Scope of pointer to class instance
-
If I define a variable in a function, its scope remains local to that function. Suppose I define a new instance of a class and a pointer in a function. Is the pointer local to the function? Can it be referenced outside the function? Do I have to declare the pointer in the header if I want to use it outside the function? For example:
X::SomeFunction() { int ABC = 0; Y* pYY = new Y;//Creates new instance of class Y}IntegerABCis limited in scope to function SomeFunction(). Is there any limitation to the scope ofpYY? CanpYYbe used in any other function other thanSomeFunction()? Thanks -
If I define a variable in a function, its scope remains local to that function. Suppose I define a new instance of a class and a pointer in a function. Is the pointer local to the function? Can it be referenced outside the function? Do I have to declare the pointer in the header if I want to use it outside the function? For example:
X::SomeFunction() { int ABC = 0; Y* pYY = new Y;//Creates new instance of class Y}IntegerABCis limited in scope to function SomeFunction(). Is there any limitation to the scope ofpYY? CanpYYbe used in any other function other thanSomeFunction()? ThanksThe scope of the variable
pYYitself is limited toX::SomeFunction. However, given that the object thatpYYis pointing to has been allocated on the heap that object continues to exist beyondX::SomeFunction. You could for instance returnpYYfromX::SomeFunctionand use it at the caller site.-- gleat http://blogorama.nerdworks.in[^] --