Do you like spaced repetition learning? Have you used Anki or Quizlet? Whether or not spaced repetition works for you, periodically working on flash-card like questions can be a lot of fun, and just may help you retain information. Here are a few problems tied to the course material. Visit them periodically, and feel free to use them in your own spaced repetition learning practice!
Most of the reinforcement questions here deal with language-independent concepts. You should also practice with the language-dependent questions as well.
var x = 2; function f() { print x; } function g() { var x = 5; f(); print x; } g(); print x;
var x = 1; function f() {return x;} function g() {var x = 2; return f();} print g() + x;
var x = 100; function setX(n) {x = n;} function printX() {console.log(x);} function first() {setX(1); printX();} function second() {var x; setX(2); printX();} setX(0); first(); printX(); second(); printX();
let x = 3; let x = 3;
-2**2
in JavaScript and Python. Explain why the evaluation produced the value it did in each language.a+(b-c)
and (a+b)-c
produce different results?let x = 5 function f() { x = x * 3 } function g() { x = x * 5 } function h(a, b) { return a + x } print(h(f(), g()))
Array
but in other languages like Java, Rust, and Swift there exist many parameterized array types?x ? y : z
of a typical statically-typed language, what type checking and inference rules would a compiler be required to enforce? x
must be boolean and the types of y
and z
must be compatible. Inference: the type of the entire expression is the least general type of both y
and z
.Here are some problems that require some thinking, and some actual work. They may involve writing little scripts, or making sketches. They aren’t exactly short-answer problems.
(-b + sqrt(4 × a × c)) / (2 × a)Do you need a special symbol for unary negation? Why or why not?
function f() { return { x: 5 } }
let b = 8 let a = b + b (4 + 5).toString(16)
let place = "mundo" ["Hola", "Ciao"].forEach((command) => { console.log(command + ", " + place) })
const sayHello = function () { console.log("Hello") } (function() { console.log("Goodbye") }())What is being illustrated in each of the above? Go, Python, Scala, and Ruby all allow line endings to end statements and you don't hear people complaining about them the way they do about JavaScript. Pick one of these four languages and show why they don’t have problems with the four "problematic" cases of JavaScript.
f(2)
should return 2, but if local variables were allocated statically, it would return 3.
void f(int x) { int a = 2; int b; if (x < 0) { a = 3; return 0; } else { b = f(-x); return b + a; } }
public void fail() { class Failure extends RuntimeException {} throw new Failure(); }
function g(h) { var x = 2; h() } function main() { var x = 5 function f() { print x + 3 } g(f) } main()
function f(a) { let x = a - 1 function g() { print x - 17 } h(g) } function h(p) { let x = 13 p() } f(18)
null
and undefined
do not get coerced to objects, and non-functions do not get coerced to functions. I would say it's roughly 90% weakly typed. It’s not 100% weakly typed because its designer thought that coercing things to functions and even objects would be going too far, since there is no consensus on the obvious coercion to make.Dog
were a subclass of class Animal
, then objects of class Dog[]
would be compatible with the type Animal[]
. Write a fragment of Java code that shows that this means that Java is not completely statically typed. Include in your answer a well-written explanation that shows you truly understand the difference between static and dynamic typing. Dog
and Rat
are subclasses of Animal, this code
Animal[] pets = new Dog[4]; pets[0] = new Rat();compiles fine but when executed throws an
ArrayStoreException
, that's right, a run-time typecheck error. This means Java is NOT 100% statically typed because this typecheck occurs at run time. A language can only be called 100% statically typed if all type conflicts are detected at compile time.local fun split [] = ([],[]) | split [h] = ([h], []) | split (x::y::t) = let val (s1,s2) = split t in (x::s1,y::s2) end fun merge c ([], x) = x | merge c (x, []) = x | merge c (h1::t1, h2::t2) = if c(h1,h2)<0 then h1::merge c(t1,h2::t2) else h2::merge c(h1::t1,t2); in fun sort c [] = [] | sort c x = let val (p, q) = split x in merge c(sort c p, sort c q) end; end;
c
within sort
split
merge
y
in the third clause of split
?
sort
and explain why it is not what you would
expect.
struct A {B* x; int y;}; struct B {A* x; int y;};Suppose the rules for this language stated that this language used structural equivalence for types. How would you feel if you were a compiler and had to typecheck an expression in which an A was used as a B? What problem might you run into?
struct { int n; char c; } A[10][10];If the address of
A[0][0]
is 1000 (decimal), what is the
address of A[3][7]
?
double *a[n]; double (*b)[n]; double (*c[n])(); double (*d())[n];
double *a[n]; double (*b)[n]; double (*c[n])(); double (*d())[n];
double (*foo(double (*)(double, double[]), double)) (double, ...);Describe rigorously, in English, the type of foo.
x
, make it so that *x == x
. If this is not possible, state why it is not possible.x
, make it so that *x == x
. If this is not possible, state why it is not possible.p == *p
. If it is not possible to do so, state why no such variable can be defined.
(x += 7) *= z
but you can't say this in C. Explain the reason why, using precise, technical terminology. See if this same phenomenon holds for conditional expressions, too. What other languages behave like C++ in this respect?
a–f(b)–c*d
can produce different values depending on how a compiler decides to order, or even parallelize operations. Give a small program in the language of your choice (or even one of your own design) that would produce different values for this expression for different evaluation orders. Please note that by "different evaluation orders" we do not mean that the compiler can violate the precedence and associativity rules of the language.while (true) line := readLine(); if isAllBlanks(line) then exit end; consumeLine(line); end;Show how you might accomplish the same task using a while or repeat loop, if mid-test loops were not available. (Hint: one alternative duplicates part of the code; another introduces a Boolean flag variable.) How do these alternatives compare to the mid-test version?
If
in Java or C or
JavaScript, such that the call If(c, e1, e2)
would return e1
if c evaluated to true, and e2 if it evaluated to false.
Show why, in these languages, such a function is absolutely not the same as
the conditional expression c?e1:e2
. You can show a code fragment
that would return different results based on whether the function
were called versus the conditional expression were evaluated.
(p==null ? null : p.value)
is null-safe, whereas
the call If(p==null, null, p.value)
is not: it throws a NullPointerException
if p is null. A more blatant example can be seen in the difference between
(true ? 1 : formatMyHardDrive())
and
If(true, 1, formatMyHardDrive())
.
int first_zero_row = -1; /* assume no such row */ int i, j; for (i = 0; i < n; i++) { /* for each row */ for (j = 0; j < n; j++) { /* for each entry in the row */ if (A[i][j]) goto next; /* if non-zero go on to the next row */ } first_zero_row = i; /* went all through the row, you got it! */ break; /* get out of the whole thing */ next: ; } /* first_zero_row is now set */The intent of the code is to set
first_zero_row
to the index of the first all-zero row,
if any, of an n × n matrix, or -1 if no such row exists. Do you find the example convincing? Is there a good structured alternative in C? In any language? Give answers in the form of a short essay. Include a good introductory section, a background section describing views on the goto
statement throughout history, a beefy section analyzing alternatives to Rubin's problem, and a good concluding section. Talk about solutions in at least three languages.[a0, a1, ..., an-1]
and returns the list
[a0, f(a1), f(f(a2)),
f(f(f(a3))), ...]
.
For example, if you pass as arguments the function that
doubles its inputs, and the list [4, 3, 1, 2, 2]
, then
the return value would be [4, 6, 4, 16, 32]
.
Hint: Do the f(f(f...))
as a separate function. Also
you do not have to make your function tail recursive.
function ss(a) { function s(i, acc) { return i == a.length ? acc : s(i+1, a[i]*a[i]+acc) } return s(0, 0); }
def ss(a) s = Proc.new{|i,acc| i==a.length ? acc : s[i+1, a[i]*a[i]+acc]} s[0, 0] end
Given: a list [a0, a1, ..., an-1]
,
Return: a0*a1 + a2*a3
+ a4*a5 + ....
For example, if given [3, 5, 0, 28, 4, 7]
we return
15 + 0 + 28 = 43. If there are an odd number of
elements in the list, assume there is an extra 1
for padding.
Here is, by the way, a non-tail-recursive formulation:
fun sum_of_prods [] = 0 | sum_of_prods (x::nil) = x | sum_of_prods (x::y::t) = x * y + sum_of_prods t;
eval
, that accepts an
array of integers a, and a function f, and returns an array of
functions, each of which, when called, invokes f on the
corresponding element of a.
For example, if your function was called g
, then calling
g([6,3,1,8,7,9], dog)
would return an array of functions p
such that, for example, calling p[3]()
would invoke dog(8)
.
function g(a, f) { var b = []; for (var i = 0; i < a.length; i++) { b[i] = function(i){return function(){f(a[i])};}(i); } return b; }
square
and
addSix
the obvious way, and we called this new method
weird
, then:
[4, 6, 7, 3, 5, 2, 4].weird(addSix, square)would return the function
z
such that
z(2) == 55because the element at index 2 within the array is 7 and $7^2 + 6 = 55$.
Weekdays := Day_Set(False, True, True, True, True, True, False);
we could construct the aggregate directly in the variable Weekdays
. Give an example of an assignment statement that illustrates the necessity of constructing an aggregate in temporary storage (before copying to the target variable).
AngleReader
which displays a picture of a circle and notifies all its listeners
of the angle, in degrees, that the mouse cursor makes with the horizontal
axis of the circle as the mouse moves over it.
val dot = let fun transpose ... = in .... end;The transpose function should work like this
transpose ([x1,...,xn],[y1,..,yn]) = [(x1,y1),...,(xn,yn)]raising
Domain
if the arrays have different lengths. The body
of the definition of dot (between the in
and end
) should
contain only instances of the functions transpose
, o
, foldr
,
map
, op*
, op+
, and the value 0.
x = 1; y = [2, 3, 4]; function f(a, b) {b++; a = x + 1;} f(y[x], x); print x, y[0], y[1], y[2];
2 2 2 4
.
y[1]
, and that becomes 2 + 1 = 3, so it prints
2 2 3 4
.
b++
changes x
immediately
so x becomes 2. Then, since a refers to the expression
{y[x]
}, it will need to compute y[2]
which is 3.
The script prints 2 2 3 3
.
x = 1; y = 2; function f(a, b) {a = 3; print b, x;} f(x, x + y); print x;
void foo() { int i; printf("%d ", i++); } int main() { int j; for (j = 1; j <= 10; j++) foo(); }Local variable i in subroutine
foo
is never initialized. On many systems, however,
the program will display repeatable behavior, printing 0 1 2 3 4 5 6 7 8
9. Suggest an explanation. Also explain why the behavior on other systems might
be different, or nondeterministic.
with Ada.Text_IO, Ada.Integer_Text_IO; use Ada.Text_IO, Ada.Integer_Text_IO; procedure P is A: Integer := 4; type T is access Integer; B: T := new Integer'(4); C: T := new Integer'(4); procedure Q (X: in out Integer; Y: T; Z: in out T) is begin X := 5; Y.all := 5; Z.all := 5; end Q; begin Q (A, B, C); Put (A); Put (B.all); Put (C.all); end P;
call foo(2) print* 2 stop end subroutine foo(x) x = x + 1 return end
nextOdd
. The first time
you call this subroutine you get the value 1. The next time, you get
a 3, then 5, then 7, and so on. Show a snippet of code that uses this
subroutine from outside the module. Is it possible to
make this module hack-proof? In other words, once you compile this module,
can you be sure that malicious code can't do something to disrupt the
sequence of values resulting from successive calls to this function?
m
, s
, or y
?
Container
:
template <class Item> class Container { public: unsigned numberOfItems() {return currentSize;} bool isEmpty() {return currentSize == 0;}; virtual void flush() = 0; ~Container() {flush();} private: unsigned currentSize; };
Here the idea is that each particular (derived) container class shall
implement its own flush()
operation (which makes sense because
different containers are flushed in different ways: there may be
arrays, linked lists, rings or hashtables used in the representation),
and when a container is destroyed its flush()
operation will
be automatically invoked. However, the idea is flawed and the
code as written causes a terrible thing to happen. What happens?
Person
and
subclasses for different jobs (e.g., programmer, manager,
ticket agent, flight attendant, supervisor, student, etc.)
is a lousy design. Give an alternative.
class Item { private int id; private static int numberOfItemsCreated = 0; public Item() {id = numberOfItemsCreated++;} // pretend that there are more members here... };As you can see, every item that gets created will get a unique id. Because this pattern occurs frequently, it might be nice to generalize this and make make something reusable out of it so we don't have to write this code inside every class that needs ids. Perhaps we need an interface or abstract class. Tell me why these two suggestions won't work with a detailed, technical answer. Then tell me something that will work. (Note: there is nothing wrong with the access modifiers above; the problems with my two suggestions have to do with the nature of interfaces and abstract classes.)
class C {protected: int x; ...}; C c;
then your job is to assign a new value to c.x
.
Assume there are no public operations of C
that
modify x
that you know of. Also, do not use
any preprocessor tricks (like #define protected
public
).
class C {int x;}; C c; C* p = &c; cout << p;and there is no compile-time nor link time error, despite the fact that
operator<<(C*)
is not a member of ostream
(since that class was declared before you declared C
), nor for
that matter did anyone declare the global function
ostream& operator<<(ostream&, C*);So why does it all work? Explain exactly what gets printed and why.
class Foo { public int a; public String b; } ... class Bar extends Foo { public float c; public int b; }Does the representation of a Bar object contain one b field or two? If two, are both accessible, or only one? Under what circumstances? Answer for C++, Java, Python, and Scala.
Foo
is an abstract class in a C++ program, why is it
acceptable to declare variables of type Foo*
, but not of
type Foo
? Does this problem even make sense to ask in
Java or Python?
Put
procedure at the same time, their outputs may be interleaved. (This
could produce amusing and even distasteful results, e.g. writing
"sole" in parallel with "ash") Show how to set
things up so only one task is writing at a time.
if not T'Terminated then T.E; end if;
let countDown = function () { let i = 10; let update = function () { document.getElementById("t").innerHTML = i; if (i-- > 0) setTimeout(update, 1000); } update(); }The second argues that this is better:
let countDown = function () { let update = function (i) { document.getElementById("t").innerHTML = i; if (i-- > 0) setTimeout(function () {update(i)}, 1000); } update(10); }Your role is to figure out which programmer is right, if any. Make a fairly extensive list of the pros and cons of each approach. Your list will be graded on completeness, neatness, correct usage of terminology (don't forget to mention "anonymous function" and "closure"), and how articulately you express yourself. Bad grammar will impact your grade negatively. You may want to consider readability and (especially) performance.
Keep sharp by practicing your programming skills.
#
character.divmod
operator, write two solutions, one using the operator and one that does not.fringe
that takes a tree as
argument and returns an ordered list of its leaves.
Then we can say
def same_fringe(t1, t2): return fringe(t1) == fringe(t2)
fringe
in Python or JavaScript.
same_fringe
when
the trees differ in their first few leaves?
same_fringe
in Python.
same_fringe
in JavaScript or Go.
std:array
, and using std::vector
.dog="spike"; class <<dog; def bark; "arf"; end; end; dog.bark
"arf"
, since we essentially gave this one
dog its own bark
method, by attaching it to its own
singleton class.
class <<"sparky"; def bark; "woof"; end; end; "spraky".bark
class <<"sparky"; def bark; "woof"; end; end; "sparky".bark
dog="sparky"; class <<dog; def bark; "woof"; end; end; dog.bark
"woof"
, because we attached the bark method
to the object referenced by the variable dog, through
its metaclass.
class <<"sparky"; def bark; "woof"; end; end; "sparky".bark
NoMethodError
because the object in whose
metaclass we added the bark method WAS NOT the same
string object we tried to call bark
on.
class <<:sparky; def bark; "woof"; end; end; :sparky.bark
TypeError
because Ruby does not allow metaclasses
on symbol objects. Maybe this is because Ruby treats
symbols in such a super-special way an never puts them
in the object pool. It probably holds them as plain
old small integers.class <<2; def bark; "woof"; end; end; 2.bark
TypeError
for the same reason as for symbols.
class Counter attr_reader :val; @val = 0; def bump(); @val += 1; end end Counter.new.bump; Counter.new.val
def dot(a, b) a.zip(b).map{|x,y| x*y}.inject{|x,y| x+y} end
assert_equal(dot([], []), 0)
and it failed because her
method returned nil
, not 0.
def dot(a, b) a.zip(b).inject(0){|x,y| x+y[0]*y[1]} endEventually she wrote a unit test with the first array longer than the second and got a
TypeError
. She fixed that by raising
an ArgumentError
if the arguments to dot had different
lengths. Show her fixed up method.
def dot(a, b) raise ArgumentError if a.length != b.length a.zip(b).inject(0) {|x,y| x+y[0]*y[1]} end
class Array def *(a) ..............her code here.............. end endAnd the unit tests worked:
[3,4,2] * [1,5,0] == 23
for example.
But this fancy *
definition had a very nasty side effect,
which, if she did this in real production code, would have
probably broken something big time. What went down with
this definition?
*
operator in the array class; any new definition replaces
the old one. Old code using Array * int and Array * string will break badly,
because ints and strings can’t be converted to arrays.
class Pair implements Cloneable { private Object first, second; public Pair(Object x, Object y) {first = x; second = y;} public first() {return first;} public second() {return second;} }