Perfect numbers: Difference between revisions

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 =={{header|D}}==
+===Functional Version===
+<lang d>import std.stdio, std.algorithm, std.range;
+bool isPerfectNumber1(in uint n) pure nothrow
+in {
+    assert(n > 0);
+} body {
+    return n == iota(1, n - 1).filter!(i => n % i == 0).sum;
+}
+void main() {
+    iota(1, 10_000).filter!isPerfectNumber1.writeln;
+}</lang>
+{{out}}
+<pre>[6, 28, 496, 8128]</pre>
 ===Faster Imperative Version===
 {{trans|Algol}}
 <lang d>import std.stdio, std.math, std.range, std.algorithm;
-bool isPerfectNumber(in int n) pure nothrow {
+bool isPerfectNumber2(in int n) pure nothrow {
     if (n < 2)
         return false;
-    int sum = 1;
+    int total = 1;
     foreach (immutable i; 2 .. cast(int)real(n).sqrt + 1)
         if (n % i == 0) {
             immutable int q = n / i;
-            sum += i;
+            total += i;
             if (q > i)
-                sum += q;
+                total += q;
         }
-    return sum == n;
+    return total == n;
 }
 void main() {
-_000.iota.filter!isPerfectNumber.writeln;
+_000.iota.filter!isPerfectNumber2.writeln;
 }</lang>
 {{out}}
@@ Line 485: / Line 501: @@
 With a <code>33_550_337.iota</code> it outputs:
 <pre>[6, 28, 496, 8128, 33550336]</pre>
-===Functional Style===
-Same output.
-<lang d>import std.stdio, std.algorithm, std.range;
-bool isPerfect(in uint n) pure nothrow
-in {
-    assert(n > 0);
-} body {
-    return n == reduce!((s, i) => n % i ? s : s + i)(0, iota(1, n-1));
-}
-void main() {
-    iota(1, 10_000).filter!isPerfect.writeln;
-}</lang>
 =={{header|E}}==