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How do I subtract IEEE 754 numbers?
For example: 0,546875 – 32.875…
-> 0,546875 is 0 01111110 10001100000000000000000 in IEEE-754
-> -32.875 is 1 10000111 01000101111000000000000 in IEEE-754
So how do I do the subtraction? I know I have to to make both exponents equal but what do I do after that? 2’Complement of -32.875 mantissa and add with 0.546875 mantissa?
Really not any different than you do it with pencil and paper. Okay a little different
the bigger number dominates, shift the smaller number’s mantissa off into the bit bucket until the exponents match
then perform the subtraction with the mantissas
Then normalize (which in this case it is)
That was with base 10 numbers.
In IEEE float, single precision
normalize that we have to shift the decimal place 16 places to the left so
The exponent is biased so we add 127 to 16 and get 143 = 0x8F. It is a positive number so the sign bit is a 0 we start to build the IEEE floating point number the leading
1 before the decimal is implied and not used in single precision, we get rid of it and keep the fraction
sign bit, exponent, mantissa
And if you write a program to see what a computer things 123400 is you get the same thing:
So we know the exponent and mantissa for the first operand’
Now the second operand
Normalize, shift decimal 12 bits left
The exponent is biased add 127 and get 139 = 0x8B = 0b10001011
Put it all together
And a computer program/compiler gives the same
Now to answer your question. Using the component parts of the floating point numbers, I have restored the implied 1 here because we need it
We have to line up our decimal places just like in grade school before we can subtract so in this context you have to shift the smaller exponent number right, tossing mantissa bits off the end until the exponents match
Now we can subtract the mantissas. If the sign bits match then we are going to actually subtract if they dont match then we add. They match this will be a subtraction.
computers perform a subtraction by using addition logic, inverting the second operator on the way into the adder and asserting the carry in bit, like this:
And now just like with paper and pencil lets perform the add
or do it with hex on your calculator
A little bit about how the hardware works, since this was really a subtract using the adder we also invert the carry out bit (or on some computers they leave it as is). So that carry out of a 1 is a good thing we basically discard it. Had it been a carry out of a zero we would have needed more work. We dont have a carry out so our answer is really 0xE66800.
Very quickly lets see that another way, instead of inverting and adding one lets just use a calculator
By trying to visualize it I perhaps made it worse. The result of the mantissa subtracting is 111001100110100000000000 (0xE66800), there was no movement in the most significant bit we end up with a 24 bit number in this case with the msbit of a 1. No normalization. To normalize you need to shift the mantissa left or right until the 24 bits lines up with the most significant 1 in that left most position, adjusting the exponent for each bit shift.
Now stripping the 1. bit off the answer we put the parts together
If you have been following along by writing a program to do this, I did as well. This program violates the C standard by using a union in an improper way. I got away with it with my compiler on my computer, dont expect it to work all the time.
And our result matches the output of the above program, we got a 0x47E66800 doing it by hand
If you are writing a program to synthesize the floating point math your program can perform the subtract, you dont have to do the invert and add plus one thing, over complicates it as we saw above. If you get a negative result though you need to play with the sign bit, invert your result, then normalize.
So:
1) extract the parts, sign, exponent, mantissa.
2) Align your decimal places by sacrificing mantissa bits from the number with the smallest exponent, shift that mantissa to the right until the exponents match
3) being a subtract operation if the sign bits are the same then you perform a subtract, if the sign bits are different you perform an add of the mantissas.
4) if the result is a zero then your answer is a zero, encode the IEEE value for zero as the result, otherwise:
5) normalize the number, shift the answer to the right or left (The answer can be 25 bits from a 24 bit add/subtract, add/subtract can have a dramatic shift to normalize, either one right or many bits to the left) until you have a 24 bit number with the most significant one left justified. 24 bits is for single precision float. The more correct way to define normalizing is to shift left or right until the number resembles 1.something. if you had 0.001 you would shift left 3, if you had 11.10 you would shift right 1. a shift left increases your exponent, a shift right decreases it. No different than when we converted from integer to float above.
6) for single precision remove the leading 1. from the mantissa, if the exponent has overflowed then you get into building a signaling nan. If the sign bits were different and you performed an add, then you have to deal with figuring out the result sign bit. If as above everything fine you just place the sign bit, exponent and mantissa in the result
Multiply and divide is different, you asked about subract, so that is all I covered.