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Code details (cont)

Break up complex expressions, e.g. 

if (2*k < n-m)
    *xp = c[k+1];
else
    *xp = d[k-1];
*x += *xp;

... is better than ...

*x += (*xp = ((2*k < n-m) ? c[k+1] : d[k-1]))

Especially applies to complex conditional exprs Cond?Expr1:Expr2