mjhjmtu

I want to print the size of a file from some folder /etc/*.conf in this case.

When I use:

cd /etc

du -ch $(ls | grep .conf) | tail -1 | cut -f1 

I get 120K.

When I use:

du -bch $(ls | grep .conf) | tail -1 | cut -f1

I get 46K. and this should be the same size but in bytes right? so it should be some kind of 120000, right?

When i use:

du -bsh $(ls | grep .conf) | tail -1 | cut -f1

I get 1.3K, what is this man?

du -hc rounds out the size of the smaller files to the block size of the filesystem, typically 4K.

[root@testvm1 etc]# du -hc *.conf
4.0K asound.conf
4.0K chrony.conf
4.0K dracut.conf
....
4.0K vconsole.conf
4.0K yum.conf
104K total

du -bch avoids the rounding, which results in a lower total:

[root@testvm1 etc]# du -bch *.conf
55 asound.conf
1.1K chrony.conf
1.3K dracut.conf
....
41 vconsole.conf
970 yum.conf
32K total

du -sh does not print a total value at the end. The output of your du -bsh command will most likely be the size of the last file in list.

The -b option is equivalent to --apparent-size --block-size=1. To use block-size calculations while keeping the unit as bytes, use only the --block-size option.

[root@testvm1 etc]# du --block-size=1 -c *.conf
4096 asound.conf
4096 chrony.conf
4096 dracut.conf
...
4096 vconsole.conf
4096 yum.conf
106496 total

You should check the result of du without the | tail -1 | cut -1

(ignoring the -h option which just add the k, M...)

(based on http://man7.org/linux/man-pages/man1/du.1.html)

-c will print the disk usage of all files plus a total. note that depending of your filesystem format disk usage of a file will be bigger than its real size)

-bc will print the "real" size rather than the size its use on the disk.

-bs will only print the total "real" size of each file/folder given to du. Since you dive du each file it will calculate the size of each *.conf file and your last line is the size of the last *.conf file you've given him.

PS: you can probably do: du -bch *.conf rather than the grep on ls result.