...to overrun compiler resources for a program that would not exceed runtime resources. One would hope that the code for most programs would fit in 2GB even integer x(10000000000) x=1 write(*,*) x(987654) end will compile in 32bit g95, but won't run. integer x(1000000000) data x/1000000000*1/ write(*,*) x(987654) end the...

... arguments and they can be passed on to a user subroutine, provided the subroutine does the right thing. The standard doesn't require checking for undefined variables. Given something like DIMENSION A(10000000000) it's hard to imagine an efficient way to check for undefinedness. If a value is required, the variable must be in a defined state, but it's a programmer requirement....

... } 0. 0. { D: usertime0 D: usertime1 } 0. 0. { D: systime0 D: systime1 } 0. 10000000000. { D: utmax D: utmin } 0. 10000000000. { D: usermax D: usermin } 0. 10000000000. { D: sysmax D: sysmin } begin utime to utime0 cputime ( -- duser0 dsystem0 ) to systime0 to usertime0 ...

... utime1 } 0. 0. { D: usertime0 D: usertime1 } 0. 0. { D: systime0 D: systime1 } 0. 10000000000. { D: utmax D: utmin } 0. 10000000000. { D: usermax D: usermin } 0. 10000000000. { D: sysmax D: sysmin } begin utime to utime0 cputime ( -- duser0 dsystem0 ) to systime0 to usertime0 1000000 for next...

... a bit to the run time. (hence 1.25s of system time.) ; time rc -c 'primes 1 10000000000 | sed 205963q | tail -5' 2837629 2837633 2837677 2837693 2837711 3.64u 1.25s 4.29r rc -c primes 1 10000000000 | sed 205963q | tail -5 # status= rc 42288: primes 42289: sys: write on closed pipe pc=0x00005ef5|| i ...

> Note that due to primes' syntax I can't time that until I know what the 205,963rd prime number is. erm, what about time rc -c 'primes 1 10000000000 | sed 205963q | tail -5' takes just over 1 second on my laptop.

...INCLUDE ../bignum/factor.frt (* The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317. Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000. *) : INIT ( -- ) S" 10000000000" s2 GIANT S" 0" s3 GIANT ; : TERM ( n -- ) DUP 0 <# #S #> s1 GIANT s1 SWAP s2 GS^MOD s3 s1 GG+ ; : 1000T ( -- g ) INIT #1001 1 DO I TERM LOOP s3 s2 GGMOD...

...INCLUDE ../bignum/factor.frt (* The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317. Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000. *) : INIT ( -- ) S" 10000000000" s2 GIANT S" 0" s3 GIANT ; : TERM ( n -- ) DUP 0 <# #S #> s1 GIANT s1 SWAP s2 GS^MOD s3 s1 GG+ ; : 1000T ( -- g ) INIT #1001 1 DO I TERM LOOP s3 s2 GGMOD s3 ....

...> your machine. I had it run with the normal gforth. Gforth-fast on the 2GHz Athlon64: 955µs, on a 2.6GHz Core 2 Quad 650µs. So actually iForth64 must be under one 1ms, too ;-). &10000000000 Constant mod# : *mod ( a b -- c ) um* mod# um/mod drop ; Wasn't there a way to do integer division (and consequently compute the modulus) by multiplication with the inverse? Won'...

...modulus might be the limiting factor. If you timed it with the development version of gforth-fast, I doubt it. I guesstimate that the program spends more then 1ms on UM/MOD on your machine. &10000000000 Constant mod# : *mod ( a b -- c ) um* mod# um/mod drop ; Wasn't there a way to do integer division (and consequently compute the modulus) by multiplication with the inverse? ...