Why does
Block[{$Pre = ( 1 + #) &}, 1]
produce 1 while
Block[{$PreRead = Print["hi"]}, 1]
produces
"hi"
1
?
In the Block block, shouldn't the expression have it's Pre evaluation as well?
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1 Answer
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5
No, $Pre and $PreRead cannot be set within a "line" of code and apply to that same "line" as well. Your example with $PreRead is a false result because you did not make it a function.
Consider:
Block[{$PreRead = (Print["hi"]; #) &}, 1]
1
If outside Block you set:
$PreRead = (Print["<<", #, ">>"]; #) &;
and then evaluate a Block:
Block[{x},
1;
2;
3;
]
<<RowBox[{Block,[,RowBox[{RowBox[{{,x,}}],,,RowBox[{1,;,2,;,3,;}]}],]}]>>
You see that the entire "line" (I can't think of a better term at the moment) of code is sent to $PreRead, not individual parts of it.
answered Oct 23, 2012 at 3:54
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$\begingroup$ That is unfortunate. If I ever have to use
$Preor$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? $\endgroup$– VF1Commented Oct 23, 2012 at 4:11 -
$\begingroup$ @VF1 Because the
"hi"you're seeing is just because thePrintstatement evaluated, and not as a result of$PreReadbeing correctly set. If it indeed were set correctly, as you had hoped, you should've seen two"hi"statements for the followingBlock[{$PreRead = Print["hi"]}, 1; 1;]$\endgroup$– rm -rf ♦Commented Oct 23, 2012 at 4:38 -
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$\begingroup$ @rm-rf Actually, if the rhs of the expression
$PreReadis evaluated, I would assume that$PreReadhas been set. Anyways, regarding the fact that setting it doesn't change the first line -$Pre = 1 + # &; 1returns2. Why isCompoundExpressionspecial andBlocknot? $\endgroup$– VF1Commented Oct 23, 2012 at 14:18 -
$\begingroup$ @VF1 I suspect you already had
$Predefined... try evaluating$Pre =.and then repeat your experiment. I get 1. $\endgroup$– rm -rf ♦Commented Oct 23, 2012 at 14:24