Chapter 7 Answers

Nama : Christian Gunawan
NIM : 1801384174

Kali ini saya akan menjawab assigntment #7 dari chapter 7 Programming Language Concepts R Sebesta E-book :


Review Questions #6-10 :

6. What associativity rules are used by APL?

In APL, all operators have the same level of precedence. Thus, the order of evaluation of operators in APL expressions is determined entirely by the associativity rule, which is right to left for all operators.

For example, in the expression : A × B + C

the addition operator is evaluated first, followed by the multiplication operator (* is the APL multiplication operator). If A were 3, B were 4, and C were 5, then the value of this APL expression would be 27.

7. What is the difference between the way operators are implemented in C++ and Ruby?

What sets Ruby apart from the C-based languages in the area of expressions is that all of the arithmetic, relational, and assignment operators, as well as array indexing, shifts, and bitwise logic operators, are implemented as methods. For example, the expression a + b is a call to the + method of the object referenced by a, passing the object referenced by b as a parameter.

8. Define functional side effect.

A side effect of a function, naturally called a functional side effect, occurs when the function changes either one of its parameters or a global variable. (A global variable is declared outside the function but is accessible in the function.)

9. What is a coercion?

Coercion was defined as an implicit type conversion that is initiated by the compiler. 

Type conversions explicitly requested by the programmer are referred to as explicit conversions, or casts, not coercions.

10. What is a conditional expression?

If-then-else statements can be used to perform a conditional expression assignment. For example, consider :

if (count == 0)

       average = 0;

else

       average = sum / count;

In the C-based languages, this code can be specified more conveniently in an assignment statement using a conditional expression, which has the form

expression_1 ? expression_2 : expression_3.

Problem Set #6-10 :

6. Should C’s single-operand assignment forms (for example, ++count) be included in other languages (that do not already have them)? Why or why not?

Yes C should, because it will ease the increment or even decrement while we use in looping rather than manually by the assigning, and also by using that we can easily know that it is not operation, instead it is an increment or decrement which is usually used in repetition.

7. Describe a situation in which the add operator in a programming language would not be

commutative.

It wouldn’t be commutative when it deals with the negative integers. Recall that we can consider subtraction as addition in which one of two operator is a negative integer.

8. Describe a situation in which the add operator in a programming language would not be associative.

It is not associative when it includes the other operator with higher precedence like the multiplication and division.

9. Assume the following rules of associativity and precedence for expressions:

Precedence Highest *, /, not

+, –, &, mod

– (unary)

=, /=, < , <=, >=, >

Lowest or, xor

Associativity Left to right

Show the order of evaluation of the following expressions by parenthesizing all sub expressions and placing a superscript on the right parenthesis to indicate order. For example, for the expression

a + b * c + d

the order of evaluation would be represented as

((a + (b * c)1)2 + d)3

a. a * b - 1 + c                  ((( a * b )1 - 1)2 + c )3
b. a*(b-1)/c mod d           ((( a * ( b - 1 )1 )2 / c )3 mod d )4
c. (a-b)/c&(d*e/a-3)         (((a - b)1 / c)2 & ((d * e)3 / a)4 - 3)5)6
d. -a or c = d and e           (( -a )1 or ( ( c = d )2 and e )3 )4
e. a>b xor c or d<=17      (((a > b)1 xor c)3 or (d <= 17)2 )4
f. –a + b                            (–( a + b )1 )2

10. Show the order of evaluation of the expressions of Problem 9, assuming that there are no precedence rules and all operators associate right to left.

(a) ( a * ( b – ( 1 + c )1 )2 )3
(b) ( a * ( ( b – 1 )2 / ( c mod d )1 )3 )4
(c) ( ( a – b )5 / ( c & ( d * ( e / ( a – 3 )1 )2 )3 )4 )6
(d) ( – ( a or ( c = ( d and e )1 )2 )3 )4
(e) ( a > ( xor ( c or ( d <= 17 )1 )2 )3 )4
(f) ( – ( a + b )1 )2