Item 5: Use the same form in corresponding uses of new and delete.
What's wrong with this picture?
string *stringArray = new string[100];
...
delete stringArray;
Everything here appears to be in order ? the use of new is matched with a use of delete ? but something is still
quite wrong: your program's behavior is undefined. At the very least, 99 of the 100 string objects pointed to by
stringArray are unlikely to be properly destroyed, because their destructors will probably never be called.
When you use new, two things happen. First, memory is allocated (via the function operator new, about which
I'll have more to say in Items 7-10 as well as Item M8). Second, one or more constructors are called for that
memory. When you use delete, two other things happen: one or more destructors are called for the memory, then
the memory is deallocated (via the function operator delete ? see Items 8 and M8). The big question for delete is
this: how many objects reside in the memory being deleted? The answer to that determines how many
destructors must be called.
Actually, the question is simpler: does the pointer being deleted point to a single object or to an array of
objects? The only way for delete to know is for you to tell it. If you don't use brackets in your use of delete,
delete assumes a single object is pointed to. Otherwise, it assumes that an array is pointed to:
string *stringPtr1 = new string;
string *stringPtr2 = new string[100];
...
delete stringPtr1;
delete [] stringPtr2;
// delete an object
// delete an array of
// objects
What would happen if you used the "[]" form on stringPtr1? The result is undefined. What would happen if you
didn't use the "[]" form on stringPtr2? Well, that's undefined too. Furthermore, it's undefined even for built-in
types like ints, even though such types lack destructors. The rule, then, is simple: if you use [] when you call
new, you must use [] when you call delete. If you don't use [] when you call new, don't use [] when you call
delete.
This is a particularly important rule to bear in mind when you are writing a class containing a pointer data
member and also offering multiple constructors, because then you've got to be careful to use the same form of
new in all the constructors to initialize the pointer member. If you don't, how will you know what form of delete
to use in your destructor? For a further examination of this issue, see Item 11.
This rule is also important for the typedef-inclined, because it means that a typedef's author must document
which form of delete should be employed when new is used to conjure up objects of the typedef type. For
example, consider this typedef:
typedef string AddressLines[4];
// a person's address
// has 4 lines, each of
// which is a string
Because AddressLines is an array, this use of new,
string *pal = new AddressLines;
// note that "new
// AddressLines" returns
// a string*, just like
// "new string[4]" would
must be matched with the array form of delete:
delete pal; // undefined!
delete [] pal; // fine
To avoid such confusion, you're probably best off abstaining from typedefs for array types. That should be easy,
however, because the standard C++ library (see Item 49) includes string and vector templates that reduce the
need for built-in arrays to nearly zero. Here, for example, AddressLines could be defined to be a vector of
strings. That is, AddressLines could be of type vector.
What's wrong with this picture?
string *stringArray = new string[100];
...
delete stringArray;
Everything here appears to be in order ? the use of new is matched with a use of delete ? but something is still
quite wrong: your program's behavior is undefined. At the very least, 99 of the 100 string objects pointed to by
stringArray are unlikely to be properly destroyed, because their destructors will probably never be called.
When you use new, two things happen. First, memory is allocated (via the function operator new, about which
I'll have more to say in Items 7-10 as well as Item M8). Second, one or more constructors are called for that
memory. When you use delete, two other things happen: one or more destructors are called for the memory, then
the memory is deallocated (via the function operator delete ? see Items 8 and M8). The big question for delete is
this: how many objects reside in the memory being deleted? The answer to that determines how many
destructors must be called.
Actually, the question is simpler: does the pointer being deleted point to a single object or to an array of
objects? The only way for delete to know is for you to tell it. If you don't use brackets in your use of delete,
delete assumes a single object is pointed to. Otherwise, it assumes that an array is pointed to:
string *stringPtr1 = new string;
string *stringPtr2 = new string[100];
...
delete stringPtr1;
delete [] stringPtr2;
// delete an object
// delete an array of
// objects
What would happen if you used the "[]" form on stringPtr1? The result is undefined. What would happen if you
didn't use the "[]" form on stringPtr2? Well, that's undefined too. Furthermore, it's undefined even for built-in
types like ints, even though such types lack destructors. The rule, then, is simple: if you use [] when you call
new, you must use [] when you call delete. If you don't use [] when you call new, don't use [] when you call
delete.
This is a particularly important rule to bear in mind when you are writing a class containing a pointer data
member and also offering multiple constructors, because then you've got to be careful to use the same form of
new in all the constructors to initialize the pointer member. If you don't, how will you know what form of delete
to use in your destructor? For a further examination of this issue, see Item 11.
This rule is also important for the typedef-inclined, because it means that a typedef's author must document
which form of delete should be employed when new is used to conjure up objects of the typedef type. For
example, consider this typedef:
typedef string AddressLines[4];
// a person's address
// has 4 lines, each of
// which is a string
Because AddressLines is an array, this use of new,
string *pal = new AddressLines;
// note that "new
// AddressLines" returns
// a string*, just like
// "new string[4]" would
must be matched with the array form of delete:
delete pal; // undefined!
delete [] pal; // fine
To avoid such confusion, you're probably best off abstaining from typedefs for array types. That should be easy,
however, because the standard C++ library (see Item 49) includes string and vector templates that reduce the
need for built-in arrays to nearly zero. Here, for example, AddressLines could be defined to be a vector of
strings. That is, AddressLines could be of type vector
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