pow x 0 = 1 pow x n = if even n then pxn2 * pxn2 else x * pow x (n-1) where pxn2 = pow x (n/2) f x = pow x 5Since n is known we can specialise pow in its second argument and unfold the recursive calls:
pow5 x = x * x4 where x4 = x2 * x2 x2 = x * x f x = pow5 xpow5 is known as the residual. We could now also unfold pow5 giving:
f x = x * x4 where x4 = x2 * x2 x2 = x * xIt is important that the partial evaluation algorithm should terminate. This is not guaranteed in the presence of recursive function definitions. For example, if partial evaluation were applied to the right hand side of the second clause for pow above, it would never terminate because the value of n is not known. Partial evaluation might change the termination properties of the program if, for example, the expression (x * 0) was reduced to 0 it would terminate even if x (and thus x * 0) did not. It may be necessary to reorder an expression to partially evaluate it, e.g.
f x y = (x + y) + 1 g z = f 3 zIf we rewrite f:
f x y = (x + 1) + ythen the expression x+1 becomes a constant for the function g and we can say
g z = f 3 z = (3 + 1) + z = 4 + zPartial evaluation of built-in functions applied to constant arguments is known as constant folding. See also full laziness.
Last updated: 1999-05-25