Sum the numbers on standard in

Question

Consider a stream/file with one integer per line. For example:

123
5
99

Your code should output the sum of these numbers, that is 227.

The input format is strictly one integer per line. You cannot, for example, assume the input is on one line as an array of integers.

You can take input either from STDIN, in form of a filename, or a file with a name of your choice; you can choose which one. No other ways of getting input are allowed.

The input will contain at least one integer. You can assume all integers are non-negative and that their total sum is less than 232.

Answer 1

Bash + coreutils, 16 bytes

xargs|tr \  +|bc

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There are two spaces after the \. This works for negative numbers as well.

Explanation:

xargs             # known trick to turn newlines into spaces, while adding a
                  #trailing newline when printing the result (needed for bc)
|tr \  +          # turn spaces into '+'s
|bc               # calculates the sum

You may wonder why tr \\n +|bc isn't better, since it turns newlines directly into '+'s. Well, that has 2 unforeseen errors:

• if the input has a trailing newline, then it is converted to a trailing '+', hence there is no number after it to perform the addition
• and the most weird issue is that bc requires a trailing newline after the input, but you just replaced all of the input newlines with '+'s.

Answer 2

05AB1E, 2 bytes

|O

Explanation:

|   Get input as array
 O  Sum

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Answer 3

MATL, 2 bytes

Us

This expects the input in a text file called defin.

Gif or it didn't happen:

Or try it online! (thanks to Dennis for the set-up!)

Explanation

When a MATL program is run, if a file called defin is found (the name refers to "default input"), its contents are automatically loaded as text and pushed to the stack as a string before executing the code.

Function U evaluates the string to convert it to a column vector of numbers, and s computes the sum, which is implicitly displayed.

Answer 4

Japt, 2 bytes

Nx

Explanation

     Implicit: parse STDIN into array of numbers, strings, and arrays
N    Get the resulting parsed array.
 x   Sum.
     Implicit: output result of last expression

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Answer 5

Paste + bc, 13 bytes

paste -sd+|bc

Explanation:

paste -s        Take one line at a time from input
        d+      Joining by '+'
          |bc   Pass as expression to bc

Another shell answer!

Answer 6

Perl 6, 13 bytes

say sum lines

Try it

Explanation

• lines() returns a list of lines from $*IN or $*ARGFILES a “magic” command-line input handle.
• sum(…) was added to Perl 6 to allow [+] List to be optimized for Positionals that can calculate their sum without generating all of their values like 1..100000
  (I just thought sum was just too cute here to use [+] like I normally would)
• say(…) call the .gist method on its input, and prints it with an additional newline.

Answer 7

C, 53 bytes

r;main(i){for(;~scanf("%d",&i);r+=i);printf("%d",r);}

Answer 8

Python 3, 28 bytes

print(sum(map(int,open(0))))

Taken from this tip. I've been told this won't work on Windows.

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Answer 9

Retina, 11 7 bytes

-4 thanks to Martin Ender

.*
$*
1

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---

Convert to unary:

.*
$*

Count the number of 1s:

1

Answer 10

Brain-Flak, 20 bytes

(([]){[{}]{}([])}{})

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Explanation

This is a golf off of a solution made by Riley in chat. His solution was:

([])({<{}>{}<([])>}{})

If your familiar with Brain-Flak this is pretty self-explanatory. It pushes the stack height and pops one value as it counts down, at the end it pushes the sum of all the runs.

It is a pretty good golf but he zeros both {} and ([]) however these will have a values that only differ by one so if instead we remove the masks and make one of the two negative they should nearly cancel out.

([])({[{}]{}([])}{})

Since they always differ by one we have the unfortunate circumstance where our answer is always off by the stack height. In order to remedy this we simply move the beginning of the push to encompass the first stack height.

(([]){[{}]{}([])}{})

Answer 11

Python 2, 40 bytes

import sys;print sum(map(int,sys.stdin))

Answer 12

Perl 5, 9 bytes

8 bytes of code + -p flag.

$\+=$_}{

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With -p, the input is read one line at a time, stored in $_ each time. We use $\ as accumulator, because thanks to -p flag, it's implicitly printed at the end. The unmatched }{ are used so -p flag only prints $\ once at the end instead of printing $_ and $\ at each line it reads like it normally does.

Answer 13

R,11 bytes

sum(scan())

scan takes the input, one number per line. And sum, well, sums.

Answer 14

Pure Bash, 37 36 bytes

Thanks to @KevinCruijssen for a byte!

while read a;do((b+=a));done;echo $b

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Answer 15

Haskell, 32 bytes

interact$show.sum.map read.lines

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interact collects the whole input from stdin, passes it to the function given as its argument and prints the string it gets back from this function. The function is:

            lines   -- split input into list of lines at nl
      map read      -- convert every line to a number (read is polymorphic,
                    -- but as want to sum it later, the type checker knows
                    -- it has to be numbers)
    sum             -- sum the list of numbers
show                -- convert back to string

Answer 16

PHP, 22 bytes

<?=array_sum(file(t));

This assumes there is a file named "t" with a list of integers.

file() opens a file and returns an array with each line stored a separate element in the array. array_sum() sums all the elements in an array.

Answer 17

Awk, 19 bytes

{s+=$1}END{print s}

Explanation:

{s+=$1}                For all lines in the input, add to s
        END             End loop
           {print s}    Print s

Answer 18

dc, 14 bytes

0[+?z2=a]dsaxp

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Explanation:

 [      ] sa   # recursive macro stored in register a, does the following:
  +            # - sum both numbers on stack
               #   (prints to stderr 1st time since there's only 1)
   ?           # - read next line, push to stack as number
    z          # - push size of stack
     2         # - push 2
      =a       # - if stack size = 2, ? yielded something, so recurse
               # - otherwise end macro (implicit)
0              # push 0 (accumulator)
         d     # duplicate macro before storing it
            x  # Call macro
             p # The sum should be on the stack now, so print it

Answer 19

CJam, 5 bytes

q~]1b

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How it works

q     e# Read all input from STDIN.
 ~    e# Evaluate that input, pushing several integers.
  ]   e# Wrap the entire stack in an array.
   1b e# Convert from base 1 to integer.
      e# :+ (reduce by sum) would work as well, but 1b handles empty arrays.

Answer 20

Python, 38 30 bytes

lambda n:sum(map(int,open(n)))

In python, files are opened by open('filename') (obviously). They are, however, iterables. Each time you iterate through the file, you get the next line. So map iterates over each list, calling int on it, and then sums the resulting list.

Call with the filename as input. (i.e. f('numbers.txt'))

8 bytes saved by using map(int, open(n)) instead of a list comprehension. Original code:

lambda n:sum([int(i)for i in open(n)]) 

Answer 21

Mathematica, 19 bytes

Assumes Mathematica's notebook environment.

Tr[#&@@@Import@"a"]

Expects the input to be in a file a.

Answer 22

Jelly, 9 8 bytes

ƈFпFỴVS

STDIN isn't really Jelly's thing...

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How it works

ƈFпFỴVS  Main link. No arguments. Implicit argument: 0

  п      While loop; while the condition returns a truthy value, execute the body
          and set the return value to the result. Collect all results (including 0,
          the initial return value) in an array and return that array.
ƈ           Body: Yield a character from STDIN or [] if the input is exhausted.
 F          Condition: Flatten, mapping 0 to [], '0' to "0", and [] to [] (falsy).
    F     Flatten the result.
     Ỵ    Split at newlines.
      V   Evaluate the resulting strings.
       S  Take the sum.

Answer 23

Brachylog, 4 bytes

ṇịᵐ+

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Explanation

ṇ         Split the Input on linebreaks
 ịᵐ       Map: String to Integer
   +      Sum

Answer 24

Pure bash, 30

read -d_ b
echo $[${b//'
'/+}]

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• reads the input file in one go into the variable b. -d_ tells read that the line delimiter is _ instead of newline
• ${b//'newline'/+} replaces the newlines in b with +
• echo $[ ... ] arithmetically evaluates the resulting expression and outputs it.

Answer 25

Vim, 16 bytes/keystrokes

:%s/\n/+
C<C-r>=<C-r>"<C-h>

Since V is backwards compatible, you can Try it online!

Answer 26

Pyth, 3 bytes

s.Q

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 .Q  Read all of standard input, evaluating each line.
s    Take the sum.

Answer 27

jq, 5 bytes

add, plus the command line flag -s.

For example:

% echo "1\n2\n3\n4\n5" | jq -s add
15

Answer 28

Actually, 2 bytes

kΣ

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Explanation:

kΣ
    (implicit input - read each line, evaluate it, and push it to the stack)
k   pop all stack elements and push them as a list
 Σ  sum
    (implicit output)

Answer 29

dc, 22

[pq]sq0[?z2>q+lmx]dsmx

This seems rather longer than it should be, but it is tricky to decide when the end of the file is reached. The only way I can think of to do this is check the stack length after the ? command.

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[pq]                    # macro to `p`rint top-of-stack, then `q`uit the program
    sq                  # save the above macro in the `q` register
      0                 # push `0` to the stack.  Each input number is added to this (stack accumulator)
       [         ]      # macro to:
        ?               # - read line of input
         z              # - push stack length to stack
          2             # - push 2 to the stack
           >q           # - if 2 > stack length then invoke macro stored in `q` register
             +          # - add input to stack accumulator
              lmx       # - load macro stored in `m` register and execute it
                  d     # duplicate macro
                   sm   # store in register `m`
                     x  # execute macro

Note the macro m is called recursively. Modern dc implements tail recursion for this sort of thing, so there should be no worries about overflowing the stack.

Answer 30

Ruby, 19 15 bytes

Loving that new #sum in Ruby 2.4 :)

p$<.sum(&:to_i)

This program accepts input from either stdin or any filename(s) given as command line argument