Thread sends notify() before waiting thread says wait()

Question

In the code below the ref book I'm using "Sun certified Programmer for Java 2" had the calculator thread starting before the Reader thread was even created and hence the calculator thread was finished it's task and sending a notify() before the Reader even had it's socks on and saying wait()
( And this cost me a lot of time tryin to figure this problem out as the code must have being written for a Z80 processor and not a 2.4GHz P4)
To fix this I put the calculator starting after the reader thread, but .......
I would have thought that java would have some protection against this happening, ie, like a notify flag which a thread can set if it has done it's job BEFORE the waiting thread even starts waiting, ie a preemptive mechanism to help speed things up ?
package MyPackage;
public class Reader extends Thread {
Calculator MyCalculator;
public Reader(Calculator calc){
MyCalculator = calc;
public void run(){
synchronized(MyCalculator){
try{
System.out.println("Waiting for calculation ... ");
MyCalculator.wait();
}catch (InterruptedException e){}
System.out.println("Total is: " + MyCalculator.total);
public static void main(String args []){
Calculator calculator = new Calculator();
//calculator.start();
new Reader( calculator ).start();
calculator.start();
package MyPackage;
public class Calculator extends Thread{
int total;
public void run(){
synchronized(this) {
for(int i=0; i<100; i++){
total += i;
System.out.println("Calculation complete, notifying waiting thread..");
notify();

Answer 1

It's up to the 'umble programmer to code it right. The code you quoted was crap.
The most common way to wait is on a condition:
while (hotAndBotherd())
    obj.wait();And then the notifying thread will take steps to change that:
//code that cools off the hot and bothered
obj.notify[all]();If the initial state of things is hot and bothered then there is no race condition: the first thread will wait.