Prevent duplicate entry into vector? C++ [closed] - c++

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I have a program that prompts a user to enter a value. Each value that the user enters gets placed into a vector 'other' which is there only for validation. If a duplicate value has been entered, the user will get a prompt until they enter a unique value.
The main problem I am facing is that for some reason when running the code and printing out the results of the vector, there appears to be a duplicate entry. Could anyone please tell me why that is?
See below for my code:
// prompt to continue
cout << "Would you like to continue? [Y]es, [N]o: ";
cin >> toContinue;
while (toContinue == 'Y')
{
bool isDuplicate = 0;
// prompt for product no.
cout << "Please enter product number: ";
cin >> productB;
// Validation check for duplicate entries
for (size_t i = 0; i < other.size(); i++)
{
if (productB == other[i])
isDuplicate = 1;
while (isDuplicate == 1)
{
cout << "You have already entered this product number!" << endl;
cout << "Please enter correct product number: ";
cin >> productB;
if (productB != other[i])
other.push_back(productB);
isDuplicate = 0;
}
}
// prompt to continue
cout << "Would you like to continue? [Y]es, [N]o: ";
cin >> toContinue;
}

Answer 1

While its common to use std::set for unique elements, if you function must return a vector for some reasons, I used such approach:
std::set<int> my_set;
my_set.insert(1);
my_set.insert(2);
my_set.insert(1);
// ... insert more
std::vector<int> my_vector(my_set.size());
std::copy(my_set.begin(), my_set.end(), my_vector.begin());
assert(my_vector.size()==2);
Note that the vector my_vector will be sorted.

Answer 2

Once you enter a duplicate, you let the user re-enter a number; then you check only if the new entered number is the same as the duplicate entered before; but you do not check if the user entered a different but still duplicate value.
Generally, you mix user input with program logic; unbundling this makes the code more readable and less error prone. See, for example, the following fragments showing how one could separate these concerns:
bool isContained(const vector<int> &v, int value) {
// your code to check for duplicates goes here
}
int main() {
...
while (toContinue == 'Y') {
// prompt for product no.
cout << "Please enter product number: ";
cin >> productB;
if (isContained(other, productB)) {
cout << "You have already entered this product number!" << endl;
}
else {
other.push_back(productB);
}
// prompt to continue
cout << "Would you like to continue? [Y]es, [N]o: ";
cin >> toContinue;
}
}
Also a general hint: using appropriate data structures can also help to avoid unnecessary lines of codes; A container avoiding duplicates, for example, is std::set.

Answer 3

You can help yourself a great deal by breaking up the components of your logic into smaller functions.
Most of what I have done here is tidy up, but note the encapsulation of the contains function.
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
bool contains(std::vector<int> const& vals, int val)
{
return std::count(std::begin(vals), std::end(vals), val) != 0;
}
bool shouldContinue()
{
char toContinue;
cout << "Would you like to continue? [Y]es, [N]o: ";
cin >> toContinue;
return toContinue == 'Y';
}
int getProduct(bool again)
{
int productB;
if (again)
{
cout << "You have already entered this product number!" << endl;
}
cout << "Please enter correct product number: ";
cin >> productB;
return productB;
}
void printProducts(std::vector<int> const& vals)
{
std::cout << "You have selected:";
const char* sep = " ";
for(int p : vals)
{
std::cout << sep << p;
sep = ", ";
}
std::cout << std::endl;
}
int main()
{
std::vector<int> other;
while (shouldContinue())
{
int productB = getProduct(false);
while(contains(other, productB))
{
productB = getProduct(true);
}
other.push_back(productB);
}
printProducts(other);
}