Running `df -B...` with a SIZE which is greater than the total disk space returns the same value in the 'Used' and 'Available' column - see example below with T, P and E:
$ for f in 1 K M G T P E Z Y ; do ( set -x && df -B$f . ) ; done + df -B1 . Filesystem 1B-blocks Used Available Use% Mounted on D: 596087726080 100868861952 495218864128 17% /cygdrive/d + df -BK . Filesystem 1K-blocks Used Available Use% Mounted on D: 582116920K 98504748K 483612172K 17% /cygdrive/d + df -BM . Filesystem 1M-blocks Used Available Use% Mounted on D: 568474M 96197M 472278M 17% /cygdrive/d + df -BG . Filesystem 1G-blocks Used Available Use% Mounted on D: 556G 94G 462G 17% /cygdrive/d + df -BT . Filesystem 1T-blocks Used Available Use% Mounted on D: 1T 1T 1T 17% /cygdrive/d + df -BP . Filesystem 1P-blocks Used Available Use% Mounted on D: 1P 1P 1P 17% /cygdrive/d + df -BE . Filesystem 1E-blocks Used Available Use% Mounted on D: 1E 1E 1E 17% /cygdrive/d + df -BZ . df: -B argument `Z' too large + df -BY . df: -B argument `Y' too large Apart from saying that 1[TPE] is still available ... shouldn't it return 0[TPE] for the "Used" column, as rounding a value of 17% should lead to zero (mathematically)? BTW: why are `Z' and `Y' "too large"? $ df --version df (GNU coreutils) 8.10 Packaged by Cygwin (8.10-1) It's on a 32-bit XP. Have a nice day, Berny
