Archive for November 2014
Chapter 5
Name : Romy Hermawan
NIM : 1801378902
Class : LD001
Lecturer: Ir. Tri Djoko Wahjono,M.Sc
REVIEW QUESTION
6. What is the l-values of a variables? What is the r-values?
Answer:
The l-values of a variable is its address while he r-values of a variable is its value.
7. Define binding and binding time.
Answer:
Binding is an association between an attribute and an entity, such as between a variable and its type/ value, or between an operation and a symbol. Binding time is the time at which a binding takes place.
8. After language design and implementation [what are the four times bindings can take place in a program?]
Binding is an association between an attribute and an entity, such as between a variable and its type/ value, or between an operation and a symbol. Binding time is the time at which a binding takes place.
8. After language design and implementation [what are the four times bindings can take place in a program?]
Answer:
compile time: a variable in java bound to a particular data type
load time: a variable bound to a storage cell when a program is loaded into memory
link time: a call to a library subprogram is bound to the subprogram code
run time: certain variables declared in pascal and in c++ functions
9. Define static binding and dynamic binding.
compile time: a variable in java bound to a particular data type
load time: a variable bound to a storage cell when a program is loaded into memory
link time: a call to a library subprogram is bound to the subprogram code
run time: certain variables declared in pascal and in c++ functions
9. Define static binding and dynamic binding.
Answer:
A static binding is if it first occurs before run time and remains unchanged throughout program execution. A dynamic binding is if it first occurs during execution or can change during execution of the program.
10. What are the advantages and disadvantages of implicit declarations?
A static binding is if it first occurs before run time and remains unchanged throughout program execution. A dynamic binding is if it first occurs during execution or can change during execution of the program.
10. What are the advantages and disadvantages of implicit declarations?
Answer:
The advantage is write ability.
The advantage is write ability.
PROBLEM SET
6. Consider the following JavaScript skeletal program:
// The main program
var x;
function sub1() {
var x;
function sub2() {
. . .
}
}
function sub3() {
. . .
}
Assume that the execution of this program is in the following unit order:
main calls sub1
sub1 calls sub2
sub2 calls sub3
a. Assuming static coping, in the following, which declaration of x is the correct one for a reference to x?
i. Sub1
ii. Sub2
iii. Sub3
b. Repeat part , but assume dynamic scoping.
i. Sub1
ii. Sub2
iii. Sub3
b. Repeat part , but assume dynamic scoping.
Answer:
a. Static :
1. Sub1 : sub1
2. Sub2 : sub1
3. Sub3: main
b. Dynamic :
1. Sub1 : sub1
2. Sub2 : sub1
3. Sub3: sub1
1. Sub1 : sub1
2. Sub2 : sub1
3. Sub3: main
b. Dynamic :
1. Sub1 : sub1
2. Sub2 : sub1
3. Sub3: sub1
7. Assume the following JavaScript program was interpreted using
static-scoping rules. What value of x is displayed in function sub1?
Under dynamic-scoping rules, what value of x is displayed in function
sub1?
var x;
function sub1() {
document.write("x = " + x + "<br />");
}
function sub2() {
var x;
x = 10;
sub1();
}
x = 5;
sub2();
Answer:
Static scope: x=5, Dynamic scoping: x=10.
8. Consider the following JavaScript:
var x, y, z;
function sub1() {
var a, y, z;
function sub2() {
var a, b, z;
...
}
...
}
function sub3() {
var a, x, w;
...
}
List all the variables, along with the program units where they are declared, that are visible in the bodies of sub1, sub2 and sub3, assuming static scoping is used.
Answer:
Sub1: a(sub1), y(sub1), z(sub1), x(main).
Sub2: a(sub2), b(sub2), z(sub2), y(sub1), x(main)
Sub3: a(sub3), x(sub3), w(sub3), y(main), z(main)
9. Consider the following Python program:
x = 1;
y = 3;
z = 5;
def sub1():
a = 7;
y = 9;
z = 11;
. . .
def sub2():
global x;
a = 13;
x = 15;
w = 17;
. . .
def sub3():
nonlocal a;
a = 19;
b = 21;
z = 23;
. . .
. . .
List all the variables, along with the program units where they are
declared, that are visible in the bodies of sub1, sub2, and sub3, assumingstatic scoping is used.
x = 1;
y = 3;
z = 5;
def sub1():
a = 7;
y = 9;
z = 11;
. . .
def sub2():
global x;
a = 13;
x = 15;
w = 17;
. . .
def sub3():
nonlocal a;
a = 19;
b = 21;
z = 23;
. . .
. . .
List all the variables, along with the program units where they are
declared, that are visible in the bodies of sub1, sub2, and sub3, assumingstatic scoping is used.
Answer:
point 1 : x = 1(main), y = 9 (sub1), z = 11(sub1) ,a = 7(sub1);
point 2 : x =15(sub2), w = 17(sub2), a = 13(sub2), y = 9(sub1);
point 3 : x = 15(sub2), b = 21(sub3), a = 19(sub1), z = 23(sub3), w = 17(sub 2);
point 4 : x = 15(sub2), b = 21(sub3), a = 19(sub1), z = 23(sub3), w = 17(sub 2);
point 2 : x =15(sub2), w = 17(sub2), a = 13(sub2), y = 9(sub1);
point 3 : x = 15(sub2), b = 21(sub3), a = 19(sub1), z = 23(sub3), w = 17(sub 2);
point 4 : x = 15(sub2), b = 21(sub3), a = 19(sub1), z = 23(sub3), w = 17(sub 2);
10. Consider the following C program:
void fun(void) {
int a, b, c; /* definition 1 */
. . .
while (. . .) {
int b, c, d; /*definition 2 */
. . . 1
while (. . .) {
int c, d, e; /* definition 3 */
. . . 2
}
. . . 3
}
. . . 4
}
For each of the four marked points in this function, list each visible variable,
along with the number of the definition statement that defines it.
int a, b, c; /* definition 1 */
. . .
while (. . .) {
int b, c, d; /*definition 2 */
. . . 1
while (. . .) {
int c, d, e; /* definition 3 */
. . . 2
}
. . . 3
}
. . . 4
}
For each of the four marked points in this function, list each visible variable,
along with the number of the definition statement that defines it.
Answer:
Point 1: a:1, b:2, c:2, d:2
Point 2: a:1, b:2, c:3, d:3, e:3
Point 3: a:1, b:2, c:2, d:2
Point 1: a:1, b:2, c:2, d:2
Point 2: a:1, b:2, c:3, d:3, e:3
Point 3: a:1, b:2, c:2, d:2
Chapter 6
Name : Romy Hermawan
NIM : 1801378902
Class : LD001
Lecturer: Ir. Tri Djoko Wahjono,M.Sc
REVIEW QUESTION:
6. What are the advantages of user-defined enumeration types?Answer:
Improved writability and readibility, No arithmetic operations are legal on enum types and no enumeration types can be assigned values outside its defined range.
7. In what ways are the user-defined enumeration types of C# more reliable than those of C++?
Answer:
C# enumeration types are like those of C++, except that they are never coerced to integer. So, operations on enumeration types are restricted to those that make sense. Also, the range of values is restricted to that of the particular enumeration type.
8.What are the design issues for array
9.What happens when an nonexistent element of an array is referenced in Perl ?
10. What happens when a nnexistent element of an array is referenced in Perl?
7. In what ways are the user-defined enumeration types of C# more reliable than those of C++?
Answer:
C# enumeration types are like those of C++, except that they are never coerced to integer. So, operations on enumeration types are restricted to those that make sense. Also, the range of values is restricted to that of the particular enumeration type.
8.What are the design issues for array
Answer:
What types are legal for subscripts?
Are subscripting expressions in element references range checked?
When are subscript ranges bound?
When does array allocation take place?
Are ragged or rectangular multidimensioned arrays allowed, or both?
Can arrays be initialized when they have their storage allocated?
What kinds of slices are allowed, if any?
What types are legal for subscripts?
Are subscripting expressions in element references range checked?
When are subscript ranges bound?
When does array allocation take place?
Are ragged or rectangular multidimensioned arrays allowed, or both?
Can arrays be initialized when they have their storage allocated?
What kinds of slices are allowed, if any?
9.What happens when an nonexistent element of an array is referenced in Perl ?
Answer:
– If you try to append non-existent elements from an array to another one, the initial array will grow as needed, even though the elements to append do not exist.10. What happens when a nnexistent element of an array is referenced in Perl?
Answer:
A reference to a nonexistent ele-ment in Perl yields undef, but no error is reported.
A reference to a nonexistent ele-ment in Perl yields undef, but no error is reported.
SET PROBLEM:
6. Explain all of the differences between Ada’s subtypes and derived types.
Answer:
8. What are all of the differences between the enumeration types of C++ and those of Java?
An Ada subtype is a possibly range-constrained version of an existing type. A subtype is type equivalent with its parent type. while a derived type is a new type that is based on some previously defined type with which it is not equivalent, although it may have identical structure. Derived types inherit all the properties of their parent types.
7. What significant justification is there for the -> operator in C and C++?
Answer:
The only justification for the -> operator in C and C++ is writability. It is slightly easier to write p -> q than (*p).q.
Answer :
In Java they can include fields, constructors, and methods. The possible values of an enumeration are the only possible instances of the class. All enumeration types inherit to String, as well as a few other methods. An array of the instances of an enumeration type can be fetched with the static method values. The internal numeric value of an enumeration variable can be fetched with the ordinal method. No expression of any other type can be assigned to an enumeration variable. Also, an enumeration variable is never coerced to any other type.
Answer :
Advantage : Unconstrained variant records in Ada allow the values of their variants to change types during execution.
Disadvantage: The type of the variant can be changed only by assigning the entire record, including the discriminant.
10. Multidimensional arrays can be stored in row major order, as in C++, or
in column major order, as in Fortran. Develop the access functions for
both of these arrangements for three-dimensional arrays
Answer:
Let the subscript ranges of the three dimensions be named min(1), min(2), min(3), max(1), max(2), and max(3). Let the sizes of the subscript ranges be size(1), size(2), and size(3).
Assume the element size is 1.
Assume the element size is 1.
Row Major:
location(a[i,j,k]) = (address of a[min(1),min(2),min(3)]) +((i-min(1))*size(3) + (j-min(2)))*size(2) + (k-min(3))
location(a[i,j,k]) = (address of a[min(1),min(2),min(3)]) +((i-min(1))*size(3) + (j-min(2)))*size(2) + (k-min(3))
Column Major:
location(a[i,j,k]) = (address of a[min(1),min(2),min(3)]) +((k-min(3))*size(1) + (j-min(2)))*size(2) + (i-min(1))
location(a[i,j,k]) = (address of a[min(1),min(2),min(3)]) +((k-min(3))*size(1) + (j-min(2)))*size(2) + (i-min(1))
