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Math.round()

 
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public class Example {
public static void main(String [] args) {
double values[] = {-2.3, -1.0, 0.25, 4};
int cnt = 0;
for (int x=0; x < values.length; x++) {
if (Math.round(values[x] + .5) == Math.ceil(values[x])) {
++cnt;
}
}
System.out.println("same results " + cnt + " time(s)");
}
}

what is the result?
A. same results 0 time(s)
B. same results 2 time(s)
C. same results 4 time(s)
D. Compilation fails.
E. An exception is thrown at runtime.

Answer : B

Explanation Given In K&B Book:

Math.round() adds .5 to the argument then performs a floor(). Since the code adds an additional .5 before round() is called, it�s as if we are adding 1 then doing a floor(). The values that start out as integer values will in effect be incremented by 1 on the round() side but not on the ceil() side, and the noninteger values will end up equal.


i didn't understand.can anybody explain me please?
 
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the most crucial point in understanding is the behaviour of round method when the arguement is negative.

rest of the code can be understood by draeing a table and filling in aprropriate values as per the methods
we call this handrun
 
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Niranjan. i didnt understand anything you said.
What do you mean by crucial point.
please do explain.
 
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public static int round(float a)
Returns the closest int to the argument. The result is rounded to an integer by adding 0.5, taking the floor of the result, and casting the result to type int.


Eg1: Math.round(35.6)
Step 1: add 0.5 i.e. 35.6 + 0.5 = 36.1
Step 2: find the floor ( i.e. value towards �ve infinity i.e. greatest integer less than
or equal to given value)
for 36.1 greatest integer, which is less than or equal to 36
hence Math.round(35.6) = 36

Eg2: Math.round(-37.8)
Step 1: add 0.5 i.e. �37.8 + 0.5 = -37.3
Step 2: find the floor ( i.e. value towards �ve infinity i.e. greatest integer less than
or equal to given value)
for -37.3 greatest integer, which is less than or equal to �38
( -37.3 > -38 , -37.3 < -37)
hence Math.round(-37.3) = -38


Math.round() adds .5 to the argument then performs a floor(). Since the code adds an additional .5 before round() is called, it�s as if we are adding 1 then doing a floor().





Math.round(values[x] + .5) means Math. floor ( values[x]+0.5+0.5) i.e. Math.floor(values[x]+1)


Given values are {-2.3, -1.0, 0.25, 4}; to find Math.round(values[x] + .5)value add +1 and apply floor.

Regards
Naresh
[ December 21, 2005: Message edited by: Naresh Kumar ]
 
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Good explanation naresh....
regards,
sri.
 
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