Hi, by the command of isinteger I can check if it is an integer, however, when defined at first, Matlab assume it is double precision right? So even a=3, isinteger(a) returns 0.
How to solve this problem?

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per isakson
per isakson 2013년 3월 14일
편집: per isakson 2015년 1월 16일

2 개 추천

Matlab represents integers in differnt ways:
  • int8, int16, etc. see Christians answer
  • with a special use of double, which is called flint. See Floating Points
I use this function to test for flint
function isf = isflint( m )
% floating double only
try
bitand( abs( m ), 1 );
isf = true;
catch me
isf = false;
end
end
I picked up the idea from a contribution by the "Pedestrian" in the FEX.
(I stripped off comments and error handling.)

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C Zeng
C Zeng 2013년 3월 15일
Sort of lost of your code. While I am still confused, I use if floor(z)==z or ceil(z)==z to judge if it is indeed an integer.
per isakson
per isakson 2013년 3월 15일
I cannot find the reference now, however see: isodd: a pedestrian parity checker
Most important is that Matlab use flint, which is described in Floating Points
Yes, there are many ways to decide if a double represents an integer: floor, round, ceil, rem, mod and bitand. The Pedestrian, alias us, argue that bitand behaves better with overflow - I've forgotten. I trust the Pedestrian.
C Zeng
C Zeng 2013년 3월 15일
Thanks, Isakson, will take a look.
Jan
Jan 2013년 3월 16일
The problem is to decide, if 1e17 is an integer or not: Because a double can store 16 digits only, this number does not contain any information about the fractional part for reasons of the precision. bitand() detect this and throw an error, while 1e17==round(1e17) replies true, because cropping the fractional part does not influence value.
C Zeng
C Zeng 2013년 3월 17일
Thanks, I understand it. It is good to know.

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추가 답변 (2개)

Jan
Jan 2013년 3월 16일

1 개 추천

Joining the rounding with the checks for overflows:
function isf = isflint(m)
isf = (abs(m) <= bitmax && m == floor(m));
ChristianW
ChristianW 2013년 3월 14일

0 개 추천

doc isinteger
isinteger(int8(3))

댓글 수: 5

C Zeng
C Zeng 2013년 3월 15일
Thanks, however, isinteger(int8(2.5)) also returns 1, which I want to avoid.
isint = @(x) x==round(x)
C Zeng
C Zeng 2013년 3월 16일
yes, I used x==round(x) or floor(x). what do you mean by "isint = @(x)"?
Jan
Jan 2013년 3월 16일
@C Zeng: @(x) x==round(x) is an anonymous function. You find an exhaustive description ion the documentation.
C Zeng
C Zeng 2013년 3월 17일
OK

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2015년 1월 16일

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