Hi,
Prabhakar Ragde wrote:
divisors i = [j | j<-[1..i-1], i `mod` j == 0]
main = print [i | i<-[1..10000], i == sum (divisors i)]
Jerzy Karczmarczuk wrote:
My point didn't concern that point. Haskell compiler cannot change an
algorithm using lists into something which deals with indexable arrays,
usually faster. Indexing may be faster than the indirection, and the
allocation of memory costs. And there is laziness...
This may be true, but it isn't relevant in this case, since the "obvious
c program" doesn't need any arrays, only two loops:
for (int i = 1; i <= 10000; i++) {
int sum = 0;
for (int j = 1; j < i; j++)
if (i % j == 0)
sum += i;
if (sum == i)
print(i);
}
Loops can be expressed with lazy lists in Haskell. Therefore, the
presented Haskell program is perfectly equivalent to the "obvious c
program".
Tillmann Rendel
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