Floor function for int8

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

[Walter Roberson]

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]

exactly !

Answer 1

[Mike Hosea]

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

[Sean de Wolski]

Good to know!

[Jan]

@Mike: Thanks! I've learned a new command.

[Mike Hosea]

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.

Answer 2

[Paulo Silva]

a=int8(floor(8.6))

[Alexander Brodsky]

sorry, my false for bad explanation.

new example:

a=int8(3);

a1=a/4;

I want a1 to be 0.

Answer 3

[Sean de Wolski]

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;