C How to Program (8th Edition)
C How to Program (8th Edition)
8th Edition
ISBN: 9780133976892
Author: Paul J. Deitel, Harvey Deitel
Publisher: PEARSON
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Chapter 12, Problem 12.10E
Program Plan Intro

Program plan:

  1. line, ch, k variablesare used for input. There is structure stknode havingch, nextvalmember variables which represents the stack value.
  2. void push(stack **top, char alphabet) function inserts the node in the a stack.
  3. char pop(stack **top) function delete the value in the given and returns the delete value.

Program description:

The main purpose of the program is to create a stack from the input line as a character by character. Then it extracts the characters from the stack to print the reverse of the input line.

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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…
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