Pearson eText for Concepts of Programming Languages -- Instant Access (Pearson+)
12th Edition
ISBN: 9780135102268
Author: Robert Sebesta
Publisher: PEARSON+
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Chapter 5, Problem 2PS
Explanation of Solution
Advantages and disadvantages of having no types in a language:
- The advantage of having no types in a language is the flexibility and ease of
programming for the programmer, being that any variable can be used for any type value...
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make corrections of this program based on the errors shown. this is CIS 227 .
Create 6 users: Don, Liz, Shamir, Jose, Kate, and Sal.
Create 2 groups: marketing and research.
Add Shamir, Jose, and Kate to the marketing group.
Add Don, Liz, and Sal to the research group.
Create a shared directory for each group.
Create two files to put into each directory:
spreadsheetJanuary.txt
meetingNotes.txt
Assign access permissions to the directories:
Groups should have Read+Write access
Leave owner permissions as they are
“Everyone else” should not have any access
Submit for grade:
Screenshot of /etc/passwd contents showing your new users
Screenshot of /etc/group contents showing new groups with their members
Screenshot of shared directories you created with files and permissions
⚫ your circuit diagrams for your basic bricks, such as AND, OR, XOR gates and 1 bit multiplexers,
⚫ your circuit diagrams for your extended full adder, designed in Section 1 and
⚫ your circuit diagrams for your 8-bit arithmetical-logical unit, designed in Section 2.
1 An Extended Full Adder
In this Section, we are going to design an extended full adder circuit (EFA). That EFA takes 6 one bit inputs: aj, bj,
Cin, Tin, t₁ and to. Depending on the four possible combinations of values on t₁ and to, the EFA produces 3 one bit
outputs: sj, Cout and rout.
The EFA can be specified in principle by a truth table with 26 = 64 entries and 3 outputs. However, as the EFA
ignores certain inputs in certain cases, it is easier to work with the following overview specification, depending only
on t₁ and to in the first place:
t₁ to Description
00
Output Relationship
Ignored
Inputs
Addition Mode
2 Coutsjaj + bj + Cin, Tout= 0
Tin
0 1
Shift Left Mode
Sj = Cin,
Cout=bj, rout = 0
rin, aj
10
1 1
Shift Right…
Chapter 5 Solutions
Pearson eText for Concepts of Programming Languages -- Instant Access (Pearson+)
Ch. 5 - Prob. 1RQCh. 5 - Prob. 2RQCh. 5 - Prob. 3RQCh. 5 - Which category of C++ reference variables always...Ch. 5 - Prob. 5RQCh. 5 - Prob. 6RQCh. 5 - Prob. 7RQCh. 5 - Prob. 8RQCh. 5 - What are the advantages and disadvantages of...Ch. 5 - Prob. 10RQ
Ch. 5 - Prob. 11RQCh. 5 - Prob. 12RQCh. 5 - How is a reference to a nonlocal variable in a...Ch. 5 - What is the general problem with static scoping?Ch. 5 - What is the referencing environment of a...Ch. 5 - Prob. 16RQCh. 5 - What is a block?Ch. 5 - What is the purpose of the let constructs in...Ch. 5 - What is the difference between the names defined...Ch. 5 - Prob. 20RQCh. 5 - Prob. 21RQCh. 5 - What are the advantages of named constants?Ch. 5 - Which of the following identifier forms is most...Ch. 5 - Prob. 2PSCh. 5 - Write a simple assignment statement with one...Ch. 5 - Prob. 4PSCh. 5 - Describe a situation when a history-sensitive...Ch. 5 - Consider the following C program void fun (void) {...Ch. 5 - Consider the following skeletal C program: void...Ch. 5 - Which of the following identifier forms is most...Ch. 5 - Prob. 2PECh. 5 - Write a simple assignment statement with one...
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