Multiple distinct objects
You are encouraged to solve this task according to the task description, using any language you may know.
Create a sequence (array, list, whatever) consisting of n distinct items of the same type. n should be determined at runtime.
By distinct we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.
This task was inspired by the common error of intending to do this, but instead creating a sequence of nreferences to the same mutable object; it might be informative to show the way to do that as well.
This task mostly makes sense for languages operating in the pass-references-by-value style (most object-oriented or 'dynamic' languages).
Ada
<ada> A : array (1..N) of T; </ada> Here N can be unknown until run-time. T is any constrained type.
C
foo *foos = malloc(n * sizeof(*foos));
for (int i = 0; i < n; i++)
init_foo(&foos[i]);
(Or if no particular initialization is needed, skip that part, or use calloc.)
C++
Using only language primitives:
// this assumes T is a default-constructible type (all built-in types are)
T* p = new T[n]; // if T is POD, the objects are uninitialized, otherwise they are default-initialized
// ...
// when you don't need the objects any more, get rid of them
delete[] p;
Using the standard library
- include <vector>
// this assumes T is default-constructible
std::vector<T> vec1(n); // all n objects are default-initialized
// this assumes t is a value of type T (or a type which implicitly converts to T)
std::vector<T> vec2(n, t); // all n objects are copy-initialized with t
// if you want to evaluate an expression for each new object
// (e.g. if it uses randomness to initialize the state or something)
std::vector<T> vec3;
for (int i = 0; i < n; ++i)
vec3.push_back(makeDistinctObject());
Instead of vector, deque or list can be used. Note that with standard containers containing the object type itself (instead of a pointer), you don't need to explicitly destroy the objects.
For polymorphic objects:
- include <vector>
// this assumes Base is a polymorphic type which has a clone method for polymorphic copying,
// and p is a pointer to Base or to a type derived from Base.
// the following is NOT correct:
std::vector<Base*> bvec_WRONG(n, p); // create n copies of p, which all point to the same opject p points to.
// nor is this:
std::vector<Base*> bvec_ALSO_WRONG(n, p->clone()); // create n pointers to a single clone of *p
// the correct solution
std::vector<Base*> bvec(n);
for (int i = 0; i < n; ++i)
bvec[i] = p->clone(); // "p->clone()" here can be replaced with any expression returning a pointer to a new object
// another correct solution
// this solution avoids uninitialized pointers at any point
std::vector<Base*> bvec2;
for (int i = 0; i < n; ++i)
bvec2.push_back(p->clone()); // "p->clone()" here can be replaced with any expression returning a pointer to a new object
// ...
// because the container contains pointers, the objects have to be explicitly deleted
// because ordinary pointers don't have destructors, so can't delete themselves
// using a smart pointer like boost::shared_ptr instead would make this step unnecessary
for (int i = 0; i < bvec.size(); ++i)
delete bvec[i];
// alternate iteration:
for (std::vector<Base*>::iterator it = bvec2.begin(); it != bvec2.end(); ++i)
delete *i;
bvec.clear(); // make sure the dangling pointers are not used any more
bvec2.clear(); // (not necessary if bvec isn't used afterwards; alternatively,
// set the pointers to NULL after deleting; again, using a smart pointer
// would remove this need)
Of course, also in this case one can use the other sequence containers or plain new/delete instead of vector.
Common Lisp
The mistake is often written as one of these:
(make-list n :initial-element (make-the-distinct-thing))
(make-array n :initial-element (make-the-distinct-thing))
which are incorrect since (make-the-distinct-thing) is only evaluated once. A common correct version is:
(loop repeat n collect (make-the-distinct-thing))
which evaluates (make-the-distinct-thing) n times and collects each result in a list.
Haskell
If the creator of the distinct thing is in some monad, then one can write
replicateM n makeTheDistinctThing
in an appropriate do block. If it is distinguished by, say, a numeric label, one could write
map makeTheDistinctThing [1..n]
Java
simple array:
Foo[] foos = new Foo[n]; // all elements initialized to null
for (int i = 0; i < foos.length; i++)
foos[i] = new Foo();
- incorrect version:
Foo[] foos_WRONG = new Foo[n];
Arrays.fill(foos, new Foo()); // new Foo() only evaluated once
simple list:
List<Foo> foos = new ArrayList<Foo>();
for (int i = 0; i < n; i++)
foos.add(new Foo());
- incorrect:
List<Foo> foos_WRONG = Collections.nCopies(n, new Foo()); // new Foo() only evaluated once
Generic version for class given at runtime:
It's not pretty but it gets the job done. The first method here is the one that does the work. The second method is a convenience method so that you can pass in a String of the class name.
public static <E> List<E> getNNewObjects(int n, Class<? extends E> c){
List<E> ans = new LinkedList<E>();
try {
for(int i=0;i<n;i++)
ans.add(c.newInstance());
} catch (InstantiationException e) {
e.printStackTrace();
} catch (IllegalAccessException e) {
e.printStackTrace();
}
return ans;
}
public static List<Object> getNNewObjects(int n, String className)
throws ClassNotFoundException{
return getNNewObjects(n, Class.forName(className));
}
Python
The mistake is often written as:
[Foo()] * n # here Foo() can be any expression that returns a new object
which is incorrect since Foo() is only evaluated once. A common correct version is:
[Foo() for i in xrange(n)]
which evaluates Foo() n times and collects each result in a list. This last form is also discussed here, on the correct construction of a two dimensional array.
Ruby
The mistake is often written as one of these:
[Foo.new] * n # here Foo.new can be any expression that returns a new object
Array.new(n, Foo.new)
which are incorrect since Foo.new is only evaluated once. A common correct version is:
Array.new(n) { Foo.new }
which evaluates Foo.new n times and collects each result in an Array. This last form is also discussed here, on the correct construction of a two dimensional array.