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This is executed first by assigning 3.14 to Z, then to Y, and finally to X. But, as Example 9.1 shows, the assignment operator is really a function named opera tar=. In this chain, the function is called three times. On its first call, it assigns 3.14 to Z, so the input to the function is 3.14. On its second call, it assigns 3.14 to Y, so its input again must be 3.14. So that value should be the output (Le., return value) of the first call. Similarly, the output of the second call should again be 3.14 to serve as the input to the third call. The three calls to this function are nested, like this:

Fig. 6-1

fbL f(Y, f(z, 3.14)))

Fig. 6-2

The point is that the assignment operator is a function that should return the value it assigns. Therefore, instead of the return type void, the assignment operator should return a reference to the same type as the object being assigned

Fig. 5.10(a)

Rational& operator=(Rational& r)

Thus in Fig. 6-1, BC is a minor arc and BAC is a major arc. Three letters are needed to indicate a major arc.

class Rational { public: Rational(int =0, int =l); Rational(const Rational&); Rational& operator=(const Rational&); // other declarations go here private: int num; int den; // other declarations go here

A diameter of a circle is a chord through the center. A secant of a circle is a line that intersects the circle at two points. A tangent of a circle is a line that touches the circle at one and only one point no matter how far produced.

T& operator=(const T&);

Thus, CD is a diameter of circle O in Fig. 6-2, EF is a secant, and GH is a tangent to the circle at P. P is the point of contact or the point of tangency.

The return type is a reference to an object of the same class T. But then this means that the function should return the object that is being assigned, in order for the assignment chain to work. So when the assignment operator is being overloaded as a class member function, it should return the object that owns the call. Since there is no other name available for this owner object, C++ defines a special pointer, named this, which points to the owner object. We can envision the this pointer like this:

An inscribed polygon is a polygon all of whose sides are chords of a circle. A circumscribed circle is a circle passing through each vertex of a polygon.

Now we can give the correct implementation of the overloaded assignment operator:

Fig. 6-3 Thus ^ ABD, ^ BCD, and quadrilateral ABCD are inscribed polygons of circle O in Fig. 6-3. Circle O is a circumscribed circle of quadrilateral ABCD.

Write the following function that is passed an array of n pointers to floats and returns a newlv created arrav that contains those n f 1 oat values in reverse order.

Fig. 5.10(b)

float* mirror(float* p[], int n)

A circumscribed polygon is a polygon all of whose sides are tangents to a circle. An inscribed circle is a circle to which all the sides of a polygon are tangents.

Write the following function that returns the number of bytes that before it points to the null character I\ 0 I: unsigned len(const char* s)

Thus, ^ ABC is a circumscribed polygon of circle O in Fig. 6-4. Circle O is an inscribed circle of ^ ABC.

Write the following function that copies the first n bytes beginning with * ~2 into the bytes beginning with * s 1, where n is the number of bytes that ~2 has to be incremented before it points to the null character 1 0 I: \

(a) AB, BC, CD, AC, and AD. These segments may also be named by interchanging the letters; thus, BA, CB, DC, CA, and DA are also correct. (b) AB, AC, and AD (c) BD (d) D (e) C

void cpy(char* sl, const char* s2)

Fig. 1-3