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starlight

Mon Oct 31, 2005 9:44 pm


random number

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int dieroll;
dieroll=(int)(6*Math.random())+1

the above lines generate random numbers between 1-6 but i don't really get what the "+1" at the end of the statment is for. anyone can explain? Like if i want to generate a number between 1 and 6 (not including 6) why can't i just simply use  "dieroll=(int)(7*Math.random())"

and the other quesiton is 

char int d=7
int dd
dd=d++

doesn't d++ mean d=d+1 so shouldn't dd be 8 and d be 7. Why when i tested it means the other way around (d =8 while dd =7)?

and lastly one strange thing that i found in Java is why 3.0/(15/6) equals to 1.5 If 3.0/(15/6) is equal to 1.5 then 15/6 has to equal to 2 meaning the answer will round down. why is that. 

Thanks


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beard0

Mon Oct 31, 2005 10:10 pm



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When you convert (6*Math.random())+1 to an integer, you aren't rounding it, you are truncating it, meaning only if Math.Random returned exactly 1.0 (I'm not sure even if it can) would you end up with a final result of 7. Your second option would have 0 as a possible result.

dd=d++:

d++ returns the value d, and increments d


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Naveg

Mon Oct 31, 2005 10:12 pm


Re: random number

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int dieroll;
dieroll=(int)(6*Math.random())+1

the above lines generate random numbers between 1-6 but i don't really get what the "+1" at the end of the statment is for. anyone can explain? Like if i want to generate a number between 1 and 6 (not including 6) why can't i just simply use  "dieroll=(int)(7*Math.random())"

and the other quesiton is 

char int d=7
int dd
dd=d++

doesn't d++ mean d=d+1 so shouldn't dd be 8 and d be 7. Why when i tested it means the other way around (d =8 while dd =7)?

and lastly one strange thing that i found in Java is why 3.0/(15/6) equals to 1.5 If 3.0/(15/6) is equal to 1.5 then 15/6 has to equal to 2 meaning the answer will round down. why is that. 

Thanks

1) Math.random() generates a number between 0 and 1. Since the first value you are choosing from is 1, you add 1 (0+1=1)

2)This comes down to prefix and postfix operators. The prefix (++d) will increment the variable and then use the new value in the expression. However, the postfix (d++) first uses the variable and only then increments its value.

3) 15/6 is integer division. In such a case, any decimal is truncated (cut off). Be careful not to confuse this with rounding, 3.778 will become 3! So in this example, 15/6 is 2.5, but because of integer division the decimal is truncated and it becomes 2.


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[Gandalf]

Mon Oct 31, 2005 10:16 pm



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The general format for random numbers in any language is this:

RandomInteger = (int) (Math.random() * maxNumber) + minNumber;

*edit* Oops, right beard0.


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beard0

Mon Oct 31, 2005 10:26 pm



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"]The general format for random numbers in any language is this:

RandomInteger = (int) (Math.random() * maxNumber) + minNumber;

errr..... 

RandomInteger = (int) (Math.random() * (maxNumber-minNumber+1) + minNumber


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starlight

Sun Nov 06, 2005 5:03 pm



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Thanks. on the other notes. i am pretty confuse with when you can add the arithmetic and character together. for example
why does 
char c='A'+1 or
char c='A';c='A'+1 or 
char c='A'; c++; works; 
while something like
char c='A'; 
c=c+1; doesn't 


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beard0

Mon Nov 07, 2005 2:39 pm



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Thanks. on the other notes. i am pretty confuse with when you can add the arithmetic and character together. for example
why does 
char c='A'+1 or
char c='A';c='A'+1 or 
char c='A'; c++; works; 
while something like
char c='A'; 
c=c+1; doesn't 

It has to do with the way types are interpreted.  To be safe, explicitly change types, or use the increment operator:
e.g.
char c = 'A';
c = ((char) (((int) c) + 3)); //for increasing by more than 1 or
c++; //for increasing by 1