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Jensen's Device: Difference between revisions
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[http://en.wikipedia.org/wiki/Donald_Knuth Donald Knuth] later proposed the [[Man or boy test|Man or Boy Test]] as a more rigorous exercise.
=={{header|Ada}}==
<lang ada>
with Ada.Text_IO; use Ada.Text_IO;
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Put_Line (Float'Image (Sum (I'Access, 1.0, 100.0, Inv_I'Access)));
end Jensen_Device;
</
<pre>
5.18738E+00
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=={{header|C}}==
<lang cpp>#include <stdio.h>
int i;
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printf("%f\n", sum(&i, 1, 100, term_func));
return 0;
}</
Output: 5.18738
=={{header|C++}}==
<lang cpp>#include <iostream>
int i;
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std::cout << sum(i, 1, 100, term_func) << std::endl;
return 0;
}</
Output: 5.18738
=={{header|E}}==
In E, the distinct mutable locations behind assignable variables can be reified as [http://wiki.erights.org/wiki/Slot Slot] objects. The E language allows a variable name (''noun'') to be bound to a particular slot, and the slot of an already-bound noun to be extracted, using the <
(The definition of the outer <var>i</var> has been moved down to emphasize that it is unrelated to the <var>i</var> inside of <var>sum</var>.)
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}
<
The value returned by the above program (expression) is 5.187377517639621.
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=={{header|OCaml}}==
<lang ocaml>let i = ref 42 (* initial value doesn't matter *)
let sum' i lo hi term =
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let () =
Printf.printf "%f\n" (sum' i 1 100 (fun () -> 1. /. float !i))</
Output: 5.187378
=={{header|Perl}}==
<lang perl>my $i;
sub sum {
my ($i, $lo, $hi, $term) = @_;
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}
print sum(\$i, 1, 100, sub { 1 / $i }), "\n";</
Output: 5.18737751763962
Or you can take advantage of the fact that elements of the @_ are aliases of the original:
<lang perl>my $i;
sub sum {
my (undef, $lo, $hi, $term) = @_;
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}
print sum($i, 1, 100, sub { 1 / $i }), "\n";</
Output: 5.18737751763962
=={{header|PHP}}==
<lang php>$i;
function sum (&$i, $lo, $hi, $term) {
$temp = 0;
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}
echo sum($i, 1, 100, create_function('', 'global $i; return 1 / $i;')), "\n";</
Output: 5.18737751764
=={{header|Python}}==
<lang python>class Ref(object):
def __init__(self, value=None):
self.value=value
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# note the correspondence between the mathematical notation and the call to sum
# it's almost as good as sum(1/i for i in range(1,101))
print (sum (i, 1, 100, lambda: 1.0/i.value))</
Output: 5.18737751764
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