Printing the sum-size of copied files in bash

I would like from the code below to print the sum-size of the files being copied instead of the size of each copied file separately.

#!/bin/bash
input_folder=/a/b/c
output_folder=/d/e/f

cd $input_folder

for i in *.tiff; do
size=$(wc -c < $i)
cp -v $i $out | sleep 1 | echo -ne "$size%33[0Kr"
done


This bash prints every second the size of the file being copied.

bash

For example, in second 1 it copies file A and prints the size of file A. In second 2 it copies file B and prints the size of file B instead of the total size of files A+B. In second 3 it copies the file C and prints the size of file C instead of the total size of files A+B+C

1

file A

file A

2

file B

file B

A+B

3

file C

file C

A+B+C

How can I do it in bash specifically?

bash

0.1

4 Answers
4

Just sum up the single sizes and echo it afterwards.

echo

#!/bin/bash
input_folder=/a/b/c
output_folder=/d/e/f

cd $input_folder

size_sum=0
for i in *.tiff; do
size=$(wc -c < $i)
size_sum=$((size_sum + size))
cp -v $i $out | sleep 1
done

echo $size_sum


size=$(wc -c < $i)


This isn't a very smart way to get the size of a file. Running wc like that will involve reading the whole file, which really is completely unnecessary since all you want is the size which the filesystem already knows and can tell you.

wc

This is mitigated by the fact that you copy the file afterwards, and so need to read it in any case. A moderately-sized file will probably stay in the cache for the duration, so the reads don't actually hit the disk. But still, it would seem sensible to just read the size.

With GNU utilities, you can get the size of the file with stat -c %s "$filename". Sadly, the options stat takes are different on other systems, and I don't think there's a better, more portable way (apart from running Perl)

stat -c %s "$filename"

stat

cp -v $i $out | sleep 1 | echo -ne "$size%33[0Kr"


Here, the pipeline seems odd. cp doesn't produce any output (to standard output), at least not without -i, so there's nothing to pipe. Even if it did, sleep doesn't read anything so any output from cp would be wasted. The same thing happens between sleep and echo. This is basically the total opposite of what pipelines are usually used for, but it does have the effect that all the commands run in parallel, and the shell waits for all of them to complete, so that pipeline takes at least one second to run, regardless of how fast the copy would otherwise be.

cp

-i

sleep

cp

sleep

echo

If you want to sum the file sizes, the shell's arithmetic expansion $(( .. )) is what you probably want. So, without any extra sleeps:

$(( .. ))

dest=/path/to/destination
total=0
for file in *.tiff; do
size=$(stat -c %s "$file");
printf "current %d total %dn" "$size" "$((total += size))"
cp -- "$file" "$dest"
done


To save on the arithmetic in your script, you can just take the size of the concatenation of the files:

cat *.tiff | wc


The UN*X kernel will conspire with you to make this just as efficient as taking the sizes of the files separately.

wc -c

It is possible to sum with += if the variable is declared as integer:

+=

$ declare -i size
$ size+=100; echo "$size"
100
$ size+=200; echo "$size"
300


Your script (with variable expansions correctly quoted):

#!/bin/bash
input_folder=./a/b/c
output_folder=./d/e/f
declare -i size

for i in "$input_folder"/*; do
size+=$(wc -c < "$i")
cp -v "$i" "$output_folder" | sleep 1 | echo -ne "$size33[0Kr"
done


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