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Thread: How to solve this :
 03232011, 01:29 PM #1Member
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How to solve this :
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)
 03232011, 01:34 PM #2
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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.
 03232011, 01:59 PM #3Senior Member
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Yes. Review order of precedence and post/preincrement. 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.
 03242011, 01:01 PM #4Member
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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 ????
 03242011, 01:18 PM #5Senior Member
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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:
Java Code:int b; b = 0; System.out.println(b++); //will print 0 b = 0; System.out.println(++b); //will print 1
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?
 03242011, 07:19 PM #6Member
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