When I issue:

find / \( -type f -size +10M \) -and \( -type f -size -1G \)

I don't receive any results; however, when I type:

find / \( -type f -size +10M \) -and \( -type f -size -1000M \)

it returns a few files.

What's wrong? Can't I use different unit of measurement in both sides of the operator -and?


The problem isn't related to the -and operator, but to the unintuitive syntax of the argument of -size. The argument is not a maximum size given with a unit, but a maximum number and a unit in which the calculation is performed. The test -size -1k does not mean “less than 1kB” but “the number of kB (rounded up) is less than (and not equal to) 1”. Thus, -size -1k only matches files of size 0, and likewise for -size -1M, -size -1G, etc.

To match files that are 230 bytes or less, use -size -2G, or -size -1025M, etc. (observe how -2G is not equivalent to -2048M). Alternatively, use ! -size +1G — a size with the + modifier means “strictly more than this size”. To match files that are 230 -1 bytes or less, use -size -1073741824c.

I think the reason why -size -NUMBERUNIT is so weird is for historical compatibility with -size -NUMBER, which counted blocks; rounding up makes sense since a file occupying e.g. 7.5 blocks is effectively occupying 8 blocks with the last block not filled up. Hence -size -8 means “occupying 7 blocks or less”. Adding a unit works as if it was making that unit the block size, so -size -8k means “occupying 7kB or less”.

(This answer is about GNU find — some other implementations behave differently, and POSIX doesn't standardize units other than blocks and bytes.)

  • Evaluating -size -1M & -size -1G as 0 varies by platform. GNU find does this, BSD find does not. Logically, it's hard to argue that "less than 2^20" means 0!? – jeberle Aug 12 '16 at 19:06

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