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C++ Example of Passing Parameters by Reference and by Value
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void SomeRoutine( const LARGE_OBJECT &nonmodifiableObject, LARGE_OBJECT *modifiableObject );
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This approach provides the additional benefit of providing a syntactic differentiation within the called routine between objects that are supposed to be treated as modifiable and those that aren t. In a modifiable object, the references to members will use the object->member notation, whereas for nonmodifiable objects references to members will use object.member notation. The limitation of this approach is difficulties propagating const references. If you control your own code base, it s good discipline to use const whenever possible
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13. Unusual Data Types
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(Meyers 1998), and you should be able to declare pass-by-value parameters as const references. For library code or other code that you don t control, you ll run into problems using const routine parameters. The fallback position is still to use references for read-only parameters but not declare them const. With that approach, you won t realize the full benefits of the compiler checking for attempts to modify non-modifiable arguments to a routine, but you ll at least give yourself the visual distinction between object->member and object.member.
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Use auto_ptrs If you haven t developed the habit of using auto_ptrs, get into the habit! auto_ptrs avoid many of the memory-leakage problems associated with regular pointers by deleting memory automatically when the auto_ptr goes out of scope. Scott Meyers More Effective C++, Item #9 contains a good discussion of auto_ptr (Meyers 1996). Get smart about smart pointers Smart pointers are a replacement for regular pointers or dumb pointers (Meyers 1996). They operate similarly to regular pointers, but they provide more control over resource management, copy operations, assignment operations, object construction, and object destruction. The issues involved are specific to C++. More Effective C++, Item #28, contains a complete discussion.
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Here are a few tips on using pointers that apply specifically to the C language.
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Use explicit pointer types rather than the default type C lets you use char or void pointers for any type of variable. As long as the pointer points, the language doesn t really care what it points at. If you use explicit types for your pointers, however, the compiler can give you warnings about mismatched pointer types and inappropriate dereferences. If you don t, it can t. Use the specific pointer type whenever you can.
The corollary to this rule is to use explicit type casting when you have to make a type conversion. For example, in the fragment below, it s clear that a variable of type NODE_ PTR is being allocated:
C Example of Explicit Type Casting
NodePtr = (NODE_PTR) calloc( 1, sizeof( NODE ) );
Avoid type casting Avoiding type casting doesn t have anything to do with going to acting school or getting out of always playing the heavy. It has to do with avoiding squeezing a variable of one type into the space for a variable of another type. Type casting
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turns off your complier s ability to check for type mismatches and therefore creates a hole in your defensive-programming armor. A program that requires many type casts probably has some architectural gaps that need to be revisited. Redesign if that s possible; otherwise, try to avoid type casts as much as you can.
Follow the asterisk rule for parameter passing You can pass an argument back from a routine in C only if you have an asterisk (*) in front of the argument in the assignment statement. Many C programmers have difficulty determining when C allows a value to be passed back to a calling routine. It s easy to remember that, as long as you have an asterisk in front of the parameter when you assign it a value, the value is passed back to the calling routine. Regardless of how many asterisks you stack up in the declaration, you must have at least one in the assignment statement if you want to pass back a value. For example, in the following fragment, the value assigned to parameter isn t passed back to the calling routine because the assignment statement doesn t use an asterisk:
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