This matches the de-facto behavior based on constant folding
and the default lowering to fmin/fmax.
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docs/LangRef.rst | ||
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11486 | I don't understand this text because it seems like if the standard says that a QNaN is returned if either input is a SNaN, then why would the lowering need to additionally canonicalize the inputs? It seems like that would not change the result. |
docs/LangRef.rst | ||
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11486 | The standard says minnum(x, snan) -> qnan. This differs from the behavior from fmin, where fmin(x, snan) -> x. You need to quiet to get the other operand instead of qnan |
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Try to be clearer.
One thing I am still unclear ones is if the standard requires payload bits to be zero or preserved
I don't understand this text because it seems like if the standard says that a QNaN is returned if either input is a SNaN, then why would the lowering need to additionally canonicalize the inputs? It seems like that would not change the result.