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I'm mystified by this:

g[a_, b_, c_] := 
  Module[{d}, 
    d = If[(b < 1), b, b + 1]; 
    If[a < b, d = 0.]; 
    d*c]
g[a, 20, 40]

840

I expected to see an output involving symbolic argument a !?? Now I change the second If so that d is set as returned by If, rather than by above's If argument statement, and now things work as expected:

g[a_, b_, c_] := 
  Module[{d}, 
    d = If[(b < 1), b, b + 1]; 
    d = If[a < b, 0., d]; 
    d*c]
g[a, 20, 40]

40 If[a < 20, 0., d$132874]

Just for a bit more evidence, here is a 3rd casting, which also behaves as expected, although evaluation differs as expected:

g[a_, b_, c_] := 
  Module[{}, 
    If[(b < 1), b, b + 1]*If[a < b, 0., 1.]*c]
g[a, 20, 40]

840 If[a < 20, 0., 1.]

How does one explain the result of the first expression form?

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  • 2
    $\begingroup$ I expected to see an output involving symbolic argument a !?? why? Also note that since a is a symbol, then If[a < b, d = 0.] does nothing since you do not have 3rd argument to If. Check the third argument to If in help for more info. $\endgroup$
    – Nasser
    Sep 20, 2019 at 23:42

1 Answer 1

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The explanation is simple. In

g[a_, b_, c_] :=
  Module[{d},
    d = If[(b < 1), b, b + 1];
    If[a < b, d = 0.];
    d*c]
g[a, 20, 40]

the 1st line of the body of the module sets d to 21. The 2nd line can not be evaluated since the argument symbol a is not bound to a number, so it is left unevaluated and has no effect on d. The 3rd line, therefore, evaluates to 21 * 40, so 840 is returned.

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