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Help on converting dubble to uint8 ?

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Henry Buck
Henry Buck le 18 Sep 2016
Commenté : Henry Buck le 18 Sep 2016
Hey,
I have wave signal that change between 2-20 and it's type of dubble. I need to convert it to type uint8, for example:
The wave signale is INDX(change from 2 to 20) type of dubble. Convert signal is: ADR = uint8(INDX);
the result of that conversion gives me values from 0 to 20. Even when ADR is runs from 0 to 18 or 1 to 19 I get the same result : ADR = 0 to 20 ?
Is it possible to get the same values of INDX after using INDX input ? for example, if INDX is running from 3 to 17, the result (ADR=uint8(INDX)) will be 3 to 17 instead 0f 0 to 20 ? or, if INDX is running from 2 to 18, the result (ADR=uint8(INDX)) will be 2 to 18 instead 0f 0 to 20 ? or, if INDX is running from 2 to 20, the result (ADR=uint8(INDX)) will be 2 to 20 instead 0f 0 to 20 ?
I need to keep the same values of the start and end(of INDX) after using uint8(INDX). uint8 type ,must be use.
Thanks, Henry

Réponse acceptée

Walter Roberson
Walter Roberson le 18 Sep 2016
You must have made a mistake somewhere.
>> uint8(3:17)
ans =
3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
In the expression 3:17 the values are double precision values, so this shows that converting double precision values to uint8 works as expected for small non-negative values.
  5 commentaires
Walter Roberson
Walter Roberson le 18 Sep 2016
Not being a vector is irrelevant.
>> A = 53*(reshape(3:17, 3, 5) + rand(3,5)); B = 48; INDX = uint8(A/B)
INDX =
4 8 10 14 17
5 9 12 15 19
6 9 13 16 19
Henry Buck
Henry Buck le 18 Sep 2016
hey guys,
I have to be honest. It is my mistake.
Thanks again.
Henry

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