CSE2050 Fall 2004 Exam #1, open book, open notes. Name ____________________

1.  True or false.  Circle T or F (2 pts each)

ANSWERS
  F  Indexing an array out of bounds throws an exception.     (no bounds check)
  T  == has higher precedence than =
  T  cout is an object.
  F  "dog" > "cat"                                            (comparing array addresses, not contents)
  F  100 / 7 * 7 == 100                                       (integer division drops remainder)
  T  A pointer is a special case of an iterator.
  T  Pass by reference is faster than pass by value.
  F  Expressions must be passed to a function by reference.   (they cannot be)
  T  **argv == argv[0][0]
  T  a->b == (*a).b

2. Write templated function v2m(v, m) to copy the contents of a vector<T> v
to map<int, T> m such that v[i] == m[i] for all i from 0 to v.size()-1
as in this example (30 pts):

  vector<string> v(5);
  map<int, string> m;
  v[3] = "hi";
  v2m(v, m);
  cout << m[3] << m.size();  // hi5

  // ANSWER
  template <class T>
  void v2m(const vector<T>& v, map<int, T>& m) {
    for (int i=0; i<int(v.size()); ++i)
      m[i] = v[i];
  }


3.  Write code to print the contents of map<int, string> m in the format
"key = value", one line per pair, when the value is not "".  In the above
example, only the line "3 = hi" (without the quotes) would be printed.
Do not assume the keys are contiguous from 0 to m.size()-1 (20 pts):

  // ANSWER
  for (map<int, string>::iterator p=m.begin(); p!=m.end(); ++p)
    if (p->second != "")
      cout << p->first << " = " << p->second << endl;


4.  The following templated function prints a sequence.  What type of
iterators does it expect? (10 pts)

  template <class T>
  void print(T b, T e) {
    for (int i=0; i < e-b; ++i)
      cout << b[i] << "\n";
  }

  ANSWER: random access (because of e-b and b[i])

5.  Rewrite print() to accept input iterators. (20 pts)

  // ANSWER
  template <class T>
  void print(T b, T e) {
    while (b!=e)
      cout << *b++ << "\n";
  }

  // OR
  template <class T>
  void print(T b, T e) {
    for (; b!=e; ++b)
      cout << *b << "\n";
  }