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Jun 28, 2017 at 3:31 history edited xzczd
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Sep 29, 2015 at 12:41 comment added Leonid Shifrin @Szabolcs This is one of them. Another one, which is IMO no less important, is that keeping l.h.s. entirely unevaluated would often lead to discrepancy between l.h.s. in definition and actual l.h.s. of a typical function call involving a given symbol, in case if such l.h.s. evaluates non-trivially - because otherwise in one case, it would evaluate, and in the other, it wouldn't. In other words, the current policy tries to make evaluation during function calls be maximally consistent with evaluation during assignments.
Sep 29, 2015 at 10:32 comment added Szabolcs Check out the updated answer. I believe this is the reason behind the design.
Sep 27, 2015 at 22:37 history tweeted twitter.com/#!/StackMma/status/648265277633859584
Sep 26, 2015 at 21:16 answer added Szabolcs timeline score: 6
Sep 26, 2015 at 16:52 answer added Michael E2 timeline score: 10
Sep 26, 2015 at 16:23 answer added Rojo timeline score: 7
Sep 26, 2015 at 16:22 history edited Szabolcs
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Sep 26, 2015 at 16:20 comment added Leonid Shifrin I have a rather detailed discussion of evaluation during assignments here. Basically, HoldAll simply means that arguments are passed to the function in unevaluated form, but does not restrict what functions decide to do with them. And Set and SetDelayed do evaluate their first arguments, albeit in a special way. An example very similar to yours I also considered in this answer, in the section named "Evaluation: OwnValues".
Sep 26, 2015 at 16:04 comment added Rolf Mertig Because HoldAll is documented to not do nothing, but something. If you do SetDelayed[Unevaluated[f][x_], x^2]; you will get what you want.
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Sep 26, 2015 at 15:16 history edited bbgodfrey CC BY-SA 3.0
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Sep 26, 2015 at 15:17
Sep 26, 2015 at 14:54 history asked XiaoaiX CC BY-SA 3.0