Whether for nested Ifs or within scoping constructs like Module and Block, I never know whether to end the final expression with a semicolon. I know the semicolon suppresses output, but, for example, in a series of nested Ifs, one needs only to semicolon the outermost If to suppress output.
So, is it better practice to semicolon everything unless output is needed? Or is it better to semicolon nothing until the last point where output is unwanted?
I may be missing the point of this question, but I think it is important to note that ; is the short form of CompoundExpression, and it is not primarily for suppressing output.
You can see how ; is interpreted using one of the methods I described here:
HoldForm @ FullForm[a; b; c] HoldForm @ FullForm[a; b;] CompoundExpression[a, b, c] CompoundExpression[a, b, Null]
Note that when a final expression is omitted Null is inserted, just like this behavior with ,:
{1, 2,} {1, 2, Null}
The suppression of output is not a behavior of CompoundExpression but rather Null, which when returned as output (alone) is not printed. For example 2 Null/2 evaluates to Null, therefore when given as input no output is printed.
There is little point in sprinkling ; around your code without need, unless perhaps as a visual reminder that the evaluated form of a given expression is not directly used by the surrounding head. One exception is when writing code that will eventually be made into a function or Module as line breaks are not valid separators in this case, therefore even functions that already return Null such as SetDelayed should be terminated with a ;, e.g.:
foo[bar_Integer] := bar^2; foo[bar_Real] := bar/2; This avoids the error seen with:
Module[{foo}, foo[bar_Integer] := bar^2 foo[bar_Real] := bar/2 ] ; is really forThe actual function of CompoundExpression is described in the documentation:
expr1;expr2; ... evaluates the expri in turn, giving the last one as the result.
CompoundExpression, while more succinct and canonical than other methods, is not the only way to accomplish this. For example you could use
Last[{expr1, expr2, ...}] in the same manner, e.g.: a = 5; b = a^2; Binomial[b, a] Last[{a = 5, b = a^2, Binomial[b, a]}] 53130 53130
And suppression of output:
x = Range@500; Last[{x = Range@500,}] (* no output printed *)
You can also return a specific expression using Slot in a simple Function:
#2 &[a = 5, b = a^2, Binomial[b, a]] 25
This is specifically appropriate when you need to perform some action after generating an expression, such as closing a stream:
str = StringToStream["abcdefg 123456"]; # &[Read[str, Word], Close @ str] "abcdefg"
I would be remiss not to mention that these alternative methods are not actually equivalent to CompoundExpression because even though they only return one expression (possibly Null) they still accumulate all of them in memory. Compare these, each pair of lines run in a fresh Kernel:
Range@1*^7; Range@1*^7; Range@1*^7; Range@1*^7; MaxMemoryUsed[] 94923152
Last[{Range@1*^7, Range@1*^7, Range@1*^7, Range@1*^7,}] MaxMemoryUsed[] 174925336
To get the memory performance of ; one would need something more complex, such as:
SetAttributes[ce, HoldAll] ce[x__] := Fold[#2 &, , Hold @ x] ce[Range@1*^7, Range@1*^7, Range@1*^7, Range@1*^7,] MaxMemoryUsed[] 93506816
Additional reading:
Nullcome from" and the "surprise multiplications" problems... $\endgroup$If[x < 1, y = 1 z = 3 , y = 4 ]without a;betweeny=1andz=3. Even if you have thez=3in a new line in the notebook. One still needs a;$\endgroup$