Am 15.11.2012 19:44, schrieb Allin Cottrell: > On Thu, 15 Nov 2012, Pindar wrote: > >> And in terms of 'the loop line' string-substitution generates also an error: >> (my challenge solution of the interaction stuff) > True, up till now $-string substitution has not been available > in the line that starts a loop. This is now relaxed. In > current CVS both of the following should work OK, regardless > of the language in which gretl is running: That's nice!
Cheers Leon > > <hansl> > # Using numerical values > loop i=1..4 -q > printf "i=%d\n", i > loop j=i+1..i+2 -q > printf " j=%d\n", j > endloop > endloop > > # Using string substitution > loop i=1..4 -q > printf "i=$i\n" > loop j=$i+1..$i+2 -q > printf " j=$j\n" > endloop > endloop > </hansl> > > Allin Cottrell > _______________________________________________ > Gretl-users mailing list > Gretl-users(a)lists.wfu.edu > http://lists.wfu.edu/mailman/listinfo/gretl-users
Am 15.11.2012 19:44, schrieb Allin
Cottrell:
That's nice!On Thu, 15 Nov 2012, Pindar wrote:And in terms of 'the loop line' string-substitution generates also an error: (my challenge solution of the interaction stuff)True, up till now $-string substitution has not been available in the line that starts a loop. This is now relaxed. In current CVS both of the following should work OK, regardless of the language in which gretl is running: Cheers Leon <hansl> # Using numerical values loop i=1..4 -q printf "i=%d\n", i loop j=i+1..i+2 -q printf " j=%d\n", j endloop endloop # Using string substitution loop i=1..4 -q printf "i=$i\n" loop j=$i+1..$i+2 -q printf " j=$j\n" endloop endloop </hansl> Allin Cottrell _______________________________________________ Gretl-users mailing list gretl-us...@lists.wfu.edu http://lists.wfu.edu/mailman/listinfo/gretl-users |