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

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MIKE JOHN on 30 Jun 2021
Commented: MIKE JOHN on 1 Jul 2021
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?

Divyam Gupta on 30 Jun 2021
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.
MIKE JOHN on 1 Jul 2021

Andy Bartlett on 30 Jun 2021
Edited: Andy Bartlett on 30 Jun 2021
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.
MIKE JOHN on 1 Jul 2021
I have learned a lot!

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