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Floor function for int8

8 次查看（过去 30 天）
Alexander Brodsky
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
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Walter Roberson
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) ?

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采纳的回答

Mike Hosea
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
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Jan
Jan 2011-8-1
@Mike: Thanks! I've learned a new command.
Mike Hosea
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.

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更多回答（2 个）

Paulo Silva
Paulo Silva 2011-8-1
a=int8(floor(8.6))
  1 个评论
Alexander Brodsky
Alexander Brodsky 2011-8-1
sorry, my false for bad explanation.
new example:
a=int8(3);
a1=a/4;
I want a1 to be 0.

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Sean de Wolski
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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