I have only a little experience in C++. I'm trying to write a program to print each element of the 'sales' array:
#include <iostream>
#include <iomanip>
using namespace std;
void printArray(int, int);
int main()
{
char chips[5][50] = {"mild", "medium", "sweet", "hot", "zesty"};
int sales[5] = {0};
int tempSales, counter;
const int i = 5;
for (counter = 0; counter < i; counter++)
{
cout << "Please enter in the sales for " << chips[counter] << ": ";
cin >> tempSales;
tempSales >> sales[counter][5];
}
cout << "{";
for (int counter = 0; counter < i; counter++)
{
cout << chips[counter] << ", ";
}
cout << "}" << endl;
cout << "{";
for (int counter = 0; counter < i; counter++)
{
cout << sales[counter] << ", ";
}
cout << "}" << endl;
return 0;
}
To solve this problem, I need to have the same commands and keywords I still have, and it can't be any advanced or weird syntax. For some reason, my input from the:
cin >> tempSales
Is not functioning. Here are the results:
{mild, medium, sweet, hot, zesty, }
{0,0,0,0,0, }
Whereas I just want to see 1, 2, 3, 4, and 5 for the second array. Why is it only print 0 and not reading my input? Please help!
Like rranjik stated, you shouldn't need a 2D array if you're only listing the number of sales, which it appears you're doing from what you provided, is this not the case?
Is it necessary for you to use the bitshift operator >> for your assignment? For a simple integer assignment it isn't really necessary, and you could do:
int sales[5] = {0}; Change the array to a simple array instead of 2D.
sales[counter] = tempSales; Use standard assignment for the integer on line 19
cout << sales[counter] << ", "; Change your output accordingly.
Hope this helps!
I don't think you need a 2D array for sales. Try cout << sales[counter][5] << ", "; or change int sales[5][6] = {0}; to int sales[5] = {0};. As Luke mentioned, use standard assignment sales[counter] = tempSales;.
Related
Hye, Im a beginner trying to learn C++ language. This is my code that I tried to find reverse input numbers using array. Can help me point my mistakes since I always got infinite loop.
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
const int ARRAY_SIZE=50;
int size[ARRAY_SIZE];
unsigned short int i;
cout << "You may enter up to 50 integers:\n";
cout << "\nHow many would you like to enter? ";
cin >> size[ARRAY_SIZE];
cout << "Enter your number: \n";
for (int i = 0; i < ARRAY_SIZE; i++)
{
cin >> size[i];
}
cout << "\nYour numbers reversed are:\n";
for (i = size[ARRAY_SIZE] - 1; i >= 0; i++)
cout << " size[i]" << " ";
}
Your infinite loop is because i is unsigned, so i >= 0 is always true.
Here's a C++-ified version:
#include <iostream>
#include <vector>
int main() {
std::cout << "You may enter up to 50 integers:\n";
std::cout << "\nHow many would you like to enter? ";
int count;
std::cin >> count;
// Use a std::vector which can be extended easily
std::vector<int> numbers;
for (int i = 0; i < count; ++i) {
std::cout << "Enter your number: \n";
int v;
std::cin >> v;
// Add this number to the list
numbers.push_back(v);
}
std::cout << "\nYour numbers reversed are:\n";
// Use a reverse iterator to iterate through the list backwards
for (auto i = numbers.rbegin(); i != numbers.rend(); ++i) {
// An iterator needs to be de-referenced with * to yield the value
std::cout << *i << " ";
}
std::cout << std::endl;
return 0;
}
There's many problems in your original code, but the clincher is this:
for (i = size[ARRAY_SIZE] - 1; i >= 0; i++)
cout << " size[i]" << " ";
}
Since you keep adding to i through each cycle you'll never go below zero, especially not for an unsigned short int. This should be:
for (int i = count - 1; i > 0; --i) {
std::cout << numbers[i];
}
Presuming you have a thing called numbers instead of the bizarrely named size and the array size is count, not i, as i is generally reserved for iterators and loop indexes.
I know this isnt the right kind of question to be asking, but for the life of me I could not figure out what is causing this problem.
I need to write a problem that takes a set number of integers or doubles and returns their sum.
I have written the code to make this work, making sure to check each time I changed something.
#include<iostream>
using namespace std;
template <class T>
class totalClass
{
private:
T *p;
T Total;
T sum;
int size;
public:
T total(int x)
{
size = x;
p = new T[x];
for (int i = 0; i < size; i++)
p[i] = T();
if (size > 1)
{
for (int i = 0; i < size; ++i)
{
cin >> sum;
Total += sum;
}
}
return Total;
}
};
int main()
{
int size, result1;
double result2;
cout << "Enter: ";
cin >> size;
cout << "the number of ints you wish to enter: Enter: " << size << " integers:";
totalClass<int> test;
result1 = test.total(size);
cout << " Total = " << result1 << endl;
cout << "Enter: ";
cin >> size;
cout << "the number of doubles you wish to enter: Enter: " << size << " doubles:";
totalClass<double> test2;
result2 = test2.total(size);
cout << " Total = " << result2 << endl;
}
My doubles are getting added up correctly but my integer addition always seems to add up to some crazy number. Is there something wrong with my problem that I cannot see?
If you forget to initialize a variable and attempt to use it or do math with it, you might end up with "crazy numbers." Make sure all of your variables are Initialized.
I am messing around with dynamic arrays for a user defined amount of inputs for an and gate.
The issue I am running into is that I don't know how many inputs the user is going to test and I need to be able to have an if-else statement that tests each input.
#include <iostream>
#include <iomanip>
#include <string>
using namespace std;
class logic_gate {
public:
int x = 0;
};
int main() {
int userInput = 0;
cout << "How many inputs do you want on your and gate?: ";
cin >> userInput;
cout << endl;
logic_gate *and_gate = new logic_gate[userInput];
cout << endl << "Please enter the values of each bit below . . ." << endl <<
endl;
int userTest1 = 0;
for (int i = 0; i < userInput; i++) {
cout << "#" << i + 1 << ": ";
cin >> userTest1;
and_gate[i].x = userTest1;
}
return 0;
}
Here is the code that I am currently trying to find a solution for.
To implement an AND gate with n inputs you can simply do:
int output = 1;
for (int i = 0; i < n; ++i)
{
if (!and_gate [i])
{
output = 0;
break;
}
}
// ...
Use Vector data structure, you don't need to tell its size while declaring, unlike array, and it can grow automatically.
To read input till it's arriving, put cin inside while loop condition. I used getline to read whole line and work with it, so that whenever user presses enter button at empty line, program will think that no more input is coming anymore, and will start calculating 'And' of inputs.
//don't forget to import vector
#include <iostream>
#include <vector>
#include <string>
using namespace std;
class logic_gate {
public:
int x = 0;
logic_gate(){ //default constructor
}
logic_gate(int k){ //another constructor needed
x = k;
}
};
int main(){
cout << endl << "Please enter the values of each bit below . . ." << endl;
vector<logic_gate> and_gate; //no need to tell size while declaration
string b;
while(getline(cin, b)){ //read whole line from standard input
if (b == "\0") //input is NULL
break;
and_gate.push_back(logic_gate(stoi(b))); //to convert string to integer
}
if (!and_gate.empty()){
int output = and_gate[0].x;
for (int i = 1; i < and_gate.size(); i++){
output = output & and_gate[i].x;
}
cout << "And of inputs is: " << output << endl;
}
else{
cout << "No input was given!\n";
}
return 0;
}
Feel free to ask if some doubts linger
I figured out what I wanted to do. Thanks to everyone who helped and especially Paul Sanders. Below is my final code.
#include <iostream>
using namespace std;
class logic_gate {
public:
int x = 0;
};
int main() {
int userInput;
int output = 1;
cout << "How many inputs do you want on your and gate?: ";
cin >> userInput;
cout << endl;
logic_gate *and_gate = new logic_gate[userInput];
cout << endl << "Please enter the values of each bit below . . ." << endl <<
endl;
int userTest1;
for (int i = 0; i < userInput; i++) {
cout << "#" << i + 1 << ": ";
cin >> userTest1;
and_gate[i].x = userTest1;
}
if (userInput == 1) {
output = userTest1;
cout << "The test of " << userTest1 << " is " << output << endl << endl;
}
else if (userInput > 1) {
for (int i = 0; i < userInput; i++) {
if (!and_gate[i].x)
{
output = 0;
break;
}
}
cout << "The test of ";
for (int i = 0; i < userInput; i++) {
cout << and_gate[i].x;
}
cout << " is " << output << endl << endl;
}
return 0;
}
For this program a user must enter 10 contestants and the amount of second it took for them to complete a swimming race. My problem is that I must output the 1st, 2nd and 3rd placers, so I need to get the three smallest arrays (as they would be the quickest times) but I'm unsure on how to do it. Here is my code so far.
string names[10] = {};
int times[10] = { 0 };
int num[10] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
int min1 = 0, min2 = 0, min3 = 0;
cout << "\n\n\tCrawl";
for (int i = 0; i < 10; i++)
{
cout << "\n\n\tPlease enter the name of contestant number " << num[i] << ": ";
cin >> names[i];
cout << "\n\tPlease enter the time it took for them to complete the Crawl style: ";
cin >> times[i];
while (!cin)
{
cout << "\n\tError! Please enter a valid time: ";
cin.clear();
cin.ignore();
cin >> times[i];
}
if (times[i] < times[min1])
min1 = i;
cout << "\n\n\t----------------------------------------------------------------------";
}
system("cls");
cout << "\n\n\tThe top three winners of the Crawl style race are as follows";
cout << "\n\n\t1st Place - " << names[min1];
cout << "\n\n\t2nd Place - " << names[min2];
cout << "\n\n\t3rd Place - " << names[min3];
}
_getch();
return 0;
}
As you can see, it is incomplete. I know how to get the smallest number, but its the second and third smallest that is giving me trouble.
your code is full of errors:
what do you do with min2 and min3 as long as you don't assign them?? they are always 0
try checking: cout << min2 << " " << min3;
also you don't initialize an array of strings like that.
why you use an array of integers for just printing number of input:
num? instead you can use i inside loop adding to it 1 each time
to solve your problem use a good way so consider using structs/clusses:
struct Athlete
{
std::string name;
int time;
};
int main()
{
Athlete theAthletes[10];
for(int i(0); i < 10; i++)
{
std::cout << "name: ";
std::getline(std::cin, theAthletes[i].name);
std::cin.sync(); // flushing the input buffer
std::cout << "time: ";
std::cin >> theAthletes[i].time;
std::cin.sync(); // flushing the input buffer
}
// sorting athletes by smaller time
for(i = 0; i < 10; i++)
for(int j(i + 1); j < 10; j++)
if(theAthletes[i].time > theAthletes[j].time)
{
Athlete tmp = theAthletes[i];
theAthletes[i] = theAthletes[j];
theAthletes[j] = tmp;
}
// printing the first three athletes
std::cout << "the first three athelets:\n\n";
std::cout << theAthletes[0].name << " : " << theAthletes[0].time << std::endl;
std::cout << theAthletes[1].name << " : " << theAthletes[1].time << std::endl;
std::cout << theAthletes[2].name << " : " << theAthletes[2].time << std::endl;
return 0;
}
I hope this will give u the expected output. But i suggest u to use some sorting alogirthms like bubble sort,quick sort etc.
#include <iostream>
#include<string>
using namespace std;
int main() {
int times[10] = { 0 };
int num[10] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
int min1 = 0, min2 = 0, min3 = 0,m;
string names[10] ;
cout << "\n\n\tCrawl";
for (int i = 0; i < 10; i++)
{
cout << "\n\n\tPlease enter the name of contestant number " << num[i] << ": ";
cin >> names[i];
cout << names[i];
cout << "\n\tPlease enter the time it took for them to complete the Crawl style: ";
cin >> times[i];
cout<<times[i];
while (!cin)
{
cout << "\n\tError! Please enter a valid time: ";
cin.clear();
cin.ignore();
cin >> times[i];
}
if(times[i]==times[min1]){
if(times[min1]==times[min2]){
min3=i;
}else{min2 =i;}
}else if(times[i]==times[min2]){
min3=i;
}
if (times[i] < times[min1]){
min1 = i;
cout <<i;
}
int j=0;
while(j<i){
if((times[j]>times[min1])&&(times[j]<times[min2])){
min2 =j;
j++;
}
j++;
}
m=0;
while(m<i){
if((times[m]>times[min2])&&(times[m]<times[min3])){
min3 =m;
m++;
}
m++;
}
cout << "\n\n\t----------------------------------------------------------------------";
}
cout << "\n\n\tThe top three winners of the Crawl style race are as follows";
cout << "\n\n\t1st Place - " << names[min1];
cout << "\n\n\t2nd Place - " << names[min2];
cout << "\n\n\t3rd Place - " << names[min3];
return 0;
}
There is actually an algorithm in the standard library that does exactly what you need: std::partial_sort. Like others have pointed out before, to use it you need to put all the participant data into a single struct, though.
So start by defining a struct that contains all relevant data. Since it seems to me that you only use the number of the contestants in order to be able to later find the name to the swimmer with the fastest time, I'd get rid of it. Of course you could also add it back in if you like.
struct Swimmer {
int time;
std::string name;
};
Since you know that there always will be exactly 10 participants in a race, you can also go ahead and replace the C-style array by a std::array.
The code to read in the users then could look like this:
std::array<Swimmer, 10> participants;
for (auto& participant : participants) {
std::cout << "\n\n\tPlease enter the name of the next contestant: ";
std::cin >> participant.name;
std::cout << "\n\tPlease enter the time it took for them to complete the Crawl style: ";
while(true) {
if (std::cin >> participant.time) {
break;
}
std::cout << "\n\tError! Please enter a valid time: ";
std::cin.clear();
std::cin.ignore();
}
std::cout << "\n\n\t----------------------------------------------------------------------";
}
Partial sorting is now essentially a one-liner:
std::partial_sort(std::begin(participants),
std::begin(participants) + 3,
std::end(participants),
[] (auto const& p1, auto const& p2) { return p1.time < p2.time; });
Finally you can simply output the names of the first three participants in the array:
std::cout << "\n\n\tThe top three winners of the Crawl style race are as follows";
std::cout << "\n\n\t1st Place - " << participants[0].name;
std::cout << "\n\n\t2nd Place - " << participants[1].name;
std::cout << "\n\n\t3rd Place - " << participants[2].name << std::endl;
The full working code can be found on coliru.
This is not a full solution to your problem, but just meant to point you into the right direction...
#include <iostream>
#include <limits>
#include <algorithm>
using namespace std;
template <int N>
struct RememberNsmallest {
int a[N];
RememberNsmallest() { std::fill_n(a,N,std::numeric_limits<int>::max()); }
void operator()(int x){
int smallerThan = -1;
for (int i=0;i<N;i++){
if (x < a[i]) { smallerThan = i; break;}
}
if (smallerThan == -1) return;
for (int i=N-1;i>smallerThan;i--){ a[i] = a[i-1]; }
a[smallerThan] = x;
}
};
int main() {
int a[] = { 3, 5, 123, 0 ,-123, 1000};
RememberNsmallest<3> rns;
rns = std::for_each(a,a+6,rns);
std::cout << rns.a[0] << " " << rns.a[1] << " " << rns.a[2] << std::endl;
// your code goes here
return 0;
}
This will print
-123 0 3
As you need to know also the names for the best times, you should use a
struct TimeAndName {
int time;
std::string name;
}
And change the above functor to take a TimeAndName instead of the int and make it also remember the names... or come up with a different solution ;), but in any case you should use a struct similar to TimeAndName.
As your array is rather small, you could even consider to use a std::vector<TimeAndName> and sort it via std::sort by using your custom TimeAndName::operator<.
So, I'm creating a coin change algorithm that take a Value N and any number of denomination and if it doesn't have a 1, i have to include 1 automatically. I already did this, but there is a flaw now i have 2 matrix and i need to use 1 of them. Is it possible to rewrite S[i] matrix and still increase the size of array.... Also how can i find the max denomination and the second highest and sooo on till the smallest? Should i just sort it out in an highest to lowest to make it easier or is there a simpler way to look for them one after another?
int main()
{
int N,coin;
bool hasOne;
cout << "Enter the value N to produce: " << endl;
cin >> N;
cout << "Enter number of different coins: " << endl;
cin >> coin;
int *S = new int[coin];
cout << "Enter the denominations to use with a space after it" << endl;
cout << "(1 will be added if necessary): " << endl;
for(int i = 0; i < coin; i++)
{
cin >> S[i];
if(S[i] == 1)
{
hasOne = true;
}
cout << S[i] << " ";
}
cout << endl;
if(!hasOne)
{
int *newS = new int[coin];
for(int i = 0; i < coin; i++)
{
newS[i] = S[i];
newS[coin-1] = 1;
cout << newS[i] << " ";
}
cout << endl;
cout << "1 has been included" << endl;
}
//system("PAUSE");
return 0;
}
You could implement it with std::vector, then you only need to use push_back.
std::sort can be used to sort the denominations into descending order, then it's just a matter of checking whether the last is 1 and adding it if it was missing. (There is a lot of error checking missing in this code, for instance, you should probably check that no denomination is >= 0, since you are using signed integers).
#include <iostream> // for std::cout/std::cin
#include <vector> // for std::vector
#include <algorithm> // for std::sort
int main()
{
std::cout << "Enter the value N to produce:\n";
int N;
std::cin >> N;
std::cout << "Enter the number of different denominations:\n";
size_t denomCount;
std::cin >> denomCount;
std::vector<int> denominations(denomCount);
for (size_t i = 0; i < denomCount; ++i) {
std::cout << "Enter denomination #" << (i + 1) << ":\n";
std::cin >> denominations[i];
}
// sort into descending order.
std::sort(denominations.begin(), denominations.end(),
[](int lhs, int rhs) { return lhs > rhs; });
// if the lowest denom isn't 1... add 1.
if (denominations.back() != 1)
denominations.push_back(1);
for (int coin: denominations) {
int numCoins = N / coin;
N %= coin;
if (numCoins > 0)
std::cout << numCoins << " x " << coin << '\n';
}
return 0;
}
Live demo: http://ideone.com/h2SIHs