Re^6: Odometer pattern iterator (in C).

Cool! It even runs in the same direction as my original example.

Though I suspect that once the slice and map operations have been unwound in C, it'll look a lot more complicated.

With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'

Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.

"Science is about questioning the status quo. Questioning authority". I'm with torvalds on this

In the absence of evidence, opinion is indistinguishable from prejudice. Agile (and TDD) debunked

Comment on Re^6: Odometer pattern iterator (in C).

Replies are listed 'Best First'.
Re^7: Odometer pattern iterator (in C). by hdb (Monsignor) on May 29, 2015 at 13:13 UTC
I have now compared it to the solution you have posted and it probably amounts to almost the same algorithm...	[reply]
Re^8: Odometer pattern iterator (in C). by BrowserUk (Patriarch) on May 29, 2015 at 14:34 UTC
I attempted to adapt your Perl to C, but I made a mistake somewhere. I don't know if you can see the problem, but I can't and I need to move on: `U8 firstIdx( U8 N, U8 M ) { U32 i; U8 a = malloc( M ); for( i = N - M; i < N; ++i ) a[ i - M + 1 ] = i; return a; } U8 nextIdx( U8 N, U8 M, U8 idx ) { U32 i = M - 1, j; if( idx[ i ] == i ) return NULL; while( i > 0 && idx[ i ] - idx[ i - 1 ] == 1 ) i--; --idx[ i ]; for( j = i+1; j < M; ++j ) idx[ i ] += N - idx[ M - j + i ]; return idx; }` [download] With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday' Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error. "Science is about questioning the status quo. Questioning authority". I'm with torvalds on this In the absence of evidence, opinion is indistinguishable from prejudice. Agile (and TDD) debunked	[reply] [d/l]
Re^9: Odometer pattern iterator (in C). by hdb (Monsignor) on May 29, 2015 at 15:12 UTC
You cannot do this inplace: `for( j = i+1; j < M; ++j ) idx[ i ] += N - idx[ M - j + i ];` [download] as you are overwriting values that are still needed. This corresponds to the "reversing" the string part of the regex. In addition, you misread the formula, there is no `+=` involved.	[reply] [d/l] [select]