Use awk to have a rounded percentage
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python is slow
bash supports only integer arithmetics
"bc" can't round to the ceiling
This commit is contained in:
Jérémy Lecour 2019-03-24 21:10:47 +01:00
parent 4136b310c1
commit a7abd80f93

View file

@ -914,7 +914,7 @@ if is_debian; then
# Check if reserved blocks for root is at least 5% on every mounted partitions.
if [ "$IS_TUNE2FS_M5" = 1 ]; then
BC_BIN=$(command -v bc)
min=5
parts=$(grep -E "ext(3|4)" /proc/mounts | cut -d ' ' -f1 | tr -s '\n' ' ')
for part in $parts; do
blockCount=$(dumpe2fs -h "$part" 2>/dev/null | grep -e "Block count:" | grep -Eo "[0-9]+")
@ -923,15 +923,12 @@ if is_debian; then
continue
fi
reservedBlockCount=$(dumpe2fs -h "$part" 2>/dev/null | grep -e "Reserved block count:" | grep -Eo "[0-9]+")
if [ -x "$BC_BIN" ]; then
# Use bc to have a rounded percentage
percentage=$(echo "(${reservedBlockCount} * 100 / ${blockCount}) + 1" | $BC_BIN)
else
# fallback to python
percentage=$(python -c "print(int(round(float(${reservedBlockCount})/${blockCount}*100)))")
fi
if [ "$percentage" -lt 5 ]; then
failed "IS_TUNE2FS_M5" "Partition ${part} has less than 5% reserved blocks (${percentage}%)"
# Use awk to have a rounded percentage
# python is slow, bash is unable and bc rounds weirdly
percentage=$(awk "BEGIN { pc=100*${reservedBlockCount}/${blockCount}; i=int(pc); print (pc-i<0.5)?i:i+1 }")
if [ "$percentage" -lt "${min}" ]; then
failed "IS_TUNE2FS_M5" "Partition ${part} has less than ${min}% reserved blocks (${percentage}%)"
fi
done
fi