AwakenToMe@aol.com, in a profound display of stubbornness, continues to insist that his program to enumerate all possible words of length N (that is, aaaaa, aaaab, aaaac, etc.) is somehow interesting. I am therefore forced to drive in the nail with a sledgehammer. Forgive me. He writes:
It's [...] trivial enough to be done by 99% of the people on cypherpunks in their sleep.
Yes. But let me ask you this. Have you done it yet??
Most of us don't bother writing up four line programs and shipping them out, no.
really? Wow. four lines of code? You must be a really good programmer. duh.
Hardly. A ten year old could do it. I know, since I wrote substantially more sophisticated stuff when I was ten. Since you insist, here is less than a minute's work. Yes, I timed it. ------Cut Here------ /* This could be more elegant, but the point is obviousness. */ #include <stdio.h> int main() { char i[6]; for (i[0] = 'a'; i[0] < 'z'; i[0]++) for (i[1] = 'a'; i[1] < 'z'; i[1]++) for (i[2] = 'a'; i[2] < 'z'; i[2]++) for (i[3] = 'a'; i[3] < 'z'; i[3]++) for (i[4] = 'a'; i[4] < 'z'; i[4]++) printf("%s\n", i); } ------Cut Here------ The operative portion of the program is six lines ling, and five of those lines are virtually identical. You can write the thing much more elegantly, without redundant code. However, I have elected to leave it as utterly brainless as possible to demonstrate that ANYONE could write the thing.
Youd be surprised at the # of requests from people who actually had a good use for it, and didnt have the time to spend writing it themselves.
Human stupidity is never a surprise. Perry