Shell command to sum integers, one per line?

Question

Bit of awk should do it?

awk '{s+=$1} END {print s}' mydatafile

Note: some versions of awk have some odd behaviours if you are going to be adding anything exceeding 2^31 (2147483647). See comments for more background. One suggestion is to use printf rather than print:

awk '{s+=$1} END {printf "%.0f", s}' mydatafile

Paste typically merges lines of multiple files, but it can also be used to convert individual lines of a file into a single line. The delimiter flag allows you to pass a x+x type equation to bc.

paste -s -d+ infile | bc

Alternatively, when piping from stdin,

<commands> | paste -s -d+ - | bc

The one-liner version in Python:

$ python -c "import sys; print(sum(int(l) for l in sys.stdin))"

I would put a big WARNING on the commonly approved solution:

awk '{s+=$1} END {print s}' mydatafile # DO NOT USE THIS!!

that is because in this form awk uses a 32 bit signed integer representation: it will overflow for sums that exceed 2147483647 (i.e., 2^31).

A more general answer (for summing integers) would be:

awk '{s+=$1} END {printf "%.0f\n", s}' mydatafile # USE THIS INSTEAD

Plain bash:

$ cat numbers.txt 
1
2
3
4
5
6
7
8
9
10
$ sum=0; while read num; do ((sum += num)); done < numbers.txt; echo $sum
55