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Why does

Block[{$Pre = ( 1 + #) &}, 1]

produce 1 while

Block[{$PreRead = Print["hi"]}, 1]



? In the Block block, shouldn't the expression have it's Pre evaluation as well?

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up vote 6 down vote accepted

No, $Pre and $PreRead cannot be set within a "line" of code and apply to that same "line" as well. Your example with $PreRead is a false result because you did not make it a function.


Block[{$PreRead = (Print["hi"]; #) &}, 1]

If outside Block you set:

$PreRead = (Print["<<", #, ">>"]; #) &;

and then evaluate a Block:


You see that the entire "line" (I can't think of a better term at the moment) of code is sent to $PreRead, not individual parts of it.

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That is unfortunate. If I ever have to use $Pre or $Post, I have to evaluate them in a separate line, do my calculation on a second, and then remove their values on a third! Also, when you say "false result," I don't quite understand. What does the fact that it isn't a function have to do with it being evaluated? – VF1 Oct 23 '12 at 4:11
@VF1 Because the "hi" you're seeing is just because the Print statement evaluated, and not as a result of $PreRead being correctly set. If it indeed were set correctly, as you had hoped, you should've seen two "hi" statements for the following Block[{$PreRead = Print["hi"]}, 1; 1;] – R. M. Oct 23 '12 at 4:38
@rm-rf Thanks for the clarification. – VF1 Oct 23 '12 at 4:42
@rm-rf Actually, if the rhs of the expression $PreRead is evaluated, I would assume that $PreRead has been set. Anyways, regarding the fact that setting it doesn't change the first line - $Pre = 1 + # &; 1 returns 2. Why is CompoundExpression special and Block not? – VF1 Oct 23 '12 at 14:18
@VF1 I suspect you already had $Pre defined... try evaluating $Pre =. and then repeat your experiment. I get 1. – R. M. Oct 23 '12 at 14:24

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