Help me please

> I am sure that if you give 10 programmers this problem and ask them what the output will be, more than 5 would probably say 2.

Oh yes -"programmers" will probably say 1 or 2. But programmers will say "Such operations like a = a++ is ambiguous and should never be used. Result of such operatons is unpredictible and depends on compiler/interpreter realisation".

> What will be printed out in the following code and what is the difference between this and your code?

> int y = n++;

This is simple and straight code for variable "n": increase n by one. Try to explain in same way step-by-step changes of variable "a" within your first piece of code.

Michael.Nazarov@sun.coma at 2007-7-13 22:43:55 >

> so that you don't know why the value it will print, why you said that

"Read mannual about prefix and postfix ?".

This is first part of selfeducation you should perform.

> I only want to know why the result is 1

Because your expression virtually equals to followed set:

a = a + 1; // ++

a = 1; // = with original a value before ++

Read ++ manual and good programming book, I insist...

Michael.Nazarov@sun.coma at 2007-7-13 22:43:55 >

> Every body knowed

> n = a++ ;

> will set

> 1. n = a;

> 2. a = a + 1; -> so a will be 2

>

> So why

> a = a++; don't set

> 1. a = a;

> 2. a = a + 1;

>

> May be i'm wrong ?

Yes you are wrong. In Java, the right side of an = is evaluated first, as BIH001 says. To use your examples.

n = a++

1. expression = a

2. a = a + 1

3. n = expression

a = a++

1. expression = a

2. a = a + 1

3. a = expression

A Java implementation that supports the Java Language Specification must work this way.

atmguya at 2007-7-13 22:43:55 >