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Java Ceil Weirdness


Java Ceil Weirdness

By : user2949214
Date : November 16 2020, 06:23 AM
will be helpful for those in need Math.ceil does, indeed, cast the integer to a double. But it only does so after the integer operation has been performed. This is the order of operations:
code :
int a = 1;
int b = 2;
int c = a / b; // now equals 0, because it's an integer operation.
double d = (double)c; // now it's a double - but equals 0.0.
double e = Math.ceil(d); // still 0.0.


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Why does 2.ceil() throws exception whereas 2.3.ceil() returns 3 in prototype.js?

Why does 2.ceil() throws exception whereas 2.3.ceil() returns 3 in prototype.js?


By : user2996962
Date : March 29 2020, 07:55 AM
should help you out To make it more generic ... I change the prototype of Number object in Javascript by doing this , The method is Math.ceil, it's not an instance method.
code :
var b = 2;
console.log(Math.ceil(b));

b = 2.3;
console.log(Math.ceil(b));
Number.prototype.ceil = function() {
    return Math.ceil(this);
};

console.log(2.3.ceil());
2.ceil(); // does not work
2.3.ceil(); // does work
2..ceil(); // does work
Bash: Using Python-function Ceil - How to do ceil of the product of two variables?

Bash: Using Python-function Ceil - How to do ceil of the product of two variables?


By : Chennuri Mounika
Date : March 29 2020, 07:55 AM
it helps some times Python uses . as the decimal point since , is the argument separator. So if you use LOSS_RATE=0.3 everything should work fine:
code :
> NUMBER=600
> LOSS_RATE=0.3
> python -c "from math import ceil; print ceil($NUMBER * $LOSS_RATE)"
180.0
> echo "from math import ceil; print ceil($NUMBER * $LOSS_RATE)"
from math import ceil; print ceil(600 * )
Why is there no ceil(float) in Java?

Why is there no ceil(float) in Java?


By : Palchuri Bhargavi
Date : March 29 2020, 07:55 AM
this will help Suppose I would like to round up float to int in Java. , You can do:
code :
value = (int) Math.ceil(value);
(Java) why is Math.ceil(15/10) is 1.0 and not 2.0?

(Java) why is Math.ceil(15/10) is 1.0 and not 2.0?


By : Nirmeen Kh Odeh
Date : March 29 2020, 07:55 AM
may help you . Because dividing an integer by another integer gives you an integer : 15/10 is 1.
From the specification :
How to find asymptotic complexity when the function has a ceil in it ? (2^(2^ceil(log2(n)))) = O( 2^n )?

How to find asymptotic complexity when the function has a ceil in it ? (2^(2^ceil(log2(n)))) = O( 2^n )?


By : 何耀文
Date : March 29 2020, 07:55 AM
it fixes the issue We need to show that, for every constant c, there exists n such that 2^(2^ceil(log2(n))) > c * 2^n. Let's consider only n = 2^k + 1 for some integer k > 1; this is our right, since we are not trying to prove the statement for all n. The desired inequality becomes
code :
2^(2^ceil(log2(2^k + 1))) >? c * 2^(2^k + 1).
ceil(log2(2^k + 1)) = k + 1
2^(2^ceil(log2(2^k + 1))) = 2^(2^(k + 1)).
2^(2^(k + 1)) >? c * 2^(2^k + 1).
2^(2^(k + 1) - 2^k - 1) = 2^(2^k - 1) >? c.
2^k - 1 >? log2(c)
2^k >? log2(c) + 1
k >? log2(log2(c) + 1).
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