node is equivalent to _asNode* node.
The class name is in front of the method name when the method is not defined inside the class { ... }. The :: is the scope operator.
David Johnstone
2010-02-19 01:21:33
+6
A:
A * in front of the data type says that the variable is a pointer to the data type, in this case, a pointer to a node. Instead of passing a copy of the entire "node" into the method, a memory address, or pointer, is passed in instead. For details, see Pointers in this C++ Tutorial.
The class name in front of the method name specifies that this is defining a method of the
CAStarclass. For details, see the Tutorial pages for Classes.
Reed Copsey
2010-02-19 01:22:01
Furthermore, you can think of it as a way of disambiguating function names. For example, say you have two classes (trivial example); one implements fractions (Fraction) and the other implements integers (Integer). Both classes may have a function called "add". By including the class name Fraction, the compiler will know that you are implementing the add function for the Fraction class, rather than the Integer class.
Ryan Rosario
2010-02-19 01:25:45
Nicely said, Ryan
Reed Copsey
2010-02-19 01:28:00
+2
A:
Are you new to programming in general, or just to C++? If you're new to programming, you probably want to take some classes. If you're just new to C++, you could try reading Practical C++ Programming an on-line C++ Primer.
Regarding your specific question: in the variable declaration, an asterisk means "this is a pointer":
int * pointer;
This also covers function declarations/prototypes, where the variables are declared,as in your example.
After the declaration, asterisk means that you're de-referencing the pointer. That is, you're getting the value at the location to which it's pointing.
printf("memory address:%d value:%d", pointer, *pointer);
You'll note that memory address will change unexpectedly, depending upon the state of the program when it's printed. In a simple program, you won't see the change, but in a complex program, you would.
atk
2010-02-19 01:26:21
I am new to C++. I've done work in mostly php, as3, and C#. but never in c++. But hearing these responses I am seeing that C++ is lil different. Like declaring methods outside of a class. But thanks for the info.
numerical25
2010-02-19 01:30:11