Error in matlab function

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.

[Leonardo Tommasini]

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]

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.

Answer 1

[Image Analyst]

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.

Answer 2

[Cam Salzberger]

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

[Leonardo Tommasini]

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]

Use double(). See my answer.

Answer 3

[Geoff Hayes]

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