Program requires user to enter non duplicate numbers c++ - c++

Question

I have written a function in a program for entering a unique number but its not working. Something is wrong with the for loop.
I need to validate that employee id is unique.
I have made a structure named employee and "emp.id" is employee id. When the user inputs an id, it should not match previous Id's which user might have entered before. This is just a function of the main program, which validates that employee id is unique.
void uniquieid()
{
int check,i;
string code;
string tempemp1;
cout<< "enter id";
cin>> code;
while(!(num-1))
{
for(i=0;i<=num-1;i++)
{
if(emp[i].id.compare(code)==0)//comparing
{
check =1;
cout<<"enter id again";
break;
}
if(check=0) //csaasc
{
emp[i].id=code;
}
}
}
getch();
}

Answer 1

If the order that the ids are entered doesn't matter, I would do something like (note: untested):
using EmpIds = std::set<std::string>;
void addUniqueId(EmpIds& ids)
{
std::pair<EmpIds::iterator, bool> inserted;
const char* again = "";
do {
std::cout << "enter id" << again;
again = " again";
std::string id;
if (!(std::cin >> id))
throw std::runtime_error("No more ids!");
inserted = ids.insert(id);
} while (!inserted.second);
}

Answer 2

There are so many things wrong with the code, but maybe it should look more like this:
void uniqueid() {
int check=1;
string code;
string tempemp1;
cout<< "enter id";
while(check) {
cin >> code;
check = 0;
for (int i = 0; i < num; ++i) {
if (emp[i].id.compare(code)==0) {
check = 1;
cout << "enter id again";
break;
}
}
if (check==0) {
/* emp[i].id=code; */
}
}
getch();
}
Note how int check=1; starts at 1 to mean that the code needs re-entering.
So while(check) means that while the code is not unique keep going.
The for loop does the compare as before, but note the idiomatic form.
The other if (check==0) is outside the for loop and this means that no duplicates were detected so code can be used. However, I'm not sure which employee the code should apply to so I've just commented out the code.

Answer 3

Can you post the employee structure?
Because from this, everything looks OK, but your if function refers to emp.
So something in your structure is causing the problem.
Without your structure, anyone answering probably won't be able to find the problem.
Right now, all i can advise you to do is store employee ids in a vector and iterate through it using a for loop.
You could do
void uniqueid() {
std::vector<std::string> empIds;
std::string code;
CODE TO STORE IDs INTO VECTOR HERE;
int vectorLength = empIds.size();
std::cout << "enter id";
std::cin >> code;
for (int i = 0; i < vectorLength; i++) {
if (empIds[i] == code) {
std::cout << "enter id again";
std::cin >> code;
} else {
empIds.push_back(code);
}
}
}

Answer 4

For a start, something like the below should work.
map <string, bool> seen;
bool isUniqueId(string id)
{
return seen[id];
}
void addId(string id)
{
seen[id] = true;
}
From main(), whenever user inputs a string id, use isUniqueId(id) to ensure its unique, and if its unique, call addId(id).
Edit : (upon request from the OP)
Your transformed code may look like below after using map.
// Global map, defaults to false
map <string, bool> seen; // seen map to store if an id is seen already or not.
void uniqueId()
{
bool good = true; // set up a good flag to check if id is good or not.
int numEmployees = 0; // Count to store number of employees with unique ids so far
string id;
cout<< "enter id\n";
cin>> id;
while(good)
{
good = false; // Assume this is unique!
if(seen[id]) // Check if we already saw this id before
{
good = true; // Alas! We already have seen this id
cout<<"enter id again\n";
continue; // If id already exists, ask for another id setting good = true;
// Note that the above continue is NOT required as loop will run again (good = true).
// Just for clarity sake.
}
else
{
// Voila, we have a new employee with unique id.
seen[id] = true; // Unique, mark as seen now
emp[numEmployees].id=code; // Note numEmployees here
numEmployees++; // Increment the count
}
}
getch();
}
At the end of while loop, you would have successfully gotten a unique id from user, otherwise it will keep asking the user for new id.