Linked Questions

I understand why $ happens here: ClearAll[foo] foo[a_] := (x_ :> a x) foo[1] ...

Using global variables the following turns an "expression" into a Function: expr = 2 x; Function[x, Evaluate[expr]] Of course ...

Many art websites allow the user to zoom into an image to see it at high resolution by using image tiles. It is great on the website, but presents problems if one wishes to compile a complete high ...

I'm confused about the renaming mechanism of With: ...

I have a question about modules and local variables. Here's my example: h = 5; Module[{a, h}, a[h_]= h^2; a[4]] (*Out[2] = 25*) I expected the module to ...

I have the following pair of things: ClearAll[foo, labeledFoo]; labeledFoo = {"FooBarBazQuux", foo}; This works like you'd expect: ...

Consider the following examples, where I have replaced dollar signs in the output by a D. ...

Why (f[1, 0] /. {x : 0 | 1 :> 1 - x}) -> 0 returns (expectedly) f[0, 1] -> 0, while ...

I want to use a replacement rule to change log[var] into var inside a function. A minimal example follows. In one case it ...

EDIT This issue is specifically different from the "duplicate" question. The creation of variables in the active context during parsing is of no surprise and is the expected behavior. What I'm ...

I've come across some weirdness in ReplaceAll. Replacement works fine outside of a module, but inside the module some replacements don't occur where one might expect them to. ...

I've run into a strange anomaly with Replace when it is performed inside the first argument of RuleDelayed within a function ...