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Asked by C Zeng on 14 Mar 2013

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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Answer by per isakson on 14 Mar 2013
Edited by per isakson on 15 Mar 2013

Matlab represent 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.)

C Zeng on 15 Mar 2013

Thanks, Isakson, will take a look.

Jan Simon on 16 Mar 2013

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 on 17 Mar 2013

Thanks, I understand it. It is good to know.

Answer by Jan Simon on 16 Mar 2013

Joining the rounding with the checks for overflows:

```function isf = isflint(m)
isf = (abs(m) <= bitmax && m == floor(m));
```

## 1 Comment

per isakson on 18 Mar 2013

I'll replace my cryptic isflint.

Answer by ChristianW on 14 Mar 2013
```doc isinteger
```
```isinteger(int8(3))
```