search for string in array of structs C++ - c++

Question

I am trying to create a function that allows me to type a name and return that employees information found in the struct. /.......................................................................
// info (ss_number,etc..)
struct employees {
int ss_number;
int dob; //date of birth
string f_name;
string l_name;
string state;
};
void print_person(employees e)
{
cout<<e.ss_number<<" "<<e.dob<<" "e.f_name<<" "<<e.l_name<<" "<<e.state;
}
void search(employees array[])
{
string first;
string last;
cout << "Enter name"; //Example Jack Patton
cin >> first >> last;
for(int i = 0; i < 10; i++) {
if(array[i].f_name == first && array[i].l_name == last) {
print_person(array[i]);
}
}
}
void main()
{
employees array[10];
search(array);
}

Answer 1

*t is better to place the request for the name outside the function. The function should do only the search.
It could be defined the following way
const employee * search( const employees array[], size_t n,
const std::pair<std::string, std::string> &name )
{
const employees *first = array;
while ( first != array + n &&
!( first->f_name == name.first && first->l_name == name.second ) )
{
++first;
}
return first;
}
And in main it could be called the following way (take into account that main shall have return type int):
int main()
{
const size_t N = 10;
employees array[N];
// filling the array
std::pair<std::string, std::string> name;
cout << "Enter name"; //Example Jack Patton
cin >> name.first >> name.second;
const employees *emp = search( array, N, name );
if ( emp != array + N )
{
print_person( *emp );
}
//...
}
Or you could use standard algorithm std::find_if with a predicate.
At the same time function print_person should be defined as
void print_person( const employees &e)
{
cout<<e.ss_number<<" "<<e.dob<<" "e.f_name<<" "<<e.l_name<<" "<<e.state;
}

Answer 2

What you have is mostly fine except that you never initialized any values for your array.
You could read them in from a file, but in the meantime write this and it will work (thanks M2tM for the data):
employees array[10] =
{ { 100, 1985, "John", "Edwards", "AB" }
, { 101, 1960, "Mike", "Thomas", "CA" }
/* fill in more rows if you want */
};
Other issues: in print_person you are missing an << before e.f_name ; and void main() should be int main().

Answer 3

This requires C++11 and fulfills your requirements. Notice I switched to vector (since you're using string you are using C++), and am using a built in algorithm with a lambda. C++ has a lot of built in functionality in the standard libraries to make this sort of thing much easier.
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;
struct Employee
{
int socialSecurtyNumber;
int yearOfBirth;
string firstName;
string lastName;
string state;
};
vector<Employee>::iterator search(vector<Employee> &employees, const string &first, const string &last)
{
return find_if(begin(employees), end(employees), [&](const Employee& employee){
return employee.firstName == first && employee.lastName == last;
});
}
int main()
{
vector<Employee> employees{
{ 100, 1985, "John", "Edwards", "AB" },
{ 101, 1960, "Mike", "Thomas", "CA" },
{ 102, 1992, "George", "Flunky", "FL" },
{ 103, 1983, "Tweet", "Johnson", "AB" },
{ 104, 1968, "Michael", "Jordan", "OH" },
};
string first;
string last;
cout << "Enter name: "; //Example Jack Patton
cin >> first >> last;
auto employee = search(employees, first, last);
if (employee != end(employees)){
cout << "FOUND! SSN: " << employee->socialSecurtyNumber << " YOB: " << employee->yearOfBirth << endl;
}else{
cout << "NOT FOUND!" << endl;
}
}