Would Like Know Excel Function Use Merge Identities Together Fill Missing Data Words Would Q43882564
I would like to know which Excel function to use to mergeidentities together to fill missing data. In other words I wouldlike to take the following 4 rows:
And end up with 2 consolidated rows like this:
What formula can I use to do this? The combined “fruit” and“initial” identities are what need to be matched against eachother, and then “color” and “quantity” need to be merged. Thanks inadvance for any help and advice!
D Quantity A B 4 Fruit Initial 5 Apples A 6 Bananas B 7 Apples A 8 Bananas B C Color Red Yellow Quantity 4 Fruit Initial 5 Apples A 6 Bananas B Color Red Yellow 8 Show transcribed image text D Quantity A B 4 Fruit Initial 5 Apples A 6 Bananas B 7 Apples A 8 Bananas B C Color Red Yellow
Quantity 4 Fruit Initial 5 Apples A 6 Bananas B Color Red Yellow 8
Expert Answer
Answer to I would like to know which Excel function to use to merge identities together to fill missing data. In other words I wou…
Would Like Output Match Correct Output See Someone Fix Code Maincpp Altered Eventcpp Modi Q43902903
I would like my output to match the correct output(see below).Can someone fix the code? main.cpp should not not be altered. onlyevent.cpp should be modified.
.This is my output:
Command Line:
1: ws
2: monday.txt
3: tuesday.txt
——————–
Day 1
——————–
1. 00:02:55 -> Computer Starting
2. 00:02:55 -> Computer Starting
3. [ No Event ]
4. 01:15:34 -> User logging in
5. 01:20:09 -> User logging in
6. [ No Event ]
7. [ No Event ]
——————–
Day 2
——————–
8. [ No Event ]
9. [ No Event ]
10. 01:20:34 -> User take a break
11. 01:22:30 -> User plays sudoku
12. [ No Event ]
——————–
Archive
——————–
13. [ No Event ]
14. 01:20:00 -> User logging in
15. [ No Event ]
16. [ No Event ]
17. [ No Event ]
18. [ No Event ]
19. [ No Event ]
20. [ No Event ]
21. [ No Event ]
——————–
*** Error in `ws’: double free or corruption (fasttop):0x0000000001a16310 ***
======= Backtrace: =========
/lib64/libc.so.6(+0x81609)[0x7f716d27c609]
ws[0x4012c3]
ws[0x401736]
/lib64/libc.so.6(__libc_start_main+0xf5)[0x7f716d21d495]
ws[0x400ec9]
======= Memory map: ========
00400000-00402000 r-xp 00000000 00:28 1077143745/home/mmaahi/ws
00601000-00602000 r–p 00001000 00:28 1077143745/home/mmaahi/ws
00602000-00603000 rw-p 00002000 00:28 1077143745/home/mmaahi/ws
01a14000-01a35000 rw-p 00000000 00:00 0 [heap]
7f7168000000-7f7168021000 rw-p 00000000 00:00 0
7f7168021000-7f716c000000 —p 00000000 00:00 0
7f716d1fb000-7f716d3bd000 r-xp 00000000 fd:00 674/usr/lib64/libc-2.17.so
7f716d3bd000-7f716d5bd000 —p 001c2000 fd:00 674/usr/lib64/libc-2.17.so
7f716d5bd000-7f716d5c1000 r–p 001c2000 fd:00 674/usr/lib64/libc-2.17.so
7f716d5c1000-7f716d5c3000 rw-p 001c6000 fd:00 674/usr/lib64/libc-2.17.so
7f716d5c3000-7f716d5c8000 rw-p 00000000 00:00 0
7f716d5c8000-7f716d5dd000 r-xp 00000000 fd:00 31805/usr/lib64/libgcc_s-4.8.5-20150702.so.1
7f716d5dd000-7f716d7dc000 —p 00015000 fd:00 31805/usr/lib64/libgcc_s-4.8.5-20150702.so.1
7f716d7dc000-7f716d7dd000 r–p 00014000 fd:00 31805/usr/lib64/libgcc_s-4.8.5-20150702.so.1
7f716d7dd000-7f716d7de000 rw-p 00015000 fd:00 31805/usr/lib64/libgcc_s-4.8.5-20150702.so.1
7f716d7de000-7f716d8df000 r-xp 00000000 fd:00 682/usr/lib64/libm-2.17.so
7f716d8df000-7f716dade000 —p 00101000 fd:00 682/usr/lib64/libm-2.17.so
7f716dade000-7f716dadf000 r–p 00100000 fd:00 682/usr/lib64/libm-2.17.so
7f716dadf000-7f716dae0000 rw-p 00101000 fd:00 682/usr/lib64/libm-2.17.so
7f716dae0000-7f716dbc9000 r-xp 00000000 fd:00 33767/usr/lib64/libstdc++.so.6.0.19
7f716dbc9000-7f716ddc8000 —p 000e9000 fd:00 33767/usr/lib64/libstdc++.so.6.0.19
7f716ddc8000-7f716ddd0000 r–p 000e8000 fd:00 33767/usr/lib64/libstdc++.so.6.0.19
7f716ddd0000-7f716ddd2000 rw-p 000f0000 fd:00 33767/usr/lib64/libstdc++.so.6.0.19
7f716ddd2000-7f716dde7000 rw-p 00000000 00:00 0
7f716dde7000-7f716de09000 r-xp 00000000 fd:00 667/usr/lib64/ld-2.17.so
7f716dfef000-7f716dff4000 rw-p 00000000 00:00 0
7f716e005000-7f716e008000 rw-p 00000000 00:00 0
7f716e008000-7f716e009000 r–p 00021000 fd:00 667/usr/lib64/ld-2.17.so
7f716e009000-7f716e00a000 rw-p 00022000 fd:00 667/usr/lib64/ld-2.17.so
7f716e00a000-7f716e00b000 rw-p 00000000 00:00 0
7fffd3449000-7fffd346a000 rw-p 00000000 00:00 0 [stack]
7fffd352b000-7fffd352d000 r-xp 00000000 00:00 0 [vdso]
ffffffffff600000-ffffffffff601000 r-xp 00000000 00:00 0[vsyscall]
Script done on Sat 18 Jan 2020 03:39:09 PM EST
This is the correct output:
Command Line:
1: ws
2: monday.txt
3: tuesday.txt
——————–
Day 1
——————–
1. 00:02:55 -> Computer Starting
2. 00:02:55 -> Computer Starting
3. [ No Event ]
4. 01:15:34 -> User logging in
5. 01:20:09 -> User logging in
6. 09:01:04 -> Browser closed
7. 17:33:33 -> User checks email
——————–
Day 2
——————–
8. 01:02:23 -> User starts working on homework
9. [ No Event ]
10. 01:20:34 -> User take a break
11. 01:22:30 -> User plays sudoku
12. 01:26:40 -> User resumes homework
Archive
——————–
13. [ No Event ]
14. 01:20:00 -> Authentication Failed
15. 01:23:20 -> Browser starts
16. 09:01:04 -> Browser closed
17. 17:33:33 -> User checks email
18. 01:02:23 -> User starts working on homework
19. 01:22:30 -> User plays sudoku
20. [ No Event ]
21. 01:26:40 -> User resumes homework
//EVENT.CPP
unsigned int g_sysClock = 0;
namespacesdds
{
//constructor
Event::Event() {
secMidnight = 0;
description=nullptr;
}
void Event::display() {
static int counter = 1;
int hour = 0;
int min = 0;
int sec = 0;
hour = secMidnight /3600;
min = (secMidnight – (hour *3600)) / 60;
sec = (secMidnight – (hour *3600) – (min * 60));
cout << setw(3)<< setfill(‘ ‘) << counter++ << “. “;
if ( description!=nullptr&& description[0] != ‘ ‘) {
cout<< setw(2) << setfill(‘0’) << hour << “:”<< setw(2) << setfill(‘0’) << min << “:”<< setw(2) << setfill(‘0’) << sec << “-> ” << description << endl;
// cout <<“[ No Event ]” << endl;
}
else {
cout<< “[ No Event ]” << endl;
}
void Event::setDescription(const char *src){
if (src != nullptr &&src[0]!=’ ‘) {
description = new char[strlen(src)+1];
strcpy(description, src);
secMidnight = g_sysClock;
}
else {
description = nullptr;
}
}
Event::~Event() {
delete[] description;
}
main.cpp
int main(int argc, char** argv) {
std::cout << “Command Line:n”;
for (int i = 0; i < argc; ++i) {
std::cout << i+1 <<“: ” << argv[i] << “n”;
}
std::cout << std::endl
sdds::Event archive[10];
size_t idxArchive = 0;
sdds::Event currentEvent;
const size_t secInDay = 60u * 60u * 24u;
for (auto day = 1; day < argc; ++day)
{
// each parameter for anapplication contains the events from one day
// process each one
std::cout <<“——————–n”;
std::cout << ” Day ” <<day << ‘n’;
std::cout <<“——————–n”;
std::ifstream in(argv[day]);
char opcode = ‘ ‘;
size_t time = secInDay + 1;
in >> opcode >>time;
// starting at midnight, untilthe end of the day
for (::g_sysClock = 0u;::g_sysClock < secInDay; ::g_sysClock++)
{
// what shouldhappen this second
while (time ==::g_sysClock)
{
// skip the delimiter
in.ignore();
// read the next opcode
in >> opcode;
// end of the file
if (in.fail())
break;
// handle the operation code
switch (opcode)
{
case ‘T’: // a new time code, this is exit thewhile loop
in >> time;
break;
case ‘S’: // start a new event, the old event isautomatically finished
char buffer[1024];
in.get(buffer, 1024,’,’);
currentEvent.setDescription(buffer);
break;
case ‘E’: // end the current event
currentEvent.setDescription(nullptr);
break;
case ‘P’: // print to scren the informationabout the current event
currentEvent.display();
break;
case ‘A’: // add a copy of the current event tothe archive
sdds::Eventcopy(currentEvent);
archive[idxArchive++] =copy;
break;
}
}
}
}
std::cout << “——————–n”;
std::cout << ” Archiven”;
std::cout << “——————–n”;
for (auto i = 0u; i < idxArchive; ++i)
archive[i].display();
std::cout << “——————–n”;
}
Expert Answer
Answer to I would like my output to match the correct output(see below). Can someone fix the code? main.cpp should not not be alte…
Would Like Print Testarray Elements 3 Elements One Line Result Like 7 12 1 0 5 8 However C Q43811675
Iwould like to print testArray elements up to 3 elements in one lineResult should be like7 12 10 5 8
However the closest result that I get is7 12 1 10 5 8
or
7 12 1 00 5 8
How can I manage to print array elements in differentlines
JAVA on Eclipse IDE
Refactor Navigate Search Project Run Window Help eclipse-workspace-workspace/src/workspace/lab3_001.java – Eclipse IDE – : Java Java Virtual Machi module-Info… Arrays Pack.java Dlab5.java 2 6 3.001.java 1 package workspace; import java.util.Arrays; 5 public class tab3001 70 public static void printArray(int[] arr, int no) { public static void main(String[] args) { // TODO Auto-generated method stub int testArray = 7, 12, 1, 0, 5, 8) intl emptyArray=0); int il anotherArray: printArray(testArray, 3), Probleme carminated Care 001 A r rivali -13.01. Contaminavea 2020 2 MacBook Pro Show transcribed image text Refactor Navigate Search Project Run Window Help eclipse-workspace-workspace/src/workspace/lab3_001.java – Eclipse IDE – : Java Java Virtual Machi module-Info… Arrays Pack.java Dlab5.java 2 6 3.001.java 1 package workspace; import java.util.Arrays; 5 public class tab3001 70 public static void printArray(int[] arr, int no) { public static void main(String[] args) { // TODO Auto-generated method stub int testArray = 7, 12, 1, 0, 5, 8) intl emptyArray=0); int il anotherArray: printArray(testArray, 3), Probleme carminated Care 001 A r rivali -13.01. Contaminavea 2020 2 MacBook Pro
Expert Answer
Answer to I would like to print testArray elements up to 3 elements in one lineResult should be like7 12 10 5 8However the closest…
Would Like Sort Data Column J Match Data Column Items Column J Event Match Simply Skips R Q43877704
I would like to sort data in column J to match the data inColumn A. I have more items in the A column than J so in the eventthere is no match it simply skips that row until there is a match.At the end I would be able to compare the tier (column B) and value(column M) of a single item number.
G H 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 – A B 1004001 Tier A 1005004 Tier C 1005048 Tier C 1005059 Tier C 1005706 Tier A 1006506 Tier C 1006801 Tier A 1008501 Tier B 1009005 Tier A 1009038 Tier B 1009049 Tier B 1009107 Tier C 1009479 Tier D 1009480 Tier B 1009901 Tier C 1009923 Tier B 1009978 Tier 1010022 Tier C 1010081 Tier D C D Units GC Units CHM Units CHM Units CHM Units BIO Units CHM GC Units CHM Units CHM Units CHM Public He: CHM Units DSHM Units FI Public He: CHM Units EXC Units EXC Units DSHM Units DSHM Units DSHM E F Melting Point Ophthamology Ophthamology Ophthamology Performance Standards Renal Residual Solvents Ophthamology Hepatic Hepatic Hepatic Amino Acids Flavors & Fragrances Oncology Coatings Coatings Amino Acids DSHM Others Amino Acids . K . L 1005004 A1-C25-SO1-RM 1005048 A1-C29-SO3-LO 1005059 R4-D13-04-LM 1005706 A1-B02-501- 1006506 A1-B29-505- 1006801 A1-B05-SO3- 1008501 R1-D18-S06-LO 1009005 R1-D06-SO1-RO 1009479 A2-E02-SO2-LO 1009901 A1-B13-501- 1009923 A1-B15-S05- 1009978 A1-C29-SO2-RM 1010022 A1-C26-SO1-RO 1010081 A1-C31-SO2-LM 1010106 R1-D38-SO3-RO 1010208 R4-D34-SO1-LO 1011007 R1-D12-S06-LO 1011018 R1-D40-SO5-RO 1011029 R5-B22-SO3- . M I $3,697.02 $324.27 $415.80 $4,481.71 $43.80 $3,509.22 $160.00 $848.00 $0.00 $377.54 $766.21 $91.20 $243.04 $10.17 $67.86 $171.10 $5,132.81 $838.04 $718.17 Show transcribed image text G H 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 – A B 1004001 Tier A 1005004 Tier C 1005048 Tier C 1005059 Tier C 1005706 Tier A 1006506 Tier C 1006801 Tier A 1008501 Tier B 1009005 Tier A 1009038 Tier B 1009049 Tier B 1009107 Tier C 1009479 Tier D 1009480 Tier B 1009901 Tier C 1009923 Tier B 1009978 Tier 1010022 Tier C 1010081 Tier D C D Units GC Units CHM Units CHM Units CHM Units BIO Units CHM GC Units CHM Units CHM Units CHM Public He: CHM Units DSHM Units FI Public He: CHM Units EXC Units EXC Units DSHM Units DSHM Units DSHM E F Melting Point Ophthamology Ophthamology Ophthamology Performance Standards Renal Residual Solvents Ophthamology Hepatic Hepatic Hepatic Amino Acids Flavors & Fragrances Oncology Coatings Coatings Amino Acids DSHM Others Amino Acids . K . L 1005004 A1-C25-SO1-RM 1005048 A1-C29-SO3-LO 1005059 R4-D13-04-LM 1005706 A1-B02-501- 1006506 A1-B29-505- 1006801 A1-B05-SO3- 1008501 R1-D18-S06-LO 1009005 R1-D06-SO1-RO 1009479 A2-E02-SO2-LO 1009901 A1-B13-501- 1009923 A1-B15-S05- 1009978 A1-C29-SO2-RM 1010022 A1-C26-SO1-RO 1010081 A1-C31-SO2-LM 1010106 R1-D38-SO3-RO 1010208 R4-D34-SO1-LO 1011007 R1-D12-S06-LO 1011018 R1-D40-SO5-RO 1011029 R5-B22-SO3- . M I $3,697.02 $324.27 $415.80 $4,481.71 $43.80 $3,509.22 $160.00 $848.00 $0.00 $377.54 $766.21 $91.20 $243.04 $10.17 $67.86 $171.10 $5,132.81 $838.04 $718.17
Expert Answer
Answer to I would like to sort data in column J to match the data in Column A. I have more items in the A column than J so in the …
Would Output Print Format 10005555 2f Q43840195
what would be the output: print(format(1000.5555,’,.2f))
Expert Answer
Answer to what would be the output: print(format(1000.5555,’,.2f))…
Would Someone Able Break C Code Line Line Tell Line Means Void Printallbooks String Titles Q43794681
Would someone be able to break this C++ code down line by lineand tell me what each line means?
void printAllBooks(string titles[],string authors[],intnumBookStored){
if(numBookStored<1){
cout<<“No books are stored”<<endl;
return ;
}
cout<<“Here is a list of books”<<endl;
int i;
for(i=0;i<numBookStored;i++)cout<<titles[i]<<” by”<<authors[i]<<endl;
}
Expert Answer
Answer to Would someone be able to break this C++ code down line by line and tell me what each line means? void printAllBooks(stri…
Would Someone Able Break C Code Line Line Tell Line Means Void Printbooksbyauthor Std Stri Q43794685
Would someone be able to break this C++ code down line by lineand tell me what each line means?
void printBooksByAuthor(std::string titles[], std::stringauthors[], int numBooks, std::string author)
{
if (numBooks == 0)
{
std::cout << “No books are stored” << std::endl;
return;
}
bool bIsBookFound = false;
std::string* foundBooks = new std::string[numBooks];
int cnt = 0;
for (int i = 0;i < numBooks;i++)
{
if (authors[i] == author)
{
foundBooks[cnt] = titles[i];
cnt++;
bIsBookFound = true;
}
}
if(bIsBookFound)
{
std::cout << “Here is a list of books by ” << author<< std::endl;
for (int i = 0;i < cnt;i++)
std::cout << foundBooks[i] << std::endl;
}
else
{
std::cout << “There are no books by ” << author;
}
delete[] foundBooks;
}
Expert Answer
Answer to Would someone be able to break this C++ code down line by line and tell me what each line means? void printBooksByAuthor…
Would Write Body Insert Deleteat Delete Locate Methods Ve Tried Writing Code T Seem Get Wo Q43904602
How would we write the body for the insert(), deleteAt(),delete(), and the locate() methods? I’ve tried writing down somecode but can’t seem to get it to work.
* This class provides a growable array for primitive long values. * @author * @version 1.0 1/16/2020 */ class BigIntArrayList { //fields private long[] list; private int count; * Constructor. Initializes the underlying array to 10 elements. */ BigIntArrayList() { list = new long[10]; count = 0; Inserts the given long value at the given index. The index is assumed to be a value between 0 and count. Existing elements are moved up as needed to make room for the new value. * @param index the index where the new value should be stored * @param value the value to be stored */ public void insert(int index, long value) { count = list.length; count++; for (int i = count – 1; i > index; i–) { list[i] = list[i – 1]; list[index] = value; * Deletes the value at the given index. The index is * assumed to be a value between 0 and count – 1. Existing elements are moved down as needed to keep all values contiguous, without any empty spaces in the array. * @param index the index of the element that should be removed * @return the value that was removed */ public long deleteAt (int index) { long[] newList = new long[list.length – 1]; for (int i = 0; i < newList.length; i++) { newList[i] = list[i]; return 0; * Deletes the first instance of the given value. Existing elements * are moved down as needed to keep all values contiguous, without * any empty spaces in the array. If the value does not exist, this * method returns without doing anything. * @param value the value to remove public void delete (long value) { //TODO complete this method * Returns the value at the given index without removing it. The index is * assumed to be a value between 0 and count – 1. * @param index the index of the element * @return the value at that index public long getValueAt (int index) { //TODO complete this method return 0; //replace this return statement * Returns the index of the first instance of the given value * in the array. If the value doesn’t exist, -1 is returned. * @param value the value to find in the array * @return the index where the value was found, or -1 if not found public int locate (long value) { int index = -1; for (int i = 0; i < list.length; i++) { if (list[i] == value) { index = i; return index; * Provides access to the number of values currently in the array. * @return the number of values in the array */ public int getCount() { return count; Show transcribed image text * This class provides a growable array for primitive long values. * @author * @version 1.0 1/16/2020 */ class BigIntArrayList { //fields private long[] list; private int count; * Constructor. Initializes the underlying array to 10 elements. */ BigIntArrayList() { list = new long[10]; count = 0; Inserts the given long value at the given index. The index is assumed to be a value between 0 and count. Existing elements are moved up as needed to make room for the new value. * @param index the index where the new value should be stored * @param value the value to be stored */ public void insert(int index, long value) { count = list.length; count++; for (int i = count – 1; i > index; i–) { list[i] = list[i – 1]; list[index] = value; * Deletes the value at the given index. The index is * assumed to be a value between 0 and count – 1. Existing elements are moved down as needed to keep all values contiguous, without any empty spaces in the array. * @param index the index of the element that should be removed * @return the value that was removed */ public long deleteAt (int index) { long[] newList = new long[list.length – 1]; for (int i = 0; i
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Answer to How would we write the body for the insert(), deleteAt(), delete(), and the locate() methods? I’ve tried writing down so…
Would Write Body Tostring Methods Using Printf Statements Language Java Q43893494
How would we write the body of the toString() methods usingprintf statements?
Language: Java
/ * * * Provides a string representation of the current Money object, in * the form $mm.mm. Negative values have a minus sign before the dollar * sign. Zero dollar amounts are represented with one zero between the * dollar sign and the decimal point. Cents that are multiples of 10 * still show two decimal places. (Examples: $12.34, -$6.50, $0.99, * -$0.99, $0.00) * @return the string representation @Override public String toString() { return null; /** * Default constructor. Creates a new Money object. * Sets both dollars and cents to zero. public Money () { dollars = 0; cents = 0; * Full constructor. Creates a new Money object. Dollars and cents * are normalized. If one is negative and the other is * positive, the resulting object will be the natural addition of the two * values. (Example: 2 dollars and -5 cents is converted to i dollar * and 95 cents.) If the absolute value of cents is greater than 99, * the resulting object will carry the extra cents into the dollars. * (Example: 1 dollar and 105 cents is converted to 2 dollars and 5 cents.) * A negative value is represented with both the dollars and cents negative. * @param dollars the number of dollars in the amount @param cents the number of cents in the amount public Money (long dollars, byte cents) { long value = dollars * 100; value += cents; this.dollars = value / 100; this.cents = (byte) (value $ 100); * Provides access to the dollars portion of the monetary value. * @return the dollars portion of this Money object public long getDollars () { return dollars; /** * Provides access to the cents portion of the monetary value. * @return the cents portion of this money object */ public byte getCents() { return cents; * Adds the current and the given Money objects together, and stores * the sum in a new Money object. The current and given Money objects * are not altered in the process. The resulting Money object’s * dollars and cents are normalized as defined in the full constructor comments. * @param other the other Money object to add to this one * @return the Money object containing the results of this addition */ public Money add (Money other) { long value = dollars * 100; value += cents; long Value = other. dollars * 100; Value += other.cents; long result = value + Value; return new Money (result / 100, (byte) (result $ 100)); * Subtracts the given Money object from this one, and stores * the result in a new Money object. The current and given Money objects * are not altered in the process. The resulting Money object’s * dollars and cents are normalized as defined in the full constructor comments. * @param other the other Money object to subtract from this one * @return the Money object containing the results of this subtraction */ public Money subtract (Money other) { long value = dollars * 100; value += cents; long Value = other.dollars * 100; oValue += other.cents; long result = value – oValue; return new Money (result / 100, (byte) (result $ 100)); Show transcribed image text / * * * Provides a string representation of the current Money object, in * the form $mm.mm. Negative values have a minus sign before the dollar * sign. Zero dollar amounts are represented with one zero between the * dollar sign and the decimal point. Cents that are multiples of 10 * still show two decimal places. (Examples: $12.34, -$6.50, $0.99, * -$0.99, $0.00) * @return the string representation @Override public String toString() { return null;
/** * Default constructor. Creates a new Money object. * Sets both dollars and cents to zero. public Money () { dollars = 0; cents = 0; * Full constructor. Creates a new Money object. Dollars and cents * are normalized. If one is negative and the other is * positive, the resulting object will be the natural addition of the two * values. (Example: 2 dollars and -5 cents is converted to i dollar * and 95 cents.) If the absolute value of cents is greater than 99, * the resulting object will carry the extra cents into the dollars. * (Example: 1 dollar and 105 cents is converted to 2 dollars and 5 cents.) * A negative value is represented with both the dollars and cents negative. * @param dollars the number of dollars in the amount @param cents the number of cents in the amount public Money (long dollars, byte cents) { long value = dollars * 100; value += cents; this.dollars = value / 100; this.cents = (byte) (value $ 100); * Provides access to the dollars portion of the monetary value. * @return the dollars portion of this Money object public long getDollars () { return dollars; /** * Provides access to the cents portion of the monetary value. * @return the cents portion of this money object */ public byte getCents() { return cents;
* Adds the current and the given Money objects together, and stores * the sum in a new Money object. The current and given Money objects * are not altered in the process. The resulting Money object’s * dollars and cents are normalized as defined in the full constructor comments. * @param other the other Money object to add to this one * @return the Money object containing the results of this addition */ public Money add (Money other) { long value = dollars * 100; value += cents; long Value = other. dollars * 100; Value += other.cents; long result = value + Value; return new Money (result / 100, (byte) (result $ 100)); * Subtracts the given Money object from this one, and stores * the result in a new Money object. The current and given Money objects * are not altered in the process. The resulting Money object’s * dollars and cents are normalized as defined in the full constructor comments. * @param other the other Money object to subtract from this one * @return the Money object containing the results of this subtraction */ public Money subtract (Money other) { long value = dollars * 100; value += cents; long Value = other.dollars * 100; oValue += other.cents; long result = value – oValue; return new Money (result / 100, (byte) (result $ 100));
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Answer to How would we write the body of the toString() methods using printf statements? Language: Java…
Would Write C Something Reads Integers Using Scanf Sentinel Value 999 Read Value 999 Read Q43884632
How would I write in C something that reads in integers usingscanf() and then when a sentinel value of -999 is read. Then oncethe value -999 is read to display the sum of all values that wereread. Also excluding the value of the -999. So I would want the sumof all values excluding -999 that were read in. If you could showme how to do this in C i would greatly appreciate it.
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Answer to How would I write in C something that reads in integers using scanf() and then when a sentinel value of -999 is read. Th…