Half precision using GPU
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Fernando
el 10 de Abr. de 2023
Comentada: Fernando
el 11 de Abr. de 2023
Hello, I was trying to see if I can run some code using half precision rather than single.
before converting my code, I was trying a very simple example.
A=gpuArray(magic(3));
A=half(A);
This gives me the error: No constructor 'half' with matching signature found.
Using the the half with the CPU works fawlessly.
Any idea if this is supported by all? Looking here, https://www.mathworks.com/help/gpucoder/ug/what-is-half-precision.html, it seems some GPU should support it?
I am using a 16 GB RTX3080 Mobile. R2022b.
2 comentarios
Walter Roberson
el 11 de Abr. de 2023
Perhaps
A=gpuArray(half(magic(3)))
??
I do not have a GPU available to test with
Respuesta aceptada
Joss Knight
el 11 de Abr. de 2023
As pointed out, gpuArray does not support half. The main reason is that half is an emulated type only meaningful for deployment to special hardware, it is not native to most processors. Feel free to investigate use of half for code generation.
Do you just want to store data in half to save space on the GPU? You can use the following code to get something like the behaviour you're after:
function u = toHalf(x)
realmaxHalf = single(65504);
x = min(max(x,-realmaxHalf),realmaxHalf);
[f,e] = frexp(abs(x));
sgn = uint16(x>=0);
sgnbit = bitshift(sgn,15);
expbits = bitshift(uint16(e+15),10);
fbits = uint16(f.*2.^10 - 1);
u = bitor(bitor(sgnbit, expbits), fbits);
end
function x = fromHalf(u)
if u == 0
x = single(0);
return
end
u = uint16(u);
sgn = single(bitshift(u,-15));
fbits = bitand(u,uint16(1023));
f = single(fbits+1)./(2.^10);
expbits = bitand(u,uint16(31744));
e = single(bitshift(expbits,-10))-15;
x = (sgn.*2-1).*f.*2.^e;
end
Note, this is a very crude implementation of fp16 that takes no account of nans, infs, correct overflow behaviour or denormals. The half version is just a uint16 with the data in it, you can't actually use it to compute anything in fp16.
4 comentarios
Joss Knight
el 11 de Abr. de 2023
'fraid not. No chance of that! Your only hope is to actually convert to int16 (by rescaling to some range), but you will find many blockers in the way such as integer overflow and unsupported mathematical operations. The code I gave you merely stores the number you have as a float into 16 bits; you can't actually do any computation with it.
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