nir: Add a bunch of b2[if] optimizations
The b2f and b2i conversions always produce zero or one which are both representable in every type and size. Since b2i and b2f support all bit sizes, we can just get rid of the conversion opcode. total instructions in shared programs: 15089335 -> 15084368 (-0.03%) instructions in affected programs: 212564 -> 207597 (-2.34%) helped: 896 HURT: 0 total cycles in shared programs: 369831123 -> 369826267 (<.01%) cycles in affected programs: 2008647 -> 2003791 (-0.24%) helped: 693 HURT: 216 Reviewed-by: Ian Romanick <ian.d.romanick@intel.com>
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@ -751,6 +751,23 @@ for left, right in itertools.combinations_with_replacement(invert.keys(), 2):
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optimizations.append((('inot', ('iand(is_used_once)', (left, a, b), (right, c, d))),
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('ior', (invert[left], a, b), (invert[right], c, d))))
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# Optimize x2yN(b2x(x)) -> b2y
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optimizations.append((('f2b', ('b2f', a)), a))
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optimizations.append((('i2b', ('b2i', a)), a))
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for x, y in itertools.product(['f', 'u', 'i'], ['f', 'u', 'i']):
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if x != 'f' and y != 'f' and x != y:
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continue
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b2x = 'b2f' if x == 'f' else 'b2i'
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b2y = 'b2f' if y == 'f' else 'b2i'
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for N in [8, 16, 32, 64]:
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if y == 'f' and N == 8:
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continue
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x2yN = '{}2{}{}'.format(x, y, N)
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optimizations.append(((x2yN, (b2x, a)), (b2y, a)))
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def fexp2i(exp, bits):
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# We assume that exp is already in the right range.
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if bits == 32:
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