Chapter 13 Answers

Nama: Christian Gunawan
NIM: 1801384174

Kali ini saya akan menjawab Assignment #13 dari Chapter 13 Programming Language Concepts R Sebesta :


Review Questions #6-10 :

6. Describe the logical architecture of a vector processor.

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. What is the difference between physical and logical concurrency?

Physical concurrency – Multiple independent processors (multiple threads of control)
Logical concurrency – 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. What is a thread of control in a program?

A thread of control in a program is the sequence of program points reached as control flows through the program.


9. Why are coroutines called quasi-concurrent?
Because they have a single thread of control.


10. What is a multithreaded program?
A program designed to have more than one thread of control.


Problem Set #6-10 :

6. 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.

The idea here is that 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. Compare the Java competition synchronization mechanism with that of Ada.

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. Accept and select statements are the two main features of the language that deal with the issue of synchronization This paper points out one major problem that arises in connection with these features and proposes a possible solution to it.


8. Compare the Java cooperation synchronization mechanism with that of Ada.

Cooperation synchronization in Java is implemented with the wait, notify, and notifyAll methods, all of which are defined in Object, the root class of all Java classes. All classes except Object inherit these methods. Every object has a wait list of all of the threads that have called wait on the object.
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. What happens if a monitor procedure calls another procedure in the same monitor?

Because the access mechanisms are part of the monitor, implementation of a monitor can be made to guarantee synchro- nized 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. Explain the relative safety of cooperation synchronization using semaphores and using Ada’s when clauses in tasks.

As soon as we start using concurrent threads, we need to think about various issues that fall under the broad description of thread safety. Generally, 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.

Chapter 12 Answers

Nama: Christian Gunawan
NIM: 1801384174

Kali ini saya akan menjawab Assignment #12 dari Chapter 12 Programming Language Concepts R Sebesta :

Review Questions #6-10 :

6. Describe a situation where dynamic binding is a great advantage over its absence.

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. With overriding, we can directly modify B’s draw function. But without it, we either make a specific function in A for B and inherit it.


7. What is a virtual method?

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. What is an abstract method? What is an abstract class?

An abstract method is a method which all descendant classes should have. An abstract class is a class which has abstract method.


9. Describe briefly the eight design issues used in this chapter for object-oriented languages.

-What non-objects are in the language?
-Are Subclasses Subtypes? If so, derived objects can be legally used wherever a parent object could be used.
-Type Checking and Polymorphism
-Single and Multiple Inheritance. Inherit from 1 (or more than 1) parent.
-Object Allocation and Deallocation. Are objects allocated from heap or stack.
-Dynamic and Static Binding. When are messages bound to methods, before or during run-time?
-Nested Classes. Can a class be nested inside another class?
-Initialization of Objects. Are objs init'd when created? Implicit or explicit?


10. What is a nesting class?

A nesting class is a class defined inside another class.


Problem Set #6-10 :

6. Compare the multiple inheritance of C++ with that provided by interfaces in Java.

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. What is one programming situation where multiple inheritance has a significant advantage over interfaces?

When two or more parent classes are derived from one grandparent class and has one child (diamond problem).


8. Explain the two problems with abstract data types that are ameliorated by inheritance.

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. Describe the categories of changes that a subclass can make to its parent class.

Subclasses can add things (variables, methods). Subclass in C++ can effectively remove a method using "private" inheritance. Inherited methods can be overridden.


10. Explain one disadvantage of inheritance.

Language & implementation complexity. The shared inheritance problem of multiple inheritance. Subclass is dependent upon its base class (which might change over time).

Chapter 11 Answers

Nama: Christian Gunawan
NIM: 1801384174

Kali ini saya akan menjawab Assignment #11 dari Chapter 11 Programming Language Concepts R Sebesta :

Review Questions #6-10 :

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

There are two approaches to hiding the representation from clients in the package specification. The first one is to include two sections in the package specification, the second one is in which entities are visible to clients and one that hides its contents.

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

The private part of an Ada package specification is visible to the compiler but not to the client programs.


8. What is the difference between private and limited private types in Ada?

Private types have built-in operations for assignment and comparison. Limited private types don’t have built-in operations.


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

Package specification, is an Ada package which provides the interface of the encapsulation (and perhaps more). Body package, is an Ada package which provides the implementation of most, if not all, of the entities named in the associated package specification.


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

The use of Ada with clause is to make the names defined in external packages visible.


Problem Set #6-10 :

6. Discuss the advantages of C# properties, relative to writing accessor methods in C++ or Java.
One of the advantage of C# properties relative writing accessor methods in C++ or Java is it can control access to the fields.


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

The main problem is that the biggest part of encapsulation is done via hidding, rather than protection. This hidding is achieved through definition hidding: 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 appart any "static" method / variable. Static is actually the only rue level of protection here, as it's the only one that another unit (file) would not be able to access even if it's aware of it's existence. This whole "protection-is-name-hiding" approach leads to a load of problems: you can access a symbol using the wrong datatype and your compiler will happily do so. To protect critical parts, you should rely on text being hidden from the compiler while it's processing certain units (via #ifdef / #define).


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

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

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. In what ways are the method calls in C++ more or less readable than those of Objective-C?

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. So sometimes methods call in C++ can be more or less readable than those of Objective-C.

Chapter 10 Answers

Nama: Christian Gunawan
NIM: 1801384174

Kali ini saya akan menjawab Assignment #10 dari Chapter 10 Programming Language Concepts R Sebesta :

Review Questions#6-10 :

6. What is the difference between an activation record and an activation record instance?

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


7. Why are the return address, dynamic link, and parameters placed in the bottom of the activation record?

*It's because the entry must appear first.


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

*RISC Machines often use registers to pass parameters.


9. What are the two steps in locating a nonlocal variable in a static-scoped language with stack-dynamic local variables and nested subprograms?

*First step, find correct activation record (the harder part) and then the second step is determine the offset within that activation record (easy part).


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

*Static chain is chain of static links connecting an activation record to all of it's static ancestors (it's enclosing subprograms).
Static depth is depth of the nesting for each enclosing static scope.
Nesting depth is the difference between the static depth of the reference and that of the scope where it was declared.
Chain offset is same as nesting depth.

Problem Set #6-10 :

6. 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?

*Each activation allocates variables in exactly the same order. Variables are not initialized to any value unless the program contains an initialization statement for the variable – they simply have whatever value is stored in the location they are allocated. If a procedure finishes executing, returns, and is immediately reinvoked, a variable would be assigned the same stack location it had on the previous invocation, and would have the last value from that previous invocation.


7. 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.

*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. 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).

*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. 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?

*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. 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.

*\\
\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}


At position 1 in procedure Sub1, the reference is to the local variable,
A, not to the nonlocal variable A from Bigsub. This reference to A has the
chain\_offset/local\_offset pair (0, 3). The reference to B is to the nonlocal B
from Bigsub. It can be represented by the pair (1, 4). The local\_offset is 4,
because a 3 offset would be the first local variable (Bigsub has no parameters). Notice that if the dynamic link were used to do a simple search for
an activation record instance with a declaration for the variable B, it would
find the variable B declared in Sub2, which would be incorrect. If the (1, 4)
pair were used with the dynamic chain, the variable E from Sub3 would be
used. The static link, however, points to the activation record for Bigsub,
which has the correct version of B . The variable B in Sub2 is not in the
referencing environment at this point and is (correctly) not accessible. The
reference to C at point 1 is to the C defined in Bigsub, which is represented
by the pair (1, 5).\\
\\

\noindent

Chapter 9 Answers

Nama : Christian Gunawan
NIM : 1801384174

Kali ini saya akan menjawab assigntment #9 dari chapter 9 Programming Language Concepts R Sebesta E-book :

Review Questions #6-10 :

6.What is a Ruby array formal parameter?

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. 

7.What is a parameter profile? What is a subprogram protocol?

The parameter profile of a subprogram contains the number, order, and types of its formal parameters. The protocol of a subprogram is its parameter profile plus, if it is a function, its return type. In languages in which subprograms have types, those types are defined by the subprogram’s protocol.

8.What are formal parameters? What are actual parameters?

The parameters in the subprogram header are called formal parameters. They are sometimes thought of as dummy variables because they are not variables in the usual sense: In most cases, they are bound to storage only when the subprogram is called, and that binding is often through some other program variables. While actual parameters is a parameter that bound to the formal parameters of the subprogram. Subprogram call statements must include the name of the subprogram and a list of parameters to be bound to the formal parameters of the subprogram.

9.What are the advantages and disadvantages of keyword parameters?

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.What are the differences between a function and a procedure?

There are two distinct categories of subprograms—procedures and functions— both of which can be viewed as approaches to extending the language. All sub- programs are collections of statements that define parameterized computations. Functions return values and procedures do not. Procedures can produce results in the calling program unit by two methods: (1) If there are variables that are not formal parameters but are still visible in both the procedure and the calling program unit, the procedure can change them; and (2) if the procedure has formal parameters that allow the transfer of data to the caller, those parameters can be changed. However Functions are called by appearances of their names in expressions, along with the required actual parameters. And functions define new user-defined operators.

Problem Sets #6-10 :

6.Present one argument against providing both static and dynamic local variables in subprograms.

In subprograms local variables can be static or dynamic; If local variable treated statically: This allows for compile-time allocation/ deallocation and ensures proper type checking but does not allow for recursion. And if local variables treated dynamically: This allows for recursion at the cost of run-time allocation/ deallocation and initialization because these are stored on a stack, referencing is indirect based on stack position and possibly timeconsuming

7. 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?
Passed by value
Passed by reference
Passed by value-result
Passed by value : list[2] = { 3 , 5 }
Passed by reference : list[2] = { 6 , 10 }
Passed by value-result : list[2] = { 6 , 10 }

8.Argue against the C design of providing only function subprograms.

If a language provides only functions, then either programmers must live with the restriction of returning only a single result from any subprogram, or functions must allow side effects, which is generally considered bad. Since having subprograms that can only modify a single value is too restrictive, C’s choice is not good.

9.From a textbook on Fortran, learn the syntax and semantics of statement functions. Justify their existence in Fortran.

The Fortran 1966 standard provided a reference syntax and semantics, but vendors continued to provide incompatible extensions. These standards have improved portability.

10.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.

One of the nice features of C++ is that you can give special meanings to operators, when they are used with user-defined classes. This is called operator overloading. You can implement C++ operator overloads by providing special member-functions on your classes that follow a particular naming convention. For example, to overload the + operator for your class, you would provide a member-function named operator+ on your class. Meanwhile for Ada, since much of the power of the language comes from its extensibility, and since proper use of that extensibility requires that we make as little distinction as possible between predefined and user-defined types, it is natural that Ada also permits new operations to be defined, by declaring new overloading of the operator symbols.

Chapter 8 Answers

Nama : Christian Gunawan
NIM : 1801384174

Kali ini saya akan menjawab assigntment #8 dari chapter 8 Programming Language Concepts R Sebesta E-book :


Review Questions #6-10 :
6. What is unusual about Python’s design of compound statements?

Ada, Python, and Ruby, the then and else clauses are statement sequences, rather than compound statements. The complete selection statement is terminated in these languages with a reserved word.1
Python uses indentation to specify compound statements. For example,

If x > y :
x = y
print "case 1"

All statements equally indented are included in the compound statement.2 Notice that rather than then, a colon is used to introduce the then clause in Python.


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

An F# if need not return a value, for example if its clause or clauses create side effects, perhaps with output statements. However, if the if expression does return a value, as in the example above, it must have an else clause.


8. What are the common solutions to the nesting problem for two-way selectors?

The else clause would be paired with the second then clause. The disadvantage of using a rule rather than some syntactic entity is that although the programmer may have meant the else clause to be the
alternative to the first then clause and the compiler found the structure syntactically correct, its semantics is the opposite.

Another way to avoid the issue of nested selection statements is to use an alternative means of forming compound statements.


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

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


10. 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?

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 #6-10 :
6. 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.


7. 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.


8. 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.


9. 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++)?

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. 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++)?

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

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