how to save variables not in scientific notation

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Nick Thomas
Nick Thomas am 22 Apr. 2018
Kommentiert: Walter Roberson am 22 Apr. 2018
I have certain variables saving in scientific notation. I am trying to use them in creating a matrix, for example:
nc = 19840; %saves the variable as 1.9840e+04
A = ones(nc,nc)
When I try to do this, I get the following error:
Error using ones
Size inputs must be integers.
I guess Matlab isn't reading it as an integer? Is there a way to make it save the variable not in scientific notation? (I don't have the problem with anything not saved in sci notation) I've tried things like sprintf and num2str without luck.
  4 Kommentare
Stephen23
Stephen23 am 22 Apr. 2018
Bearbeitet: Stephen23 am 22 Apr. 2018

@Nick Thomas: please show the output of these operations:

>> nc - 19840
>> whos nc
Nick Thomas
Nick Thomas am 22 Apr. 2018
Bearbeitet: Walter Roberson am 22 Apr. 2018
>> nc-19840
ans =
     3.6380e-12
>> whos nc
    Name      Size            Bytes  Class     Attributes
    nc        1x1                 8  double              

This is interesting, now I see that it's somehow storing a small decimal amount with the value. It doesn't make sense to me why though, here is my code for calculating nc along with the variables used.

nb = 128;
h = 0.2;
t_s = 0.004;
t_i = 0.05;
s = 500;
nc = nb+(nb*((h+2*t_s+2*t_i)*s));

So there is no reason I should be having a weird decimal at the end.

Knowing this I fixed the problem using the round(x,0) command. However, I'm still curious as to why this happened?

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the cyclist
the cyclist am 22 Apr. 2018

The reason is that the intermediate values in your calculation cannot be stored exactly. Here is one starting point for understanding why that is.

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Walter Roberson
Walter Roberson am 22 Apr. 2018
  1 Kommentar
Walter Roberson
Walter Roberson am 22 Apr. 2018
syms deltah
nb = sym(128, 'r');
h = sym(0.2, 'r') + deltah;
t_s = sym(0.004, 'r');
t_i = sym(0.05, 'r');
s = sym(500, 'r');
nc = nb+(nb*((h+2*t_s+2*t_i)*s));
nc
subs(nc, deltah, eps(0.2))
vpa(ans)
nc =
64000*deltah + 19840
ans =
1396115884484853885/70368744177664
ans =
19840.00000000000177635683940025

You can see from this that a difference of 1 bit in the representation of 0.2 makes a difference of

>> vpa(subs(nc,deltah,eps(0.2)) - 19840)
ans =
0.0000000000017763568394002504646778106689453

0.2 is not exactly representable in binary floating point: it is an infinite repeating decimal, just like 1/7 is not exactly representable in finite decimal.

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