Talk:Square root: Difference between revisions
imported>StDoodle No edit summary |
imported>Bale A function is not an operator! |
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This is, of course, because the operation has only integers to use, and thus computes an integer result; THEN it converts this to a float. If either of the two numbers is a float (ie "5.0 / 2" or "5 / 2.0") the result would be 2.5, as "expected," because when two different datatypes are used, an implicit conversion to the most precise type is the norm. So it is a bit "unexpected" that square_root() will accept an int, and make an automatic float conversion. We should probably note this somewhere... --[[User:StDoodle|StDoodle (#1059825)]] 15:32, 29 June 2010 (UTC) | This is, of course, because the operation has only integers to use, and thus computes an integer result; THEN it converts this to a float. If either of the two numbers is a float (ie "5.0 / 2" or "5 / 2.0") the result would be 2.5, as "expected," because when two different datatypes are used, an implicit conversion to the most precise type is the norm. So it is a bit "unexpected" that square_root() will accept an int, and make an automatic float conversion. We should probably note this somewhere... --[[User:StDoodle|StDoodle (#1059825)]] 15:32, 29 June 2010 (UTC) | ||
* Whoa, whoa, WHOA! What the heck are you going on about? <code>square_root()</code> is a function that returns a float. It's not a mathematical operator. Like any other function it will always return a value of the appropriate type, even if its parameters are integers. Do you think it is somehow tricky for a function that takes integer parameters to return a float? That's just silly. --[[User:Bale|Bale]] 20:10, 29 June 2010 (UTC) | |||
Revision as of 20:10, 29 June 2010
It won't truncate -- integer truncation only occurs when you have one integer divided by another. If one of the two terms is a float, then the other will be automatically converted to a float before performing the operation. --Heeheehee 17:43, 26 June 2010 (UTC)
It can be a bit confusing for some. square_root() works a little differently than most math operations. For example:
float f; f = 5 / 2;
will return 2.0
This is, of course, because the operation has only integers to use, and thus computes an integer result; THEN it converts this to a float. If either of the two numbers is a float (ie "5.0 / 2" or "5 / 2.0") the result would be 2.5, as "expected," because when two different datatypes are used, an implicit conversion to the most precise type is the norm. So it is a bit "unexpected" that square_root() will accept an int, and make an automatic float conversion. We should probably note this somewhere... --StDoodle (#1059825) 15:32, 29 June 2010 (UTC)
- Whoa, whoa, WHOA! What the heck are you going on about?
square_root()is a function that returns a float. It's not a mathematical operator. Like any other function it will always return a value of the appropriate type, even if its parameters are integers. Do you think it is somehow tricky for a function that takes integer parameters to return a float? That's just silly. --Bale 20:10, 29 June 2010 (UTC)