Load extended precision with .d operations

[phx]

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?

[Kalms]

Convert the original value to 96-bit floating-point 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 64-bit float values or to decimal numbers.

[phx]

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?

[PeterK]

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.

[phx]

Originally Posted by PeterK

looks like you are discussing a problem here which Cosmos also asked me about today.

LOL
Indeed.

He has already found a solution.

I know. But I was curious to hear other opinions.

[Kalms]

Originally Posted by phx

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?

Almost. You need to handle the implicit leading 1-bit for the mantissa. The exponent for the remaining value may then become smaller than (original exponent - 53). Extreme example: what if the lowest bits of the mantissa are all zero?

Tbh, if you have access to long double math operstions, then the Cosmos approach is less manual work and will give a near-identical decomposition.

[phx]

Originally Posted by Kalms

Extreme example: what if the lowest bits of the mantissa are all zero?

Right. It's not so trivial. I nearly forgot about the implicit 1-bit, which is always there and also determines the exponent.

Tbh, if you have access to long double math operstions, then the Cosmos approach is less manual work and will give a near-identical decomposition.

Ok. That's what I wanted to find out. Thanks.