Remove const M in macros: @zero_derivative and @from_chainrules#694
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Remove const M in macros: @zero_derivative and @from_chainrules#694
const M in macros: @zero_derivative and @from_chainrules#694Conversation
Signed-off-by: Hong Ge <3279477+yebai@users.noreply.github.com>
Signed-off-by: Hong Ge <3279477+yebai@users.noreply.github.com>
const M in @from_chainrules macroconst M in @zero_derivative macro
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Mooncake.jl documentation for PR #694 is available at: |
Signed-off-by: Hong Ge <3279477+yebai@users.noreply.github.com>
const M in @zero_derivative macroconst M in macros: @zero_derivative and @from_chainrules
Signed-off-by: Hong Ge <3279477+yebai@users.noreply.github.com>
Codecov Report✅ All modified and coverable lines are covered by tests. 📢 Thoughts on this report? Let us know! |
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Performance Ratio: |
penelopeysm
reviewed
Aug 13, 2025
Comment on lines
282
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+286
| # which does not escape the mode argument. This will work even if the names `Mooncake` | ||
| # or `Mooncake.Mode` are not available in the scope which calls this macro. | ||
| is_primitive_ex = quote | ||
| const M = $mode | ||
| function Mooncake.is_primitive( | ||
| ::Type{$(esc(ctx))}, ::Type{<:M}, ::Type{<:$(esc(sig))} | ||
| ::Type{$(esc(ctx))}, ::Type{<:($(esc(mode)))}, ::Type{<:($(esc(sig)))} |
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The comment above says to not escape mode, so I wonder if $(mode) would be better?
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I won't have time for this in the next few days. Please feel free to take over and merge it.
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Replaced by #696 |
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Fix #692 for @penelopeysm