What is shadowing, and how can it occur? Give an example.
Shadowing occurs when an identifier is bound within the scope of another binding of the same identifier. The more recent binding shadows the older one. Here is an example in C++.
{int x;
x = 3;
{int x;
x = 6;
}
}
The inner x shadows the outer one.
Given the definition
f([]) = []
f(h::t) = (h*h)::f(t) when h > 10
f(h::t) = f(t) when h <= 10
show an inside-out evaluation of expression f([4,12,15,6,11]).
Assume that arithmetic is done as soon as possible.
f([4,12,15,6,11]) = f([12,15,6,11]) (since 4 <= 10)
= 144::f([15,6,11]) (since 12 > 10 and 12*12=144)
= 144::225::f([6,11]) (since 15 > 10 and 15*15=225)
= 144::225::f([11]) (since 6 <= 10)
= 144::225::121::f([]) (since 11 > 10 and 11*11=121)
= 144::225::121::[]
= 144::225::[121]
= 144::[225,121]
= [144,225,121]
Remarks.
Be sure to follow the rules exactly. Do not forget about the [] at the end.
Write an equational definition of a function stutter where stutter([a,b,c,d]) = [a,a,b,b,c,c,d,d]. In general, stutter includes two copies of each item.
stutter([]) = []
stutter(h::t) = h::h::stutter(t)
Show an inside-out evaluation of stutter([2,3,4]) using your definition of stutter from the preceding question.
stutter([2,3,4]) = 2::2::stutter([3,4])
= 2::2:;3::3::stutter([4])
= 2::2::3::3::4::4::stutter([])
= 2::2::3::3::4::4::[]
= [2,2,3,3,4,4]
(Note: This uses the convention that :: is right-associative.)
Write an equational definition of a function called smallest so that smallest(n,x) is the smallest member of list n::x. For example, smallest(3, [6,4,7]) = 3 and smallest(8, [2,5]) = 2. You may presume that you have a function called min that takes the minimum of two numbers. For example, min(7,4) = 4.
Here are two solutions.
Definition 1: does not use min.
smallest(n,[]) = n
smallest(n,h::t) = smallest(n,t) when n <= h
smallest(n,h::t) = smallest(h,t) when n > h
Definition 2: uses min.
smallest(n,[]) = n
smallest(n,h::t) = smallest(min(n,h),t)
In a purely functional language, is it ever possible to compute the same expression twice in the same context and get different values? For example, if E is an expression, and you compute E two times, one right after another, could the first computation yield a different result from the second computation? Why or why not? (Note: Astarte is not a purely functional language. Think about the functional subset of Astarte. What is functional programming?)
Two evaluations of the same expression in the same context must always yield the same value. This is because there are no variables that can be changed.
Answer the preceding question, but instead of for a purely functional language, for an imperative language such as C++.
In a language that allows expressions to have side effects it is possible to evaluate the same expression twice and to get different answers. For example, suppose that function f is defined in C++ by
int f(int& z) {return ++z;}
Then
int z = 20;
const int x = f(z) + f(z);
does not produce the same answer as
int z = 20;
const int y = f(z);
const int x = y + y;
The two occurrences of f(z) do not produce the same answer.
By choosing among the operations lfold, rfold, map and select, write a definition for each of the following that does not use recursion or loops.
Write a function called doubleAll that takes a list x of numbers as its parameter and produces a list of the doubles numbers in list x as its result. For example, doubleAll([5,2,19,3]) = [10,4,38,6].
Remark. You can always use helper functions. This
definition still does not have any recursion, either direct
or mutual.
double(x) = x + x
doubleAll = map double
Write a function called maxlist(m,L) that computes the largest value in list m::L. That is, it computes the maximum of m and the largest value in list L. Assume that function max is available where max(x,y) is the larger of x and y.
maxlist(m,L) = lfold(m,max) L