How to solve this :

[xp_vista007]

Can anyone help me solving this :

For int a = 2 , and double b = 2.0, what are the results of the following expressions :
a = 12 * 10 % 5 * (22 * 3 % 2);
b += 1.5 * 3 + (a++);
a %= 3/a + 3 * (++b).


For x = 1 and z = 0 , what are the results of the boolean expressions :
(true) && (x >= x) || !((x > 0) && (x > y))
((x != 1) == !(x == 1)) ^ (z++ && x)

[sunde887]

Why can't you do this? We know how to do this already and we don't do peoples homework, we will, however; help you if you get stuck.

[FlipPoker@gmail.com]

Yes. Review order of precedence and post/pre-increment. Also know that a statement like a += 1 is a shortcut for a = a + 1. If I were you, I would work these out on paper rather than type them into a program. It's important to understand these rules.

[xp_vista007]

I can do this, but i have problems with statements like this ("+=", "%="),
i know that b++ is like b = b + 1; but what with ++b ????

[FlipPoker@gmail.com]

b++ is an example of postincrement. ++b is an example of preincrement. By themselves, there is no difference they just increment b:

int b = 0;
b++; //add 1 to b (b is 1)
++b; //add 1 to b (b is 2)

But the difference can be seen in this example:

int b;
b = 0;
System.out.println(b++);  //will print 0
    	 
b = 0;
System.out.println(++b); //will print 1

In the first output statement above, 0 will be printed because b is incremented after the println() method is executed (postincrement). In the second output statement, 1 will be printed because b is incremented before the println() method is executed (preincrement). You can think of postincrement and preincrement as shortcuts. You can write equivalent code by incrementing b outside of the println() method.

b += 1 is equivalent to b = b + 1
b %= 1 is equivalent to b = b % 1
b -=1 is equivalent to b = b - 1
b *= 1 is equiavlent to b = b * 1;
See the convention?

[xp_vista007]

Thnx . This was very helpfull!!