Hello,

I suspect that some similar problems are with the following example, therefore I also put it here.

The problem might be my mistake - but I can not see it and this drives me crazy at the moment

I also made another file and remained puzzled again. This time, I did the calculation, then I tried to do the same using the function - but there was an error I could not figure out again where it comes from.
Please look at the picture:



To the left is the calculation and there is a result. To the right there is a function and the same calculation - but there is an error about non matching numbers of rows/columns. Can not say what went wrong because none of the trace() functions has executed and really can not figure out what was wrong with this function.

I made another one, and the problematic function again worked now ??? .



Here is the accompanied file:

functionnotworking.sm

Could someone try to reproduce this one as well, please .

Regards,
Radovan

Edited by user 11 June 2012 15:45:43(UTC)  | Reason: Not specified

Thank you w3b5urf3r

You helped, indeed. Actually, your checking gave me the idea where to track the things. It seems there is some error inside SMath or something else I could not figure out.

Here is the the screenshot of the example which is working and there is no problem about it.


As one can see, the h-value (which is inside the if(),btw.) is calculated as a vector of numbers - as it should be expected.
On the other hand, when using the problematic example:



As one can see, the h-vector is a nested matrix???. I can not figure out why is this happening. On the other hand, if we do that outside the function everything is fine, h-vector is a vector of numbers as it should be. Therefor, if() got confused and everything else broke down.

Here is the file with both examples typeconversionerror-1.sm

Regards,
Radovan

Thanks for answering

On the other hand, I begin to doubt in my common sense
Look at this please:


Believe or not, I can not see what I was doing wrong but this seems to me quite peculiar.
Am I right or not?

Here is the file howisthisposible

Regards,
Radovan

Thank you Oscar,

I think that the point is why my original simple example gives different results. Have no answer on that. Just imagine a bit more complicated functions (like my two examples above with questionable errors - 1.type conversion and 2.wrong vector/matrix dimensions). I am not sure but I suppose there is some bug in SMath. Have no idea at the moment what is going wrong. Just suspect that this simple example has some connection with them.

Regards,
Radovan

I think that the bug is in "line" definition of function.

Look into this file, where I define f(x) without "line": 0.9_line_bug.sm

All answers are ok.

Oscar

Thank you! It's a great help for me.
Will fix this issue as soon as possible.

Regards.

hi,

it seems to work properly


regards,

w3b5urf3r

I confirm, same issue with augment()


Confirmed, but I think can be considered as a separate (unsolved) bug ( Transpose functions bug)


regards,

w3b5urf3r

Edited by user 16 June 2012 18:11:08(UTC)  | Reason: Not specified

I agree


seem right, stack() and augment() should be able to treat a scalar (or a string) as a 1x1 matrix, if other input variables are vectors, scalars or strings


regards,

w3b5urf3r

Edited by user 16 June 2012 23:00:43(UTC)  | Reason: Not specified

Hello,

Just to mention something regarding this issue. I think this has some connection with it.

In v0.94 you will have

f(a):a*mat(b1,b2,b3,3,1)

f(a)=mat(a*b1,a*b2,a*b3,3,1)

f(mat(a1,a2,a3,1,3))=mat(mat(a1*b1,a2*b1,a3*b1,1,3),mat(a1*b2,a2*b2,a3*b2,1,3),mat(a1*b3,a2*b3,a3*b3,1,3),3,1)

f(mat(a1,a2,a3,3,1))=mat(mat(a1,a2,a3,3,1)*b1,mat(a1,a2,a3,3,1)*b2,mat(a1,a2,a3,3,1)*b3,3,1) or
f(mat(a1,a2,a3,3,1))=mat(mat(a1*b1,a2*b1,a3*b1,3,1),mat(a1*b2,a2*b2,a3*b2,3,1),mat(a1*b3,a2*b3,a3*b3,3,1),3,1)

f(mat(u1,v1,z1,u2,v2,z2,u3,v3,z3,3,3))=mat(mat(u1*b1,v1*b1,z1*b1,u2*b1,v2*b1,z2*b1,u3*b1,v3*b1,z3*b1,3,3),mat(u1*b2,v1*b2,z1*b2,u2*b2,v2*b2,z2*b2,u3*b2,v3*b2,z3*b2,3,3),mat(u1*b3,v1*b3,z1*b3,u2*b3,v2*b3,z2*b3,u3*b3,v3*b3,z3*b3,3,3),3,1)

In v0.95 you will have

f(a):a*mat(b1,b2,b3,3,1)

f(a)=a*mat(b1,b2,b3,3,1)

f(mat(a1,a2,a3,1,3))=mat(a1*b1+a2*b2+a3*b3,1,1)

f(mat(a1,a2,a3,3,1))=a1*b1+a2*b2+a3*b3

f(mat(u1,v1,z1,u2,v2,z2,u3,v3,z3,3,3))=mat(u1*b1+v1*b2+z1*b3,u2*b1+v2*b2+z2*b3,u3*b1+v3*b2+z3*b3,3,1)

I think that the situation with v0.95 is much acceptable now than before. On the other hand, I think that we have to be careful now because we can not take for granted that the result of these operation are symbolic vectors, matrices - as we would expect.

Regards,
Radovan

Edited by user 18 June 2012 14:46:23(UTC)  | Reason: Not specified