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By : Xiaobo Zhou
Date : November 21 2020, 07:01 PM
help you fix your problem I am coding up an implementation of Interpolation Search in C. , They are different when the value is < 0. code : ## Why do Python's math.ceil() and math.floor() operations return floats instead of integers?

By : Jake O'Connor
Date : March 29 2020, 07:55 AM
seems to work fine The range of floating point numbers usually exceeds the range of integers. By returning a floating point value, the functions can return a sensible value for input values that lie outside the representable range of integers.
Consider: If floor() returned an integer, what should floor(1.0e30) return? ## math functions with floor

By : Nesar Ahmed
Date : March 29 2020, 07:55 AM
This might help you Floor returns a float, you're testing for an integer.
http://php.net/manual/en/function.floor.php
code :
``````<?php
\$str = "";
\$hour = 1.2;
if (floor(\$hour) > 1) {
\$str = \$str . floor(\$hour) . " hours, ";
}
else if ((int) floor(\$hour) === 1) {
\$str = \$str . floor(\$hour) . " hour ";
}
echo \$str;

?>
`````` ## Javascript pipe in math to get Math.floor without using Math.floor

By : Aamir Pervez
Date : March 29 2020, 07:55 AM
I hope this helps you . A single pipe | is BitWise OR.
Bitwise operator only allow integer values, so after decimal point value is discarded.
code :
``````1010
1100
----------bitwise or
1110
`````` ## Why do I need to do floating point arithmetic with Math.floor

By : Rrr
Date : March 29 2020, 07:55 AM
With these it helps The first line uses floating point maths, which is inaccurate. 128.766*1000 might evaluate to 128765.99999999999 or something similar.
Math.Floor rounds this down to become 128765. ## why // (division floor) produces result but math.floor() gives an OverflowError

By : Nikhil Joshi
Date : October 08 2020, 04:00 PM
I hope this helps you . That's because pow(10,1000) / 1000 is floating-point division and pow(10,1000) // 1000 is integer division.
As you can see, in your case integer division result [is] too large for a float because pow(10,1000) / 1000 attempts to produce a float, but the result will be 10**997, which won't fit even in a 64-bit float. The double-precision floating-point format (a.k.a. "binary64") is fixed-width and allows you to store numbers up to 10**308. If you wanted to store this number anyway, you'd need to use the "binary80" format, which isn't present in Python out-of-the-box and possibly still won't be able to represent the result exactly. 