Error in matlab function

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Leonardo Tommasini
Leonardo Tommasini 2017-11-7
回答： Image Analyst 2017-11-7
Hi,
does anybody knows why matlab doesn't calculate the correct value of a function?
I have the function: err_acc = err_acc + (err_corr * dt);
where: err_acc = 0; dt = 0.02; err_corr = -6
and the answer is always 0.
Everything is in int32 type.
Thanks for any help.
Best Regards.
  2 个评论
Leonardo Tommasini
Leonardo Tommasini 2017-11-7
Now I have tried to use typecast(----, 'double'); on every variable, but it goes back to int32 and set the answer to 0.
Do you maybe know why?
Steven Lord
Steven Lord 2017-11-7
Show your exact code. My guess is that you're assigning the result of that computation in double back into an element of an int32 array.

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采纳的回答

Image Analyst
Image Analyst 2017-11-7
Cast to double:
err_acc = int32(0)
dt = 0.02
err_corr = int32(-6)
err_acc = double(err_acc) + (double(err_corr) * dt)
whos err_acc
In the command window you'll see
err_acc =
int32
0
dt =
0.02
err_corr =
int32
-6
err_acc =
-0.12
Name Size Bytes Class Attributes
err_acc 1x1 8 double
Notice that err_acc is now a double, NOT an int32 anymore.

更多回答（2 个）

Cam Salzberger
Cam Salzberger 2017-11-7
编辑：Cam Salzberger 2017-11-7
Hello Leonardo,
If all variables are of int32 datatype, then they can only contain integer values. Thus, dt would contain 0, rather than 0.02, because when a floating point number is cast to an integer datatype, the non-integer part is dropped.
Even if dt is actually a floating point datatype, if you do a double*integer, MATLAB will make the result an integer datatype:
0.02*int32(-6)
ans =
int32
0
Sounds like you might want to cast everything to double before doing your computation.
-Cam
  2 个评论
Leonardo Tommasini
Leonardo Tommasini 2017-11-7
Now I have tried to use typecast(----, 'double'); on every variable, but it goes back to int32 and set the answer to 0.
Do you maybe know why?
Image Analyst
Image Analyst 2017-11-7
Use double(). See my answer.

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Geoff Hayes
Geoff Hayes 2017-11-7
Leonardo - if you are expecting a non-integer answer, then the fact that all variables are int32 will prevent this. If the variables are doubles, then
err_acc = 0;
dt = 0.02;
err_corr = -6
the answer is
err_acc = err_acc + (err_corr * dt)
err_acc =
-0.1200

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