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it1.insert(11.01); ++it1; it1.insert(22.02); ++it1; it1.insert(33.03); for (it2.reset(); !it2; ++it2) it2 = 10*it2; // multiplies each stored number by 10 it3 = it1; // replaces 110.1 with 330.3 in first item The iterators are independent of each other. While it2 traverses the list, it1 remains fixed on
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13.1 13.2 13.3 13.4 What is the difference between a function template and a template function What is the difference between a class template and a template class What are the advantages and disadvantages of using a linked list instead of a vector How is an iterator like an array subscript
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13.1 13.2 13.3 Write and test a program that instantiates a function template that returns the minimum of two values. Write and test a program that instantiates a function template that implements a binary search of a sorted array of objects. Implement and test a template for generating Queue classes. A queue works like a stack, except that insertions are made at one end of the linear structure and removed from the other end. It simulates an ordinary waiting line. Modify the Vector class template so that existing vectors can change their size. Add a constructor to the Vector class template that replicates an ordinary array as a vector. Derive an Array<T,E> class template from the Vector<T> class template, where the second template parameter E holds an enumeration type to be used for the array index.
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13.4 13.5 13.6
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13.1 A function template is a template that is used to generate functions. A template function is a function that is produced by a template. For example, swap(T&, T&) in Example 13.1 is a function template, but the call swap(m, n) generates the actual template function that is invoked by the call. A class template is a template that is used to generate classes. A template class is a class that is produced by a template. For example, Stack in Example 13.3 is a class template, but the type Stack<int> used in the declarations is an actual template class. Vectors have the advantage of direct access (also called random access ) to their components by means of the subscript operator. So if the elements are kept in order, we can locate them very quickly using the Binary Search Algorithm. Lists have the advantage of being dynamic, so that they never use more space than is currently needed, and they aren t restricted to a predetermined size (except for the size of the computer s memory). So vectors have a time advantage and lists have a space advantage.
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Both iterators and array indexes act as locators into a data structure. The following code shows that they work the same way: float a[100]; // an array of 100 floats int i = 0; // an index for the array a[i] = 3.14159; for (i = 0; i < 100; i++) cout << a[i]; List<float> list; // a list of floats ListIter<float> it(list); // an iterator for the list it = 3.14159; for (it.reset(); !it; ++it) cout << it();
Solutions to Problems
13.1 A minimum function should compare two objects of the same type and return the object whose value is smaller. The type should be the template parameter T: template <class T> T min(T x, T y) { return ( x < y x : y ); } This implementation uses the conditional expression operator: ( x < y x : y ). If x is less than y, the expression evaluates to x; otherwise it evaluates to y. Here is the test driver and a sample run: #include "Ratio.h" int main() { cout << "min(22, 44) = " << min(22, 44) << endl; cout << "min(66.66, 33.33) = " << min(66.66, 33.33) << endl; Ratio x(22, 7), y(314, 100); cout << "min(x, y) = " << min(x, y) << endl; } min(22, 44) = 22 min(66.66, 33.33) = 33.33 min(x, y) = 314/100 A search function should be passed the array a, the object key to be found, and the bounds on the array index that define the scope of the search. If the object is found, its index in the array should be returned; otherwise, the function should return -1 to signal that the object was not found: template<class T> int search(T a[], T key, int first, int last) { while (first <= last) { int mid = (first + last)/2; if (key < a[mid]) last = mid - 1; else if (key > a[mid]) first = mid + 1; else return mid; } return -1; // not found } Within the while loop, the subarray from a[first] to a[last] is bisected by mid. If key < a[mid] then key cannot be in the second half of the array, so last is reset to mid 1 to reduce the scope of the search to the first half. Otherwise, if key > a[mid], then key cannot be in the first half of the array, so first is reset to mid+1 to reduce the scope of the search to the second half. If both conditions are false, then key == a[mid] and we can return.
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