Floor function for int8
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I want to round down (floor) variable defined as int8 without converting to double !
for example: a=int8(8.6);
I want the result will equal to 8 instead of 9. Is there way to do or it impossible (without converting to double) ? The reason I need it, because I work with large matrix (25000x25000) of int8.
Thanks Alex
2 个评论
Walter Roberson
2011-8-1
To check the circumstances: do you have something like
int8(86) ./ int8(10)
and you want the result to be int8(8) instead of int8(9) ?
Alexander Brodsky
2011-8-1
采纳的回答
Mike Hosea
2011-8-1
Using your later example,
idivide(a,4,'floor')
does what you want there. I prefer to make both arguments integers of the same type, so I would probably have written
idivide(a,int8(4),'floor').
Note that the default option for idivide is 'fix', which might be what you want. Obviously, 'floor' and 'fix' are the same for non-negative quotients. -- Mike
3 个评论
Sean de Wolski
2011-8-1
Good to know!
Jan
2011-8-1
@Mike: Thanks! I've learned a new command.
Mike Hosea
2011-8-2
You're welcome. :) As I read the question and my response again, I think I should have made it more clear that both arguments need to be integers of the same type in order to avoid a cast to double inside of IDIVIDE.
更多回答（2 个）
Sean de Wolski
2011-8-1
A = int8(magic(10));
B = int8(5);
idx = (mod(A,B)>(B/2)); %elements that need to be reduced.
C = A./B;
C(idx) = C(idx)-1;
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