Why do the data become zero when using the function fi?

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fm = get_fimath();
idx = fi(1,0,1,0,fm);
a = (idx+fi(2,0,2,0,fm))*fi(1/3,0,16,17,fm);
k = fi(a,0,17,0,fm)
function fm = get_fimath()
fm = fimath('RoundingMethod', 'Floor',...
'OverflowAction', 'Wrap',...
'ProductMode','FullPrecision',...
'MaxProductWordLength', 128,...
'SumMode','FullPrecision',...
'MaxSumWordLength', 128);
end
This code is generated when using the Matlab Coder .  I want to know why is k equal to zero? Is it because of division 1/3?

采纳的回答

Divyam Gupta
Divyam Gupta 2021-6-30
Hi Mike,
In the fimath function, you've set the RoundingMethod parameter as Floor. 1/3 when floored leads to a 0 as the answer. This is why you're getting k as 0. You could consider changing the RoundingMethod parameter as per your desired result.
Hope this helps.
  3 个评论
Divyam Gupta
Divyam Gupta 2021-6-30
Hey Mike,
Kindly refer to Andy's answer below mine and let me know if that answers your questions. If not, I'd be happy to help you further.

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更多回答(1 个)

Andy Bartlett
Andy Bartlett 2021-6-30
编辑:Andy Bartlett 2021-6-30
It's just like scientific notation
is the short answer to "Why FractionLength can be bigger than WordLength?".
The long answer is the following.
The concept of a binary-point is very useful for initial understanding of fixed-point types. Similarly, the concept of a decimal-point is useful for understanding values beyond integers. But using decimal-points becomes very cumbersome for very big or very small numbers. To make it easy to represent very big or very small values, scientific notation is super valuable.
verySmallNumber = 3e-200;
veryBigNumber = 7e123;
In essence, this notation breaks the value into two parts, a mantissa and an integer exponent for the given base.
Y = mantissa .* 10.^exponent
Fixed-point follows the same concept except that
  • base is 2
  • mantissa must be an integer
  • exponent is fixed, i.e. it is part of the variables type and does not change for the life of the variable
Y = intMantissa .* 2^FixedExponent
Since FractionLength = -FixedExponent, we can also write this as follows.
Y = intMantissa .* 2^-FractionLength
A nice thing about fi is that we can let it figure out the scaling that gives the best precision for a constant.
verySmallNumberFi = fi( 3e-200, 0, 8 )
veryBigNumberFi = fi( 7e123, 0, 8 )
Notice the very big positive and negative FractionLengths of 670 and -404 that are produced.
verySmallNumberFi =
3.00068384319763e-200
numerictype(0,8,670)
veryBigNumberFi =
6.982403670347e+123
numerictype(0,8,-404)
Fi has approximated the original double values using 8-bit unsigned integer mantissas.
147 * 2^-670
169 * 2^404
It's just scientific notation in base 2.

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