使用bash命令计算文件夹大小而不使用du左右

Question

I wonder if there is some way to calculate folder size using only basic Bash commands without du. It is for very limited evironment (like rescue mode) that has no such utilities as du. Thank you in advance!

Answer 1

Theoretically, you can calculate the size of a folder using only bash built-ins. However, I doubt there is a real use-case for such a command. Even in rescue mode you should be able to run du ; maybe you just have to specify the full path: Try /bin/du or /usr/bin/du . Even if that did not work you probably have other commands (eg stat or at least ls ) which could be used to efficiently get the size of a single file, which could be used to write a simple script.

Anyways, lets sum up the apparent sizes of all (readable) files in the current/given directory recursively …

… using only bash built-ins:

(Don't use this in real live)

#! /bin/bash
dir=${1-.}

LC_ALL=C
shopt -s globstar nullglob

declare -i size
add() { records+=1; size+=${#2}+1; MAPFILE=(); }
filesize() {
  declare -i records=0
  mapfile -d '' -c 1 -C add < "$1"
  # computed size is correct iff
  # file is empty or ends with a null byte,
  # otherwise the size is off by +1.
  # heuristic to correct this:
  (( size -= records > 0 ))
}

for file in "$dir"/**; do
  [[ -f "$file" && -r "$file" && ! -L "$file" ]] || continue
  filesize "$file"
done
echo "$size $dir"

If we had a command like cat we could fix the +/-1 problem in filesize() using mapfile ... <(cat "$1"; printf .); ((size--)) mapfile ... <(cat "$1"; printf .); ((size--)) ; but as far as I know there is no way to print the exact content of a file using only bash built-ins.

Above command is basically du --appaparent-size -sb but does not include unreadable files but includes multiple hardlinks to the same file multiple times.
Also, it is super slow. On a test directory with a file generated by head -c 100M /dev/urandom the script took more than two minutes!