drawing a shape with shell script

I'm trying to draw a tree in the following form

Code:

---

    *   
  ***
  *****
 *******
*********

---

as you can see I need both spaces, and "*" stars in every line, but I can't get it quite the way I want

Code:

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#!/bin/bash

display_tree() {
        local rows=$1
        local columns=$2

        for ((i=0; i<$rows; i++))
        do
                #spaces loop
                for ((j=0; j<$columns; j++))
                do
                        echo -n " "
                        #drawing tree loop
                        for ((a=0; a<$(($i + 1)); a++))
                        do
                                echo -n "*"
                        done
                done
        echo
        done
}

if [ $# -eq 2 ]; then
        display_tree $1 $2
else
        echo "Usage: $0 rows columns"
fi

---

Code:

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$ n=6; row="";
$ for(( i=0; i<n; i++ )); do row="$row "; done; row="${row%?}*"; for(( i=0; i<n; i++ )); do echo "$row"; row="${row#?}**"; done
    *
    ***
  *****
  *******
 *********
***********

---

Quote:

---

Originally Posted by Vaphell

Code:

---

$ n=6; row="";
$ for(( i=0; i<n; i++ )); do row="$row "; done; row="${row%?}*"; for(( i=0; i<n; i++ )); do echo "$row"; row="${row#?}**"; done
    *
    ***
  *****
  *******
 *********
***********

---

---

great job!.. would you mind explaining this two sentences

Code:

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row="${row%?}*"
row="${row#?}**"

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I know it is a variable declaration, but what's on the right side of "=" sign I haven't seen it yet

Code:

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$ x="abcd"
$ echo ${x#?} ${x%?}
bcd abc

---

these parameter expansions trim chunks that match given pattern from the left (#) and from the right (%). ? symbol used here means 'single character'


first loop generates n spaces long string.
${row%?}* trims the rightmost char and adds * (last space replaced by *)
i get 1st row that way

note that (n+1)-th row has -1 space and +2 stars when compared to n-th row so in each iteration of the second loop i kill one space on the left, and add 2 *s on the right: ${row#?}**

Quote:

---

Originally Posted by Vaphell

Code:

---

$ x="abcd"
$ echo ${x#?} ${x%?}
bcd abc

---

these parameter expansions trim chunks that match given pattern from the left (#) and from the right (%). ? symbol used here means 'single character'


first loop generates n spaces long string.
${row%?}* trims the rightmost char and adds * (last space replaced by *)
i get 1st row that way

note that (n+1)-th row has -1 space and +2 stars when compared to n-th row so in each iteration of the second loop i kill one space on the left, and add 2 *s on the right: ${row#?}**

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very efficient algorithm!..
Thanks