Q&A Assignment #15 Chapter 13 Review Questions & Problem Set

Name : Vina Melinda

NIM  : 1801380106

Untuk kali ini saya akan menjawab Assignment #15 Problem Set dan Review Questions dari Chapter 13 dari Sebesta.


Review Question:

6. Q: Describe the logical architecture of a vector processor.
A: Vector processor have groups of registers that store the operands of a vector operation in which the same instruction is executed on the whole group of operands simultaneously. Originally, the kinds of programs that could most benefit from this architecture were in scientific computation, an area of computing that is often the target of multiprocessor machines.

7. Q: What is the difference between physical and logical concurrency?
A: Physical concurrency is a multiple independent processors (multiple threads of control). While logical concurrency is the appearance of physical concurrency is presented by time-sharing one processor (software can be designed as if there were multiple threads of control)

8. Q: What is a thread of control in a program?
A: A thread of control in a program is the sequence of program points reached as control flows through the program.

9. Q: Why are coroutines called quasi-concurrent?
A: They are called quasi-concurrent because they have a single thread of control.

10. Q: What is a multithreaded program?
A: A multi-threaded program is a program designed to have more than one thread of control.

Problem Set:

6. Q: Suppose two tasks, A and B, must use the shared variable Buf_Size. Task A adds 2 to Buf_Size, and task B subtracts 1 from it. Assume that such arithmetic operations are done by the three-step process of fetching the current value, performing the arithmetic, and putting the new value back. In the absence of competition synchronization, what sequences of events are possible and what values result from these operations? Assume that the initial value of Buf_Size is 6.
A: The add and subtract operations are not atomic and could be interrupted in mid-operation when the other task could then run. If A runs to completion, then B runs to completion, Buf_Size has the value 7 (6 + 2 – 1). Similarly if B runs to completion then A runs to completion. If A or B get interrupted in the middle of adding or subtracting, then whichever task finishes last will determine the value in Buf_Size. If A runs but is interrupted after it fetches Buf_Size but before it stores the modified value (allowing B to fetch Buf_Size), or if B runs first and is interrupted after the fetch but before the store, allowing A to fetch Buf_Size, then if A finishes last Buf_Size will have value 8, and if B finishes last Buf_Size will have value 5.

7. Q: Compare the Java competition synchronization mechanism with that of Ada.
A: Java methods (but not constructors) can be specified to be synchronized. A synchronized method called through a specific object must complete its execu- tion before any other synchronized method can run on that object. Competition synchronization on an object is implemented by specifying that the methods that access shared data are synchronized. The competition synchronization mechanism of the Ada Language is intended to provide a facility for tasks to synchronize their actions.

8. Q: Compare the Java cooperation synchronization mechanism with that of Ada.
A: In Java, cooperation synchronization is implemented with the wait, notify, and notifyAll methods, all of which are defined in Object, the root class of all Java classes. Every object has a wait list of all of the threads that have called wait on the object. While in Ada, cooperation synchronization is required between two tasks that when the second task must wait for the first task to finish executing before it may proceed.

9. Q: What happens if a monitor procedure calls another procedure in the same monitor?
A: Implementation of a monitor can be made to guarantee synchronized access by allowing only one access at a time. Calls to monitor procedures are implicitly blocked and stored in a queue if the monitor is busy at the time of the call.

10. Q: Explain the relative safety of cooperation synchronization using semaphores and using Ada’s when clauses in tasks.
A: We need to take steps to make sure that different threads don't interact in negative ways. If one thread is operating on some data or structure, we don't want another thread to simultaneously operate on that same data/structure and corrupt the results; when Thread A writes to a variable that Thread B accesses, we need to make sure that Thread B will actually see the value written by Thread A; we don't want one thread to hog, take or lock for too long a resource that other threads need in order to make progress.

Q&A Assignment #14 Chapter 12 Review Questions & Problem Set

Name : Vina Melinda

NIM  : 1801380106

Untuk kali ini saya akan menjawab Assignment #14 Problem Set dan Review Questions dari Chapter 12 dari Sebesta.


Review Question:

6. Q: Describe a situation where dynamic binding is a great advantage over its absence.
A: Consider the following situation: There is a base class, A, that defines a method draw that draws some figure associated with the base class. A second class, B, is defined as a subclass of A. Objects of this new class also need a draw method that is like that provided by A but a bit different because the subclass objects are slightly different. So, the subclass overrides the inherited draw method. If a client of A and B has a variable that is a reference to class A’s objects, that reference also could point at class B’s objects, making it a polymorphic reference. If the method draw, which is defined in both classes, is called through the polymorphic reference, the run-time system must determine, during execution, which method should be called, A’s or B’s.

7. Q: What is a virtual method?
A: A virtual method is a declared class method that allows overriding by a method with the same derived class signature. Virtual methods are tools used to implement the polymorphism feature of an object-oriented language, such as C#. When a virtual object instance method is invoked, the method to be called is determined based on the object's runtime type, which is usually that of the most derived class.

8. Q: What is an abstract method? What is an abstract class?
A: An abstract method is a method which all descendant classes should have. An abstract class is a class which has abstract method.

9. Q: Describe briefly the eight design issues used in this chapter for objectoriented languages.
A: 1. What non-objects are in the language?
2.Are Subclasses Subtypes? If so, derived objects can be legally used wherever a parent object could be used.
3. Type Checking and Polymorphism
4. Single and Multiple Inheritance. Inherit from 1 (or more than 1) parent.
5. Object Allocation and Deallocation. Are objects allocated from heap or stack.
6. Dynamic and Static Binding. When are messages bound to methods, before or during run-time?
7. Nested Classes. Can a class be nested inside another class?
8. Initialization of Objects. Are objects initialized when created? Implicit or explicit?

10. What is a nesting class?
A: A nesting class is a class defined inside another class.

Problem Set:

6. Q: Compare the multiple inheritance of C++ with that provided by interfaces in Java.
A: C++ inheritance is implementation inheritance. That is, a class inheriting from two of more superclasses actually inherits the code from those classes. Java’s interface mechanism is an interface inheritance. That is, a class implementing two or more interfaces simply inherits (and must provide its own implementations for) the methods of the interface.

7. Q:What is one programming situation where multiple inheritance has a significant advantage over interfaces?
A: Interface just carry "method signatures", whereas classes carry actual behaviour. Multiple inheritance greatly helps to reduce boilerplate code.

8. Q: Explain the two problems with abstract data types that are ameliorated by inheritance.
A: The problems solved are reusability of code and "extensibility". Reusability because one won't have to copy/paste his code from one data type to another, allowing for a greater readability. Extensibility because a method can accept a certain class as an argument, and get a child class of this one. This will allow the user to have a wider set of functionality, but the method will still be able to know that the entities it relies on are present.

9. Q: Describe the categories of changes that a subclass can make to its parent class.
A: Subclasses can add things (variables, methods). Subclass in C++ can effectively remove a method using "private" inheritance. Inherited methods can be overridden.

10. Q: Explain one disadvantage of inheritance.
A: Subclass is dependent upon its base class (which might change over time). Means one cannot be used independent of each other.

Q&A Assignment #13 Chapter 11 Review Questions & Problem Set

Name : Vina Melinda

NIM  : 1801380106

Untuk kali ini saya akan menjawab Assignment #13 Problem Set dan Review Questions dari Chapter 11 dari Sebesta.


Review Question:

6. Q: Explain how information hiding is provided in an Ada package.

A: Data type representations can appear in the package specification but be hidden from clients by putting them in the private clause of the package. The abstract type itself is defined to be private in the public part of the package specification. Private types have built-in operations for assignment and comparison for equality and inequality.

7. Q: To what is the private part of an Ada package specification visible?

A:  The representation of the type appears in a part of the specification called the private part, which is introduced by the reserved word private. The private clause is always at the end of the package specification. The private clause is visible to the compiler but not to client program units.

8. Q: What is the difference between private and limited private types

in Ada?

A: Types that are declared to be private are called private types. Private data types have built-in operations for assignment and comparisons for equality and inequality. Any other operation must be declared in the package specification. An alternative to private types is a more restricted form: limited private types. Nonpointer limited private types are described in the private section of a package specification, as are nonpointer private types. The only syntactic difference is that limited private types are declared to be limited private in the visible part of the package specification. that defined the type.

9. Q: What is in an Ada package specification? What about a body package?

A: The encapsulating constructs in Ada are called packages. A package can have two parts, each of which is also is called a package. These are called the package specification, which provides the interface of the encapsulation (and perhaps more), and the body package, which provides the implementation of most, if not all, of the entities named in the associated package specification.

10. Q: What is the use of the Ada with clause?

A: The with clause makes the names defined in external packages visible; in this case Ada.Text_IO, which provides functions for input and output of text.

Problem Set:

6. Q: Discuss the advantages of C# properties, relative to writing accessor

methods in C++ or Java.

A: One of the advantages of C# properties relative writing accessor methods in C++ or Java is it can control access to the fields

7. Q: Explain the dangers of C’s approach to encapsulation.

A: The main problem is that the biggest part of encapsulation is done via hiding, rather than protection. This is achieved through definition hiding: a header file is preprocessed (which is a synonym for copy-pasted) into the implementation file. Anyone with this header file will be able to access any method or public variable of a the client related to the header, left apart any "static" method / variable.

8. Q: Why didn’t C++ eliminate the problems discussed in Problem 7?

A: Because in C++ these problems can be handled by allowing nonmember functions to be “friends” of a class. Friend functions have access to the private entities of the class where they are declared to be friends. However, it didn't eliminate the problem because it evolved from C. Hence, it kept a lot of backward compatibility, and the same way of doing basic things. While some problems where solved (like the protected access, which is in-between normal and static in C), some stay, as the symbol access using wrong datatype (inherent to the linker, which doesn't do type-checking).

9. Q: What are the advantages and disadvantages of the Objective-C approach to syntactically distinguishing class methods from instance methods?

A: Instance methods use an instance of a class, whereas a class method can be used with just the class name. A class is like the blueprint of a house: You only have one blueprint and (usually) you can't do that much with the blueprint alone. An instance (or an object) is the actual house that you build based on the blueprint: You can build lots of houses from the same blueprint. You can then paint the walls a different color in each of the houses, just as you can independently change the properties of each instance of a class without affecting the other instances.

10. Q: In what ways are the method calls in C++ more or less readable than

those of Objective-C?

A: In Objective C, all method calls are essentially virtual. This makes it a bit easier on the programmer, but it comes at a possible performance decrease. Hence sometimes methods call in C++ can be more or less readable than those of Objective-C.

Q&A Assignment #12 Chapter 10 Review Questions & Problem Set

Name : Vina Melinda

NIM  : 1801380106

Untuk kali ini saya akan menjawab Assignment #12 Problem Set dan Review Questions dari Chapter 10 dari Sebesta.


Review Question:


6. Q: What is the difference between an activation record and an activation

record instance?

A: An activation record is the format, or layout, of the moncode part of a subprogram, whereas an activation record instance is a concrete example of an activation record, a collection of data in the form of an activation record.

7. Q: Why are the return address, dynamic link, and parameters placed in the

bottom of the activation record?

A: because the entried must appear first.

8. Q: What kind of machines often use registers to pass parameters?

A: RISC machines, parameters are passed in registers.

9. Q: What are the two steps in locating a nonlocal variable in a static-scoped

language with stack-dynamic local variables and nested subprograms?

A:

10. Q: Define static chain, static_depth, nesting_depth, and chain_offset.

A:

Problem Set:

6. Q: Although local variables in Java methods are dynamically allocated at the
beginning of each activation, under what circumstances could the value
of a local variable in a particular activation retain the value of the previous
activation?
A: If the variable is declared as static. Static modifier is a modifier that makes a variable history – sensitive.

7. Q: It is stated in this chapter that when nonlocal variables are accessed in a
dynamic-scoped language using the dynamic chain, variable names must
be stored in the activation records with the values. If this were actually
done, every nonlocal access would require a sequence of costly string
comparisons on names. Design an alternative to these string comparisons
that would be faster.
A: Using approach that uses an auxiliary data structure called a display. Or, to write variable names as integers. These integers act like an array. So when the activation happens, the comparisons will be faster.

8. Q: Pascal allows gotos with nonlocal targets. How could such statements

be handled if static chains were used for nonlocal variable access? Hint:

Consider the way the correct activation record instance of the static parent

of a newly enacted procedure is found (see Section 10.4.2).

A: Based on the hint statement, the target of every goto in a program could be represented as an address and a nesting_depth, where the nesting_depth is the difference between the nesting level of the procedure that contains the goto and that of the procedure containing the target. Then, when a goto is executed, the static chain is followed by the number of links indicated in the nesting_depth of the goto target. The stack top pointer is reset to the top of the activation record at the end of the chain.

9. Q: The static-chain method could be expanded slightly by using two static

links in each activation record instance where the second points to the

static grandparent activation record instance. How would this approach

affect the time required for subprogram linkage and nonlocal references?

A: Including two static links would reduce the access time to nonlocals that are defined in scopes two steps away to be equal to that for nonlocals that are one step away. Overall, because most nonlocal references are relatively close, this could significantly increase the execution efficiency of many programs.

10. Q: Design a skeletal program and a calling sequence that results in an activation

record instance in which the static and dynamic links point to different

activation-recorded instances in the run-time stack.

A:
\emph{Answer}:\\
procedure Main\_2 is\\
\verb+ + X : Integer;\\
\verb+ +procedure Bigsub is\\
\verb+ +\verb+ + A, B, C : Integer;\\
\verb+ +\verb+ + procedure Sub1 is\\
\verb+ +\verb+ +\verb+ + A, D : Integer;\\
\verb+ +\verb+ +\verb+ + begin -- of Sub1\\
\verb+ +\verb+ +\verb+ + A := B + C; $\longleftarrow$ 1\\
\verb+ +\verb+ +\verb+ + ...\\
\verb+ + end; -- of Sub1\\
\verb+ + procedure Sub2(X : Integer) is\\
\verb+ +\verb+ + B, E : Integer;\\
\verb+ +\verb+ + procedure Sub3 is\\
\verb+ +\verb+ +\verb+ + C, E : Integer;\\
\verb+ +\verb+ +\verb+ + begin -- of Sub3\\
\verb+ +\verb+ +\verb+ + ...\\
\verb+ +\verb+ +\verb+ + Sub1;\\
\verb+ +\verb+ +\verb+ + ...\\
\verb+ +\verb+ +\verb+ + E := B + A; $\longleftarrow$ 2\\
\verb+ +\verb+ + end; -- of Sub3\\
\verb+ +\verb+ + begin -- of Sub2\\
\verb+ +\verb+ + ...\\
\verb+ +\verb+ + Sub3;\\
\verb+ +\verb+ + ...\\
\verb+ +\verb+ + A := D + E; $\longleftarrow$ 3\\
\verb+ + end; -- of Sub2\\
\verb+ + begin -- of Bigsub\\
\verb+ +\verb+ + ...\\
\verb+ +\verb+ + Sub2(7);\\
\verb+ +\verb+ + ...\\
\verb+ + end; -- of Bigsub\\
begin -- of Main\_2\\
\verb+ + ...\\
\verb+ + Bigsub;\\
\verb+ + ...\\
end; -- of Main\_2\\
\\
The sequence of procedure calls is:\\
Main\_2 calls Bigsub\\
Bigsub calls Sub2\\
Sub2 calls Sub3\\
Sub3 calls Sub1\\
\\
The activation records with static and dynamic links is as follows:\\
\begin{figure}
\centering
\includegraphics[scale=0.5]{ari}

\end{figure}

Q&A Assignment #11 Chapter 9 Review Questions & Problem Set

Name : Vina Melinda

NIM  : 1801380106

Untuk kali ini saya akan menjawab Assignment #11 Problem Set dan Review Questions dari Chapter 9 dari Sebesta.

Review Questions:

6. Q: What is a Ruby array formal parameter?

A: Ruby supports a complicated but highly flexible actual parameter configuration. The initial parameters are expressions, whose value objects are passed to the corresponding formal parameters. The initial parameters can be following by a list of key => value pairs, which are placed in an anonymous hash and a reference to that hash is passed to the next formal parameter. These are used as a substitute for keyword parameters, which Ruby does not support. The hash item can be followed by a single parameter preceded by an asterisk. This parameter is called the array formal parameter.

7. Q: What is a parameter profile? What is a subprogram protocol?
A: Parameter profile of a subprogram contains the number, order, and types of its formsl parameter. A protocol of subprogam is ite parameter profile plus, if it is a function, its return type。

8. Q: What are formal parameters? What are actual parameters?
A: Formal parameter is the parameter in the subprogram header. They are sometimes thought of as dummy variables because they are not variable in the usual case. Actusl parameter is a subprogram call statement that must inckude the nameof the subprogram and a list of parameters to be bound to the formal parameter of the subprogram.

9. Q: What are the advantages and disadvantages of keyword parameters?
A: The advantage of keyword parameters is that they can appear in any order in the actual parameter list. The disadvantage to keyword parameters is that the user of the subprogram must know the names of formal parameters.

10. Q: What are the differences between a function and a procedure?
A: A function returns a value and a procedure just executes commands. The name function comes from math. It is used to calculate a value based on input. A procedure is a set of command which can be executed in order. In most programming languages, even functions can have a set of commands. Hence the difference is only in the returning a value part. But if you like to keep a function clean, (just look at functional languages), you need to make sure a function does not have a side effect.

Problem Set:

6. Q: Present one argument against providing both static and dynamic local
variables in subprograms.
A: In subprograms local variables can be static or dynamic. If local variable treated statically, this allows for compile-time allocation or deallocation and ensures proper type checking but doesn't allow recursion. However, if local variables treated dynamically, this allows for recursion at the cost of run-time allocation or deallocation and initialization because these are stored on a stack, referencing is indirect based on stack position and possibly time consuming.

7. Q:Consider the following program written in C syntax:
void fun (int first, int second) {
first += first;
second += second;
}
void main() {
int list[2] = {1, 3};
fun(list[0], list[1]);
}
For each of the following parameter-passing methods, what are the values
of the list array after execution?
a. Passed by value
b. Passed by reference
c. Passed by value-result
A: a. 1, 3
b. 2, 6
c. 2, 6

8. Q: Argue against the C design of providing only function subprograms.
A: If a language only provides functions, then programmers must live with the restriction of returning only one result from any subprogram or functions must allow side effects, which is considered bad.

9. Q: From a textbook on Fortran, learn the syntax and semantics of statement
functions. Justify their existence in Fortran.
A: The Fortran 90 standard (ISO/IEC 1539: 1991) describes the syntax and semantics of a programming language. However, the standard addresses certain aspects of the Fortran processing system, but does not address others. When specifications are not covered by the standard, the interpretation is processor dependent; that is, the processor defines the interpretation, but the interpretation for any two processors need not be the same. Programs that rely on processor-dependent interpretations typically are not portable. (Jeanne C. Adams, Walter S. Brainerd, et. al., 1992)

10. Q: Study the methods of user-defined operator overloading in C++ and Ada,
and write a report comparing the two using our criteria for evaluating
languages.
A: C was downgraded because it is essentially limited to syntactic (re)naming of predefined data types and data structures, with building-block capabilities within these limitations. Unlike Ada, neither C nor C++ allows restrictions to the range of a type. The C++ extensions to C in this category are the most significant enhancement. A class in C++ is a user-defined data type that provides data hiding, guaranteed initialization of data, implicit type conversion for user-defined types, dynamic typing, user-controlled memory management, and mechanisms for overloading operators. Ada has been downgraded slightly by the SEI for lacking a mechanism like classes that permits programming by specialization and extension. (Nelson H. Weiderman, 1991)

Q&A Assignment #10 Chapter 8 Review Questions & Problem Set

Name : Vina Melinda

NIM  : 1801380106

Untuk kali ini saya akan menjawab Assignment #10 Problem Set dan Review Questions dari Chapter 8 dari Sebesta.

Review Questions:

6. Q: What is unusual about Python’s design of compound statements?

A:  Python uses indentation to specify compound statements. For example,if x > y :
x = y
print “case 1″
equally indent statements are grouped as one compound statement.

7. Q: Under what circumstances must an F# selector have an else clause?

A: If the expression returns a value, it must have an else clause

8. Q: What are the common solutions to the nesting problem for two-way
selectors?
A: The common solution to the nesting problem is to use alternating means of forming a compound statements.

9. Q: What are the design issues for multiple-selection statements?

A:
1. What is the form and type of the expression that controls the selection?• How are the selectable segments specified?
2. Is execution flow through the structure restricted to include just a single selectable segment?
3. How are the case values specified?
4. How should unrepresented selector expression values be handled, if at all?

10. Q: Between what two language characteristics is a trade-off made when
deciding whether more than one selectable segment is executed in one
execution of a multiple selection statement?

A: The C# switch statement differs from that of its C-based predecessors in two ways. First, C# has a static semantics rule that disallows the implicit execution of more than one segment. The rule is that every selectable segment must end with an explicit unconditional branch statement: either a break,which transfers control out of the switch statement, or a goto, which can transfer control to one of the selectable segments (or virtually anywhere else).
As with C, if there is no break at the end of the selected segment, execution continues into the next segment.

Problem Set:

6. Q: Analyze the potential readability problems with using closure reserved
words for control statements that are the reverse of the corresponding
initial reserved words, such as the case-esac reserved words of
ALGOL 68. For example, consider common typing errors such as transposing
characters.

A: There's not really a readability problem, it's simply something you get used to, and after some experience the problems falls out of the process. Similar to how many Lisp users "don't see" the parentheses. They simply don't stand out in the general case for the experienced reader.
You have to recall the time of Algol, notably the "68" part, as in 1968.The bright side of the fi, esac, and od is that they clearly indicate what kind of block they're terminating, and they do it with a single token. Esac is no less clear than }, which is a meaningless bracket until you know otherwise. The {} have the benefit of consistency, while less wordy than Pascals begin - end generic block sequence.Finally, consider how dominant the English language is in computer language design, and while folks who don't speak english may have some initial issues with the languages, that clearly passes over time.So, it's a short hurdle that falls away quickly with use.

7. Q: Use the Science Citation Index to find an article that refers to Knuth
(1974). Read the article and Knuth’s paper and write a paper that summarizes
both sides of the goto issue.

A: Considering the time frame of the early to mid seventies, Knuth’s description of an advantage of the use of gotos was well understood. In fact C language which was created in the late ’60s had already dealt with the problem you’ve given as an example. With nested loops and conditions calling for exiting those loops from deep within the nesting, gotos are more efficient and even more elegant than multiple test flags and testing statements. So C incorporated the “break” statement to handle that problem. Redefining “goto” in a situation where its use is valuable to something more restricted and directed to the problem was the answer. An answer given years before.

8. Q: In his paper on the goto issue, Knuth (1974) suggests a loop control
statement that allows multiple exits. Read the paper and write an operational
semantics description of the statement

A: Work on the formal semantics of programming languages began in the 1960’s
– a useful early reference is which reports on a conference held in Badenbei-Wien in 1964. The subsequent literature on formal semantics of sequential languages is extensive. A good state of the art example of a formal definition is that of standard ML. This definition is written in Structured
Operational Semantics. It is sobering that, after a quarter century of denotational semantics, it is still found more convenient to tackle the semantics of a language like SML in an operational way. Furthermore, it must be clear that it is extremely difficult to get a formal semantics to the stage where it correctly reflects the intuitions about a language. There are some language standards like that for Modula 2 which are actually being written using formal techniques. But overall the situation is that formal semantic definitions are written only by a very small number of highly skilled people.
The situation with recording the formal semantics

9. Q: What are the arguments both for and against the exclusive use of Boolean expressions in the control statements in Java (as opposed to also
allowing arithmetic expressions, as in C++)?

A: The primary argument for using Boolean expressions exclusively as control expressions is the reliability that results from disallowing a wide range of types for this use. In C, for example, an expression of any type can appear as a control expression, so typing errors that result in references to variables of incorrect types are not detected by the compiler as errors.No , it would not be a good idea. Although this custom precedence sounds like increasing flexibility, requiring parentheses to show a custom precedence would impact in readability and writability of a program.

10. Q: In Ada, the choice lists of the case statement must be exhaustive, so that there can be no unrepresented values in the control expression. In C++,
unrepresented values can be caught at run time with the default selector.
If there is no default, an unrepresented value causes the whole
statement to be skipped. What are the pros and cons of these two designs
(Ada and C++)?

A: In Ada, the choice lists of the "case" statement must be exhaustive, so that there can be no unrepresented values in the control expression.
In C++, unrepresented values can be caught at run time with the "default" selector. If there is no default, an unrepresented value causes the whole statement to be skipped.
Ada was designed for military grade software development.
The idea is that whenever you modify code in such a way that a new case emerges (for example adding a new value for an enumeration type), you are forced to manually revisit (and therefore re-validate) all the case statements that analyze it. Having a "default" is risky: you may forget that there is a case somewhere where the new case should not have been handled by the default.

Q&A Assignment #9 Chapter 7 Review Questions & Problem Set

Name : Vina Melinda

NIM  : 1801380106

Untuk kali ini saya akan menjawab Assignment #9 Problem Set dan Review Questions dari Chapter 7 dari Sebesta.

Review Questions:

6. Q: What associativity rules are used by APL?

A: Associativity rules are used by APL: all operators have equal precedence and all operators associate right to left.

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

A: All of the arithmetic, relational, and assignment operators, as well as array indexing, shifts, and bitwise logic operators, areimplemented as methods in Ruby

8. Q: Define functional side effect.

A: It is a side effect of a function. Occurs when the function changes either one of its parameters or a global variable.

9. Q: What is a coercion?

A: Coercion or Implicit type conversion, is an automatic type conversion by the compiler. Some languages allow, some require, compilers to provide coercion

10. Q: What is a conditional expression?

A: A conditional expression is a compound expression that contains a condition that is implicity converted to type bool in C++ (operand1), an expression to be evaluated if the condition evaluates to true (operand2), and an expression to be evaluated if the condition has the value false (operand3).

Problem Set:

6. Q: 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?

A: 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. Q: Describe a situation in which the add operator in a programming language
would not be commutative.

A: In many programming languages, the operands of an addition are actually memory locations, and the result of an addition is stored in the same location as one of the operands, so the operation is not commutative. A+B may not give quite the same result as B+A, because of the storage location of the result.

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

A: Consider the integer expression A + B + C. Suppose the values of A, B, and C are 20,000, 25,000, and -20,000, respectively. Further suppose that the machine has a maximum integer value of 32,767. If the first addition is computed first, it will result in overflow. If the second addition is done first, the whole expression can be correctly computed.

9. Q: Assume the following rules of associativity and precedence for
expressions:
Precedence           Highest                      *, /, not
                                                                 +, –, &, mod
                                                                 – (unary)
                                                                 =, /=, < , <=, >=, >
                                                                 and
                              Lowest                        or, xor
Associativity         Left to right

Show the order of evaluation of the following expressions by parenthesizing
all subexpressions 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
b. a * (b - 1) / c mod d
c. (a - b) / c & (d * e / a - 3)
d. -a or c = d and e
e. a > b xor c or d <= 17

f. -a + b

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

10. Q: 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) ( 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