tricks

loghyr:xxx thomas$ ls -1 pnfswars* | sed ‘s/\(pnfswars\)\(.*\)/mv \1\2 lnfsreg\2/’ | sh

And to do this as a script:

loghyr:xxx thomas$ more reprefix.sh
#! /bin/sh
ls -1 ${1}* | sed ‘s/\(‘${1}’\)\(.*\)/mv \1\2 ‘${2}’\2/’ | sh

Get a clone and make a new branch:

290 21:08 git clone git://linux-nfs.org/~steved/cthon04.git
291 21:08 cd cthon04
292 21:08 git branch FreeBSD
293 21:08 git checkout FreeBSD
294 21:08 git branch

Edit your files and commit the changes:

299 21:09 git commit -a

Make the patch set:


303 21:15 git format-patch --cover-letter -s -n master..FreeBSD


Edit the cover letter to document your changes:

309 21:20 vi 0000-cover-letter.patch

Send the changes:


311 21:21 git send-email --smtp-encryption=tls --smtp-server=smtp.gmail.com --smtp-user=<user>@gmail.com --smtp-server-port=587 --to "<owner>" 000*


Note, while you might use a different email address inside the cover letter, it will still go out as from the user being referenced above. I.e., use the real smtp-server if you don’t want to expose your gmail account.

Note, you might have to install git-mail to do it this way. You can mail the patch set in many different ways.

If we have a binary tree, how do we determine if a key is in the tree? Or another way, how do we know it was inserted correctly? We need to visit every node.

This is a recursive problem and hard to do in gdb. The convenience variables are very simple and we can not implement a stack with them. Since they are effectively global, we can not even hide them in stack frames. What we can do is use concatenation and a node-id to generate uniqe names.

Note that you have to do all variable name declarations before you do any recursion. That statement will make sense when you violate it and wonder why your output looks strange.

define indentby
        printf "%*c", $arg0+1, 0
end

define walk_btree
        set $node_$arg0 = ($arg1)
        set $fnum_$arg0 = ($arg2)
        set $branch_$arg0 = ($arg3)
        set $space_$arg0 = ($arg4)
        set $leftid_$arg0 = $arg0 + 1
        set $rightid_$arg0 = $arg0 + 2

        indentby $space_$arg0
        printf "%d %u (btree *)%p\n", $branch_$arg0, $node_$arg0->e_type[$fnum_$arg0].key, $node_$arg0

        set $space_$arg0 = $space_$arg0 + 1
        if $node_$arg0->e_type[$fnum_$arg0].left
                walk_btree $leftid_$arg0 $node_$arg0->e_type[$fnum_$arg0].left $fnum_$arg0 0 $space_$arg0
        end
        if $node_$arg0->e_type[$fnum_$arg0].right
                walk_btree $rightid_$arg0 $node_$arg0->e_type[$fnum_$arg0].right $fnum_$arg0 1 $space_$arg0
        end
end
document walk_btree
        NODE-ID NODE FNUM BRANCH INDENTBY
        Recursively walk the binary tree printing out the key for the fnum'ed entry
end

define start_walk_btree
        walk_btree 0 $arg0 $arg1 3 1
end
document start_walk_btree
        NODE FNUM
        Recursively walk the binary tree printing out the key for the fnum'ed entry
end

Contrast that against a very simple algorithm to find if the key is in the tree (with the assumption the tree is valid).

Recursion is not needed as we lop of a branch with each iteration.

define find_btree
        set $node = $arg0
        set $fnum = $arg1
        set $key = $arg2
        set $process = 1

        while $node && $process
                if $key == $node->e_type[$fnum].key
                        set $process = 0
                else

                        if $key < $node->e_type[$fnum].key
                                $node = $node->e_type[$fnum].left
                        else
                                $node = $node->e_type[$fnum].right
                        end
                end
        end

        if $node
                printf "Found match on (btree *)%p\n", $node
        else
                printf "No matching node in the binary tree\n"
        end
end