一个重载问题

student2007 2006-05-17 07:18:52

//: C12:OverloadingUnaryOperators.cpp
// From Thinking in C++, 2nd Edition
// Available at http://www.BruceEckel.com
// (c) Bruce Eckel 2000
// Copyright notice in Copyright.txt
#include <iostream>
using namespace std;

// Non-member functions:
class Integer {
long i;
Integer* This() { return this; }
public:
Integer(long ll = 0) : i(ll) {}
// No side effects takes const& argument:
friend const Integer&
operator+(const Integer& a);
friend const Integer
operator-(const Integer& a);
friend const Integer
operator~(const Integer& a);
friend Integer*
operator&(Integer& a);
friend int
operator!(const Integer& a);
// Side effects have non-const& argument:
// Prefix:
friend const Integer&
operator++(Integer& a);
// Postfix:
friend const Integer
operator++(Integer& a, int);
// Prefix:
friend const Integer&
operator--(Integer& a);
// Postfix:
friend const Integer
operator--(Integer& a, int);
};

// Global operators:
const Integer& operator+(const Integer& a) {
cout << "+Integer\n";
return a; // Unary + has no effect
}
const Integer operator-(const Integer& a) {
cout << "-Integer\n";
return Integer(-a.i);
}
const Integer operator~(const Integer& a) {
cout << "~Integer\n";
return Integer(~a.i);
}
Integer* operator&(Integer& a) {
cout << "&Integer\n";
return a.This(); // &a is recursive!
}
int operator!(const Integer& a) {
cout << "!Integer\n";
return !a.i;
}
// Prefix; return incremented value
const Integer& operator++(Integer& a) {
cout << "++Integer\n";
a.i++;
return a;
}
// Postfix; return the value before increment:
const Integer operator++(Integer& a, int) {
cout << "Integer++\n";
Integer before(a.i);
a.i++;
return before;
}
// Prefix; return decremented value
const Integer& operator--(Integer& a) {
cout << "--Integer\n";
a.i--;
return a;
}
// Postfix; return the value before decrement:
const Integer operator--(Integer& a, int) {
cout << "Integer--\n";
Integer before(a.i);
a.i--;
return before;
}

// Show that the overloaded operators work:
void f(Integer a) {
+a;
-a;
~a;
Integer* ip = &a;
!a;
++a;
a++;
--a;
a--;
}

// Member functions (implicit "this"):
class Byte {
unsigned char b;
public:
Byte(unsigned char bb = 0) : b(bb) {}
// No side effects: const member function:
const Byte& operator+() const {
cout << "+Byte\n";
return *this;
}
const Byte operator-() const {
cout << "-Byte\n";
return Byte(-b);
}
const Byte operator~() const {
cout << "~Byte\n";
return Byte(~b);
}
Byte operator!() const {
cout << "!Byte\n";
return Byte(!b);
}
Byte* operator&() {
cout << "&Byte\n";
return this;
}
// Side effects: non-const member function:
const Byte& operator++() { // Prefix
cout << "++Byte\n";
b++;
return *this;
}
const Byte operator++(int) { // Postfix
cout << "Byte++\n";
Byte before(b);
b++;
return before;
}
const Byte& operator--() { // Prefix
cout << "--Byte\n";
--b;
return *this;
}
const Byte operator--(int) { // Postfix
cout << "Byte--\n";
Byte before(b);
--b;
return before;
}
};

void g(Byte b) {
+b;
-b;
~b;
Byte* bp = &b;
!b;
++b;
b++;
--b;
b--;
}

int main() {
Integer a;
f(a);
Byte b;
g(b);
} ///:~
--------------------------
以上是一个thinking in c++ chapter12 page265的一个程序，在该文件中我搞不懂为何
// Prefix; return decremented value
const Integer& operator--(Integer& a) {
cout << "--Integer\n";
a.i--;
return a;
}
// Postfix; return the value before decrement:
const Integer operator--(Integer& a, int) {
cout << "Integer--\n";
Integer before(a.i);
a.i--;
return before;
}
为何一个有两个参数，而另一个只有一个，在两个参数的函数中，哪个int参数有何用，整个调用过程怎样？
编译器在执行a++或++a时，如何知道该调用那一个函数？++是一元运算符，应该只有一个参数才对啊？

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