MATLAB gives the integer value of subtraction when the difference between numbers is large.

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Mustafa Duran
Mustafa Duran 2022-4-25
评论： Steven Lord 2022-4-25
Hello everyone. I have a problem in my code, i will show it with an example below:
a=[1;1;1;1];
b=[10^23;10^23;10^23;10^23];
c=a-b
fprintf("%f \n",c)
Results:
c =
1.0e+23 *
-1.000000000000000
-1.000000000000000
-1.000000000000000
-1.000000000000000
-99999999999999991611392.000000
-99999999999999991611392.000000
-99999999999999991611392.000000
-99999999999999991611392.000000
When i take action with matrix it is problematic that MATLAB give the subtraction's results as integer not float, how can i change it?
Thanks for helps.

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回答（2 个）

Sulaymon Eshkabilov
编辑：Sulaymon Eshkabilov 2022-4-25
Specify flotaing point format accordingly, e.g.:
a=[1;1;1;1];
b=[10^23;10^23;10^23;10^23];
c=a-b
fprintf("%1999.0f \n",c)
Still it does not display the correct answer. To get a correct display you would need to use these scripts posted here: mathworks.com/matlabcentral/fileexchange/22725-variable-precision-integer-arithmetic
  1 个评论
Mustafa Duran
Mustafa Duran 2022-4-25
Thanks but i will use array of "c" in another calculation in a loop, it uses false answer which is integer one.
I used vpi command but it gives that error:
If N is a double, it may be no larger than 2^53 - 1

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Steven Lord
Steven Lord 2022-4-25
Ran in:
What's the distance from the elements of b to the next larger number?
b = 10^23
b = 1.0000e+23
distance = eps(b)
distance = 16777216
If you were to add 1 to b, you don't get anywhere close to the next representable floating point number.
c = b + 1;
c == b % true, and no this is NOT a bug!
ans = logical
1
The distance between b and the next smallest representable floating point number is also much larger than 1, so if you subtract 1 you stay at b.
d = b - 1;
d == b % true and this is NOT a bug either!
ans = logical
1
Basically what you're doing is handing Elon Musk, Jeff Bezos, or Bill Gates a $1 bill and expecting his net worth to change. It's negligible compared to what they already have. See the "Accuracy of Floating-Point Data" section on this documentation page for more information.
  2 个评论
Mustafa Duran
Mustafa Duran 2022-4-25
编辑：Mustafa Duran 2022-4-25
Normally, you were right at last comment however i need this value to calculate the error term until error value is lower than 0.001 etc and after that i will break the iteration. With this rounding, MATLAB calculates error value 0 immediately, just because the not changing of values.
Steven Lord
Steven Lord 2022-4-25
This is not just MATLAB behavior. In IEEE double precision 1 is negligible compared to 10^23.
To put that in another perspective, consider something that weighs 1 kilogram, like a 1L bottle of water. Is that negligible compared to Earth's moon (with a mass of 7.3*10^22 kg according to Wikipedia)? For practical purposes, yes.
If b is just a placeholder starting value that you're using to give you an error value that's basically guaranteed to be larger than 0.001 for the first iteration, choose a smaller b.

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