shell - using expr “multiplication table”

Question

I'm learning shell to create a multiplication table, I write code like that:

#!/ in/bash

for i in 1 2 3 4 5 6 7 8 9

do

    for j in 1 2 3 4 5 6 7 8 9
    do
            if [ $j -le $i ]
            then
                    echo -ne "$i"X"$j"=`expr "$i"*"$j"` "\t"
            else
                    echo
                            break
            fi
            done
    done
echo -ne "\n"

however I get the response in console like that:

1X1=1*1
2X1=2*1         2X2=2*2
3X1=3*1         3X2=3*2         3X3=3*3
4X1=4*1         4X2=4*2         4X3=4*3         4X4=4*4
...

I hope the answer is like that "1x1=1, 2x1=2 ...". Could anyone tell me how to modify this code?

Or put a space around the * operator – Jeff Schaller Jan 01 '17 at 15:46 — Jeff Schaller, Jan 01 '17 at 15:46
... and escape/quote it I think e.g. `expr "$i" \* "$j"` – steeldriver Jan 01 '17 at 16:54 — steeldriver, Jan 01 '17 at 16:54

Wissam Roujoulah · Accepted Answer · 2017-01-01T15:23:36.820

Yes, you can use bash's built-in Arithmetic Expansion $(( )) to do some simple maths.

For Multiplication:

echo $(( 6 * 7 ))

Result

And your script would look like this:

#!/ in/bash

for i in 1 2 3 4 5 6 7 8 9

do

    for j in 1 2 3 4 5 6 7 8 9
    do
            if [ $j -le $i ]
            then
                    echo -ne "$i"X"$j"=$(($i * $j)) "\t"
            else
                    echo
                            break
            fi
            done
    done
echo -ne "\n"

score 0 · Answer 2 · answered Jan 04 '17 at 19:48

0

Simplifying the solution a bit, by removing an unnecessary test:

#!/bin/bash

for (( i = 1; i < 10; ++i )); do
  for (( j = 1; j <= i; ++j )); do
    printf '%dx%d=%-2d  ' "$i" "$j" "$(( i * j ))"
  done
  printf '\n'
done

answered Jan 04 '17 at 19:48

Kusalananda

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shell - using expr “multiplication table”

2 Answers2