Using Scooter Software Beyond Compare

Beyond Compare is a graphical file comparison tool sold by Scooter Software. Its open-source competitors are mainly vimdiff, and kdiff3. Its advantage is ease-of-use. While comparing files they can be edited instantly. You can diff complete directory trees.

It is written in Delphi Object Pascal, the source code is not open-source. It runs on Windows, x86 Linux, and OS X. It does not run on ARM, like Raspberry Pi or Odroid, see support for arm processors – like the raspberry pi. The “Standard Edition” costs $30, the “Pro Edition” costs $60. The software is in AUR.

1. Root User Problem. When using it as root-user you must use:


When running

DIFFPROG=bcompare pacdiff

the screen looks like this:

2. Git Usage. To use Beyond Compare with git difftool you have to do two things: First you must create an alias bc3 for bcompare.

[root /bin]# ln -s bcompare bc3

Second add the following lines to your ~/.gitconfig file:

        tool = bc3
        prompt = false
        bc3 = trustExitCode
        tool = bc3
        bc3 = trustExitCode

Alternatively to above changes in the ~/.gitconfig file, use the following commands:

git config --global diff.tool bc3
git config --global difftool.bc3.trustExitCode true
git config --global merge.tool bc3
git config --global mergetool.bc3.trustExitCode true

Towards web-based delta synchronization for cloud storage systems

Very interesting article.

Some remarkable excerpts:

To isolate performance issues to the JavaScript VM, the authors rebuilt the client side of WebRsync using the Chrome native client support and C++. It’s much faster.

Replacing MD5 with SipHash reduces computation complexity by almost 5x. As a fail-safe mechanism in case of hash collisions, WebRsync+ also uses a lightweight full content hash check. If this check fails then the sync will be re-started using MD5 chunk fingerprinting instead.

The client side of WebR2sync+ is 1700 lines of JavaScript. The server side is based on node.js (about 500 loc) and a set of C processing modules (a further 1000 loc).

the morning paper

Towards web-based delta synchronization for cloud storage systems Xiao et al., FAST’18

If you use Dropbox (or an equivalent service) to synchronise file between your Mac or PC and the cloud, then it uses an efficient delta-sync (rsync) protocol to only upload the parts of a file that have changed. If you use a web interface to synchronise the same files though, the entire file will be uploaded. This situation seems to hold across a wide range of popular services:

Given the universal presence of the web browser, why can’t we have efficient delta syncing for web clients? That’s the question Xiao et al. set out to investigate: they built an rsync implementation for the web, and found out it performed terribly. Having tried everything to improve the performance within the original rsync design parameters, then they resorted to a redesign which moved more of the heavy lifting back to…

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Unix Command comm: Compare Two Files

One lesser known Unix command is comm. This command is far less known than diff. comm needs two already sorted files FILE1 and FILE2. With the options

  • -1 suppress column 1 (lines unique to FILE1)
  • -2 suppress column 2 (lines unique to FILE2)
  • -3 suppress column 3 (lines that appear in both files)

For example, comm -12 F1 F2 prints all common lines in files F1 and F2.

I thought that comm had a bug, so I wrote a short Perl script to simulate the behaviour of comm. Of course, there was no bug, I just missed to notice that the records in the two files did not match due to white space.

#!/bin/perl -W
use strict;

use Getopt::Std;
my %opts = ('d' => 0, 's' => 0);
my $debug = ($opts{'d'} != 0);
my $member = defined($opts{'s'}) ? $opts{'s'} : 0;

my ($set,$prev) = (1,"");
my %H;

while (<>) {
        $prev = $ARGV if ($prev eq "");
        if ($ARGV ne $prev) {
                $set *= 2;
                $prev = $ARGV;
        $H{$_} |= $set;
        printf("\t>>\t%s: %s -> %d\n",$ARGV,$_,$H{$_}) if ($debug);

$member = 2*$set - 1 if ($member == 0);
printf("\t>>\tmember = %d\n",$member) if ($debug);
for my $i (sort keys %H) {
        printf("%s\n",$i) if ($H{$i} == $member);

Above Perl scripts does not need sorted input files, as it stores all records of the files in memory, in a hash. It uses a bitmask as a set. For example, mycomm -s2 F1 F2 prints only those records, which are only in file F2 but not in F1.

Parallelization and CPU Cache Overflow

In the post Rewriting Perl to plain C the runtime of the serial runs were reported. As expected the C program was a lot faster than the Perl script. Now running programs in parallel showed two unexpected behaviours: (1) more parallelizations can degrade runtime, and (2) running unoptimized programs can be faster.

See also CPU Usage Time Is Dependant on Load.

In the following we use the C program siriusDynCall and the Perl script siriusDynUpro which was described in above mentioned post. The program or scripts reads roughly 3GB of data. Before starting the program or script all this data has been already read into memory by using something like wc or grep.

1. AMD Processor. Running 8 parallel instances, s=size=8, p=partition=1(1)8:

for i in 1 2 3 4 5 6 7 8; do time siriusDynCall -p$i -s8 * > ../resultCp$i & done
        real 50.85s
        user 50.01s
        sys 0

Merging the results with the sort command takes a negligible amount of time

sort -m -t, -k3.1 resultCp* > resultCmerged

Best results are obtained when running just s=4 instances in parallel:

$ for i in 1 2 3 4 ; do /bin/time -p siriusDynCall -p$i -s4 * > ../dyn4413c1p$i & done
        real 33.68
        user 32.48
        sys 1.18

Continue reading

Rewriting Perl to plain C

Perl script was running too slow. Rewriting it in C made it 20 times faster.

1. Problem statement. Analyze call-tree dependency in COBOL programs. There are 77 million lines of COBOL code in ca. 30,000 files. These 30,000 COBOL programs could potentially include 74,000 COPY-books comprising 10 million lines of additional code. COBOL COPY-books are analogous to C header-files. So in total there are around 88 million lines of COBOL code. Just for comparison: the Linux kernel has ca. 20 million lines of code.

COBOL program analysis started with a simple Perl script. This Perl script is less than 200 lines, including comments. This script produced the desired dependency information.

Wading through all this COBOL code took up to 25 minutes in serial mode, and 13 minutes using 4 cores on an HP EliteBook notebook using Intel Skylake i7-6600U clocked 2.8 GHz. It took 36 minutes on an AMD FX-8120 clocked with 3.1 GHz. This execution time was deemed too long to see any changes in the output changing something in the Perl script. All runs are on Arch Linux 4.14.11-1 SMP PREEMPT.

2. Result. Rewriting the Perl script in C resulted in a speed improvement of factor 20 when run in serial mode, i.e., run times are now 110s on one core. It runs in 32s when using 8 cores on an AMD FX-8120. C program uses taylormade hashing routines.
Continue reading

Hashing with Iterator in C

Many scripting languages, like Awk, Perl, Python, Lua, and so on, have hash-tables already built-in. I needed hashing functionality similar to Perl’s hash-variables. In particular I needed lookup, insert, and iterating over all entries.

I wrote on hashing in Hashing Just Random Numbers and Hash functions: An empirical comparison — article by Peter Kankowski. See also Revisiting hash table performance, or TommyDS: a C library of hashtables and tries.

1. Structure. Each hash element contains a key and a value. Additionally there is an indicator whether storage allocation using malloc() for key and value are done during hash insert, or are done elsewhere, e.g., constant strings. All hash elements having the same hash value, see below, are inserted into a singly linked list. Additionally all elements are part of singly linked list.

struct hashElem {
        int dup;        // memory ownership: 0, 0x01, 0x02, 0x03
        char *key, *val;
        struct hashElem *next;  // next hash element with same key or NULL
        struct hashElem *nextAll;       // for iterator over all hash entries

There is a hash header which contains the start of the list, and the actual table.

typedef struct {
        int size, collisions, load;
        const char *name;       // only for diagnostic and debugging
        struct hashElem **tab;  // hash table with size times (hashElem*)
        struct hashElem *first; // first element in hash
} hash_t;

Continue reading

C Pointer Surprises

An article by Krister Walfridsson on C pointers are not hardware pointers demonstrated that even adjacent integer variables having the same hardware address may compare unequal regarding C pointers.

See the following C program:

#include <stdio.h>

int main(int argc, char *argv[]) {
        int x, y;
        int *p = &x + 1;
        int *q = &y;
        printf("%p %p %d\n", (void*)p, (void*)q, p == q);
        return 0;

You have to compile with optimization enabled, e.g., cc -O3. Otherwise gcc adds some stuff between variables. On AMD/Intel/ARM CPUs the output looks something like this:

0xbe849afc 0xbe849afc 0

I.e., the pointers point to the same address, but the pointer comparison gives “false”.

Added 06-Aug-2017: As hinted by the comment given by Ashwin Nanjappa below, the compiler actually does not generate compare instructions, but rather just adds 0=false.


$ cc -Wall -O3 -c ptrcomp.c
$ objdump -d ptrcomp.o


ptrcomp.o:     file format elf64-x86-64

Disassembly of section .text.startup:

0000000000000000 <main>:
   0:   48 83 ec 18             sub    $0x18,%rsp
   4:   48 8d 3d 00 00 00 00    lea    0x0(%rip),%rdi        # b <main+0xb>
   b:   31 c9                   xor    %ecx,%ecx
   d:   48 8d 54 24 04          lea    0x4(%rsp),%rdx
  12:   64 48 8b 04 25 28 00    mov    %fs:0x28,%rax
  19:   00 00
  1b:   48 89 44 24 08          mov    %rax,0x8(%rsp)
  20:   31 c0                   xor    %eax,%eax
  22:   48 89 d6                mov    %rdx,%rsi
  25:   e8 00 00 00 00          callq  2a <main+0x2a>
  2a:   48 8b 4c 24 08          mov    0x8(%rsp),%rcx
  2f:   64 48 33 0c 25 28 00    xor    %fs:0x28,%rcx
  36:   00 00
  38:   75 07                   jne    41 <main+0x41>
  3a:   31 c0                   xor    %eax,%eax
  3c:   48 83 c4 18             add    $0x18,%rsp
  40:   c3                      retq
  41:   e8 00 00 00 00          callq  46 <main+0x46>

Xoring oneself gives zero.