# Q: Math.round(Math.random() + 2.50001; ????

david eberhardt

Ranch Hand

Posts: 158

posted 13 years ago

I say "C" but the book says B ???

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I set up a program to test this ... one of the random numbers I got was 0.9587593517384946

so I figure I could eventually come up with a number like 0.9999993517384946

so if I add 0.9999993517384946 + 2.50001

I get something like 3.50000035

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now the documentation for

so (long)Math.floor(3.50000035 + 0.5d)

== (long)Math.floor(3.50000035 + 0.5d)

== (long)Math.floor(4.00000035)

== 4.

[B] I'd say most of the time you will get 3 as the answer EXCEPT when the random(0 number is greater than .999999...

based on this possibility, how can we say other than "impossible to say" as the right choice?

- 0

*Here's a question from the Programmer's Final Exam in the Simon Roberts "Complete Java2 Certification Study Guide"*

What is the value of the following expression?

Math.round(Math.random() + 2.50001);

A. 2

B. 3

C. It is impossible to say.

I say "C" but the book says B ???

------------------------------------------------

I set up a program to test this ... one of the random numbers I got was 0.9587593517384946

so I figure I could eventually come up with a number like 0.9999993517384946

so if I add 0.9999993517384946 + 2.50001

I get something like 3.50000035

---------------------------------------------

now the documentation for

public static long round(double a)

Returns the closest long to the argument. The result is rounded to an integer by adding 1/2, taking the floor of

the result, and casting the result to type long. In other words, the result is equal to the value of the expression:

(long)Math.floor(a + 0.5d)

so (long)Math.floor(3.50000035 + 0.5d)

== (long)Math.floor(3.50000035 + 0.5d)

== (long)Math.floor(4.00000035)

== 4.

*floor - Returns: the largest (closest to positive infinity) floating-point value that is not greater than the argument and is*

equal to a mathematical integer.

equal to a mathematical integer.

[B] I'd say most of the time you will get 3 as the answer EXCEPT when the random(0 number is greater than .999999...

based on this possibility, how can we say other than "impossible to say" as the right choice?

Dan Chisholm

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Posts: 1865

david eberhardt

Ranch Hand

Posts: 158

posted 13 years ago

- 0

Dan,

thanks.

I had set up a small loop to see if and when I got something equal to or bigger than 0.999999 (that's when the result will end up as a 4) .... didn't happen as I made the loop run only 100 times.

Usually, the random number is less than that.

I might just set up a loop to run 10,000 times and test for that number ... I imagine I will eventually see one!

Anyways - I guess that sometimes book writers are pressed for time and miss one now and then.

thanks.

I had set up a small loop to see if and when I got something equal to or bigger than 0.999999 (that's when the result will end up as a 4) .... didn't happen as I made the loop run only 100 times.

Usually, the random number is less than that.

I might just set up a loop to run 10,000 times and test for that number ... I imagine I will eventually see one!

Anyways - I guess that sometimes book writers are pressed for time and miss one now and then.

**Here's some code I wrote to show how I came up with my answer "impossible to say"**since choosing 3 is not correct for all possible values that math.random() could generate:
david eberhardt

Ranch Hand

Posts: 158

posted 13 years ago

- 0

some of the output from my test program:

C:\BIN>java OutThing

long i1 = Math.round(0.999999 + 2.50001);

i1 = 4

long i2 = Math.round(Math.random() + 2.50001);

0.052595364524934185 3

0.7981981280117669 3

0.3908890857231241 3

[ November 30, 2002: Message edited by: david eberhardt ]

C:\BIN>java OutThing

long i1 = Math.round(0.999999 + 2.50001);

i1 = 4

long i2 = Math.round(Math.random() + 2.50001);

0.052595364524934185 3

0.7981981280117669 3

0.3908890857231241 3

[ November 30, 2002: Message edited by: david eberhardt ]

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