UNIX：如何使我的脚本返回从命令行传递给它的数字的总和？

Question

Here is my current code:

for i do
    sum=$(expr $sum + $i)
done
echo =$sum

Now this works, but I want it to display all of the numbers. For example, If I enter ./sum 1 2 3, I want it to display 1+2+3=6. Right now it only displays the answer. Also, is there a way I could execute the file without ./. For instance, could I use sum 1 2 3 instead of ./sum 1 2 3. I've tried chmod 700 "myfile," but that didn't seem to work.

Answer 1

What about this?

while (( $# > 1 )); do
    printf "$1 + "
    sum=$((sum + $1 ))
    shift
done
echo "$1 = $((sum + $1))"

It loops through the arguments you provide and adds them into the variable $sum . For stylistic purposes I use printf and finally echo to have everything in the same line.

The usage of shift is explained here:

The shift builtin command is used to "shift" the positional parameters by the given number n or by 1, if no number is given.

Test

$ sh script.sh 1 2 3
1 + 2 + 3 = 6
$ sh script.sh 1 2 3 4 5
1 + 2 + 3 + 4 + 5 = 15

Answer 2

You can use IFS and $* to join all the arguments with + :

(IFS="+"; printf '%s' "$*")

sum=0
for i; do
    sum=$((sum + i))
done
printf '=%d\n' "$sum"

---

To run without the ./ prefix, you need to put the program in a directory that appears in your PATH . Unless this is intended for people other than you, you should create a bin directory in your home directory, then add this to your .bash_profile :

PATH=~/bin/:$PATH

so that your bin directory is added to the list of places to look for commands.

Answer 3

Another answer:

sum=0
for i in $@
do
    sum=$(($sum + $i))
done
echo $sum

Answer 4

Here a solution without loops:

numbers=$( printf -v str '%s+' $@; printf '%s' "${str%+}" )
sum=$(( $numbers ))
echo $numbers=$sum 

the first line joins the input arguments with +, the second line calculates the sum and the third line outputs the equation

as to the second part of your question. Just put the script somewhere in your path (for example /usr/local/bin)