06 May 2018, 15:24  #1 
Natteravn
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Load extended precision with .d operations
IIRC the 68060 has to emulate FMOVE.X #immediate,Fpn?
Let us assume I want to transfer an 80 bits extended precision constant (like Pi) into an FPU register without using the instruction above and without any memory access! What options do I have? Would it be possible with fmove.d/fadd.d, where the first fmove.d loads the most significant (52?) bits of the mantissa, while the fadd.d adds the remaining lower bits with an appropriate exponent? How to do that in a most elegant manner, i.e. how to determine the immediate values for both operations? 
06 May 2018, 17:12  #2 
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Convert the original value to 96bit floatingpoint binary representation (for example, by assembling an FMOVE.X #IMM, FPn instruction). Extract the exponent and mantissa values separately. Split the mantissa field at 53 bits. Create two new sign/mantissa pairs. Convert these pairs either to hex representations of the 64bit float values or to decimal numbers.

06 May 2018, 21:41  #3 
Natteravn
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Thanks.
Ok. So, from the 96 bits representation I take first 53 bits of the mantissa and the exponent. Then, for the FADD, I take the remaining mantissa bits, and subtract 53 from the original exponent field? Something like this, I guess? 
06 May 2018, 22:04  #4 
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Hi Frank,
looks like you are discussing a problem here which Cosmos also asked me about today. He has already found a solution. I think his method is to take a long double constant, subtract the double value of the same constant and convert the difference into a float value. This should work as long as the exponent of the long double was not too high. The missing 11 bits of the mantissa will always fit into a float value. Then use FMOVE.D #double,FPx ... and FADD.S #floatdiff,FPx. 
06 May 2018, 22:39  #5 
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07 May 2018, 09:45  #6  
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Quote:
Tbh, if you have access to long double math operstions, then the Cosmos approach is less manual work and will give a nearidentical decomposition. 

07 May 2018, 14:52  #7  
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Quote:
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