void print_string (const string& str)

cout << *itr;

Come up with a way to solve the problem by working with the solution to a smaller version of the same problem. Solve the base case.

vector ::iterator itr = vec.begin();

name_to_email.erase( "Alex Allain" );

The newline itself will be discarded—your input will contain everything up to the newline,

vector ::iterator

for ( vector ::iterator itr = vec.begin(); itr != vec.end(); ++itr )

#include #include using namespace std; map<string, string> name_to_email;