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Subdomain Scalene: Increasing size Decreasing size Largest as second Isosceles: a=b & other side larger a=c & other side larger b=c & other side larger a=b & other side smaller a=c & other side smaller b=c & other side smaller Equilateral: All sides equal Not a triangle: Largest rst Largest second Largest third Bad inputs: One bad input Two bad inputs Three bad inputs

n = n - 1 ;

they simply decrease the value of n by 1. (The increment operator ++ was used in the name C++ because it increments the original C programming language; it has everything that C has, and more.) However, when used as subexpressions (i.e., expressions within expressions), the pre-increment operation ++m is different from the post-increment operation m+ +. The pre-increment increases the variable first before using it in the larger expression, whereas the post-increment increases the value of the variable only after using the prior value of the variable within the larger expression. Since the incrementing process is equivalent to a separate assignment, there are really two statements to be executed when the increment operation is used as a subexpression: the incrementing assignment and the larger enclosing statement. The difference between the pre-increment and the post-increment is simply the difference between executing the assignment before or after the enclosing statement.

CHAP. l]

This shows the difference between the pre-increment and the post-increment:

Example Test Case (3,4,5 scalene) (5,4,3 scalene) (4,5,3 scalene)

operators:

m = 66, n; ++m; << -x m << '1, n = C-C n C-C endl; m++; << I m = m n = cc n << endl; << m cc endl; tout -x I'm = '1 C-K ++m << endl; return 0;

In the first assignment, m is pre-incremented, increasing its value to 67, which is then assigned to n. In the second assignment, m is post-incremented, so 67 is assigned to n and then m is increased to 68. In the third output statement, m is post-incremented, so the current valueof m (68) is dropped into the output stream and then m is increased to 69. In the last output statement, m is pre-incremented, so m is increased to 70 firstandthen that value is dropped into the output stream.

En dash Em dash Small tilde Trade mark sign Latin small letter s with caron Single right-pointing angle quotation mark Latin small ligature OE Latin capital letter Y with dieresis

Use of the increment and decrement operators as subexpressions can be tricky and should be used with caution. For example, the order of evaluations of expressions that involve them is not defined by the language and consequently can be unpredictable:

(5,5,8 isosceles) (5,8,5 isosceles) (8,5,5 isosceles) (8,8,5 isosceles) (8,5,8 isosceles) (5,8,8 isosceles)

EXAMPLE 1.19 The Unpredictablilty of the Order of Evaluation of Subexpressions #include main0 int n = x = ++n tout << tout << 5, x; * --n; "n = ' << n <=c ', x = ' C-K x -CC endl; ++n << u '1 << ++n << 'I ' << ++n << endl; <iostream.h>

In the assignment to x, n is first increment to 6 and then decremented back to 5 before the multiply operator is evaluated, computing 5 * 5. In the last line, the three subexpressions are evaluated from right to

int { 1

Rational&

(5,5,5 equilateral)

y.num*x.den);

Rational: :Rational(int n=O, int d=l) : num(n), den(d) -t if (d == 0) n = 0; else if (d < 0) { n *= -1; d *= -1; } reduce(); > The modification ensuring that den > 0 could instead be done in the reduce ( ) function, since that utility should be called by every member function that allows den to be changed. However, none of our other member functions allows the sign of den to change, so by requiring it to be positive when the object is constructed we don t need to check the condition again. 9.14 Implement a Vector class, with a default constructor, a copy constructor, a destructor, and

overloaded assignment operator, subscript operator, equality operator, stream insertion operator, and stream extraction operator.