I have an archive encoded with gzip 1.5. I'm unable to decode it using the C zlib library. zlib inflate() return EC -3 stream.msg = "unknown compression method".
$ gzip --list --verbose vmlinux.z
method crc date time compressed uncompressed ratio uncompressed_name
defla 12169518 Apr 29 13:00 4261643 9199404 53.7% vmlinux
The first 32 bytes of the file are:
00000000 1f 8b 08 08 29 f4 8a 60 00 03 76 6d 6c 69 6e 75 |....)..`..vmlinu|
00000010 78 00 ec 9a 7f 54 1c 55 96 c7 6f 75 37 d0 fc 70 |x....T.U..ou7..p|
I see the first 18 bytes are the RFC-1952 gzip header.
After the NULL, I expect the next byte to be RFC-1951 deflate or RFC-1950 zlib (I'm not sure which)
So, I pass zlib inflate() a z_stream:next_in pointing to to the byte #0x12.
If this were deflate encoded, then I would expect the next byte #0x12 to be 0aabbbbb (BFINAL=0 and BTYPE=some compression)
If this were zlib encoded, I would expect the next byte #0x12 to take the form 0aaa1000 bbbccccc
Instead, I see #0x12 EC = 1110 1100 Which fits neither of those.
For my code, I took the uncompress() code and modified it slightly with allocators appropriate to my environment and several different experiments with the window bits (including 15+16, -MAX_WBITS, and MAX_WBITS).
int ZEXPORT unzip (dest, destLen, source, sourceLen)
Bytef *dest;
uLongf *destLen;
const Bytef *source;
uLong sourceLen;
{
z_stream stream;
int err;
stream.next_in = (Bytef*)source;
stream.avail_in = (uInt)sourceLen;
/* Check for source > 64K on 16-bit machine: */
if ((uLong)stream.avail_in != sourceLen) return Z_BUF_ERROR;
stream.next_out = dest;
stream.avail_out = (uInt)*destLen;
if ((uLong)stream.avail_out != *destLen) return Z_BUF_ERROR;
stream.zalloc = (alloc_func)my_alloc;
stream.zfree = (free_func)my_free;
/*err = inflateInit(&stream);*/
err = inflateInit2(&stream, 15 + 16);
if (err != Z_OK) return err;
err = inflate(&stream, Z_FINISH);
if (err != Z_STREAM_END) {
inflateEnd(&stream);
return err == Z_OK ? Z_BUF_ERROR : err;
}
*destLen = stream.total_out;
err = inflateEnd(&stream);
return err;
}
How can I correct my decoding of this file?
That should work fine, assuming that my_alloc and my_free do what they need to do. You should verify that you are actually giving unzip() the data that you think you are giving it. The data you give it needs to start with the 1f 8b.
(Side comment: "unzip" is a lousy name for the function. It does not unzip, since zip is an entirely different format than either gzip or zlib. "gunzip" or "ungzip" would be appropriate.)
You are manually reading the bits in the deflate stream in the wrong order. The least significant bits are first. The low three bits of ec are 100, indicating a non-last dynamic block. 0 for non-last, then 10 for dynamic.
You can use infgen to disassemble a deflate stream. Its output for the 14 bytes provided is this initial portion of a dynamic block:
dynamic
count 286 27 16
code 0 5
code 2 7
code 3 7
code 4 5
code 5 5
code 6 4
code 7 4
code 8 2
code 9 3
code 10 2
code 11 4
code 12 4
code 16 7
code 17 7
lens 4 6 7 7 7 8 8 8 7 8
repeat 3
lens 10
Suppose I define a lazy, infinite array using a triangular reduction at the REPL, with a single element pasted onto the front:
> my #s = 0, |[\+] (1, 2 ... *)
[...]
I can print out the first few elements:
> #s[^10]
(0 1 3 6 10 15 21 28 36 45)
I'd like to move the zero element inside the reduction like so:
> my #s = [\+] (0, |(1, 2 ... *))
However, in response to this, the REPL hangs, presumably by trying to evaluate the infinite list.
If I do it in separate steps, it works:
> my #s = 0, |(1, 2 ... *)
[...]
> ([\+] #s)[^10]
(0 1 3 6 10 15 21 28 36 45)
Why doesn't the way that doesn't work...work?
Short answer:
It is probably a bug.
Long answer:
(1, 2 ... *) produces a lazy sequence because it is obviously infinite, but somehow that is not making the resulting sequence from being marked as lazy.
Putting a sequence into an array #s causes it to be eagerly evaluated unless it is marked as being lazy.
Quick fix:
Append lazy to the front.
> my #s = [\+] lazy 0, |(1, 2 ... *)
[...]
> #s[^10]
(0 1 3 6 10 15 21 28 36 45)
This is an assignment and I have been asked to implement a Semaphore in Ada as the description below.
However I have implemented the Semaphore.adb and called this Semaphore in the producerconsumer_sem.adb which I created.
I get some output which is the following.
I'm not sure if my initialization of semaphore is correct S: CountingSemaphore(1,1);.
I don't know where I call the S.wait and S.Signal now i randomly called the S.wait before Producer put item in the buffer X := I; and the S.Signal after the X := I;.
Is this the correct way?
Producer-Consumer Problem
The program producerconsumer.adb implements a non-reliable implemen-
tation of the producer-consumer problem, where data is likely be lost. In
the following, you will use three different communication mechanisms to
achieve a reliable implementation of the producer-consumer problem.
Semaphore
The Ada language does not directly provide library functions for a semaphore.
However, semaphores can be implemented by means of a protected object. Create a package specification Semaphore in the file Semaphores.ads
and the corresponding package body in the file Semaphores.adb that
implements a counting semaphore. Skeletons for the package are available on the course page.
Use the semaphore package for a reliable implementation of the producer-
consumer problem. Modify the file producerconsumer.adb and save the
final code as producerconsumer_sem.adb. In order to use the semaphore
package it shall be installed in the same directory as producerconsumer_sem.adb.
It can then be accessed by
with Semaphores;
use Semaphores;
The Output:
OutPut:
1
1
1
2
2
3
4
4
5
6
6
7
7
8
9
9
9
10
11
11
11
12
12
13
13
13
14
15
15
16
16
17
18
18
18
19
20
20
21
21
22
22
23
24
24
24
24
25
25
26
27
27
28
29
29
30
30
31
31
32
32
33
33
33
34
35
35
35
36
36
37
37
37
38
38
38
39
40
40
40
The package
package Semaphores is
protected type CountingSemaphore(Max: Natural; Initial: Natural) is
entry Wait;
entry Signal;
private
Count : Natural := Initial;
MaxCount : Natural := Max;
end CountingSemaphore;
end Semaphores;
The Semaphore I implemented semaphores.adb.
package body Semaphores is
protected body CountingSemaphore is
entry Wait when Count > 0 is
begin
Count := Count - 1;
end Wait;
entry Signal when Count < MaxCount is
begin
Count := Count + 1;
end Signal;
end CountingSemaphore;
end Semaphores;
The producerconsumer_sem.adb
with Ada.Text_IO;
use Ada.Text_IO;
with Ada.Real_Time;
use Ada.Real_Time;
with Ada.Numerics.Discrete_Random;
with Semaphores;
use Semaphores;
procedure ProducerConsumer_sem is
X : Integer; -- Shared Variable
N : constant Integer := 40; -- Number of produced and comsumed variables
S: CountingSemaphore(1,1);
--S1: CountingSemaphore(1,1);
pragma Volatile(X); -- For a volatile object all reads and updates of
-- the object as a whole are performed directly
-- to memory (Ada Reference Manual, C.6)
--Random Delays
subtype Delay_Interval is Integer range 50..250;
package Random_Delay is new Ada.Numerics.Discrete_Random
(Delay_Interval);
use Random_Delay;
G : Generator;
task Producer;
task Consumer;
task body Producer is
Next : Time;
begin
Next := Clock;
for I in 1..N loop
-- Write to X
S.Wait;
X := I;
S.Signal;
--Next 'Release' in 50..250ms
Next := Next + Milliseconds(Random(G));
Put_Line(Integer'Image(X));
delay until Next;
end loop;
end;
task body Consumer is
Next : Time;
begin
Next := Clock;
for I in 1..N loop
-- Read from X
S.Wait;
Put_Line(Integer'Image(X));
S.Signal;
Next := Next + Milliseconds(Random(G));
delay until Next;
end loop;
end;
begin -- main task
null;
end ProducerConsumer_sem;
On macOS, with FSF GCC 7.1.0 and GNAT GPL 2017, I changed your Put_Lines to Puts and got pretty-much the answer you state in the question.
The question says to create Semaphore.ads, .adb. This will work on Windows, and may work on macOS, but won’t work on Linux, because of GNAT’s file naming convention (see the end of this; it’s a good idea to get into the habit of using lower-case file names).
If you want to ensure that only one task has access to X at a time, I don’t think there’s much wrong with your Wait, Signal calls, though when I put a delay 0.1 at the beginning of Producer, the first value output was 151619216 (because X isn’t initialized). However! if the point is to communicate one update to X at a time (as implied by the names producer/consumer), you should
initialize the semaphore with a count of 0 (and a max of 1). This makes it a binary semaphore.
in Consumer, Wait only (i.e. remove the Signal)
in Producer, Signal only (i.e. remove the Wait). Also, remove the Put to avoid confusion!
I am trying to get used to iSpin/Promela. I am using:
Spin Version 6.4.3 -- 16 December 2014,
iSpin Version 1.1.4 -- 27 November 2014,
TclTk Version 8.6/8.6,
Windows 8.1.
Here is an example where I try to use LTL. The verification of the LTL property should produce an error if the two steps in the for loop are non-atomic:
1 #define ten ((n !=10) && (finished == 2))
2
3 int n = 0;
4 int finished = 0;
5 active [2] proctype P() {
6 //assert(_pid == 0 || _pid == 1);
7
8 int t = 0;
9 byte j;
10 for (j : 1 .. 5) {
11 atomic {
12 t = n;
13 n = t+1;
14 }
15 }
16 finished = finished+1;
17 }
18
19 ltl alwaysten {[] ! ten }
In the verification tap I just want to test the LTL property, so I disable all safety properties and activate "use claim". The claim name is "alwaysten".
But it seems that the LTL property is just evaluated if I activate "assertion violations". Why? A collegue is using iSpin v1.1.0 and he does not need to activate this? What am I doing wrong? I want to prove assertions and LTL properties independently...
Here is the trace:
pan: elapsed time 0.002 seconds
To replay the error-trail, goto Simulate/Replay and select "Run"
spin -a 1_2_ConcurrentCounters_8.pml
ltl alwaysten: [] (! (((n!=10)) && ((finished==2))))
C:/cygwin/bin/gcc -DMEMLIM=1024 -O2 -DXUSAFE -w -o pan pan.c
./pan -m10000 -E -a -N alwaysten
Pid: 6980
warning: only one claim defined, -N ignored
(Spin Version 6.4.3 -- 16 December 2014)
+ Partial Order Reduction
Full statespace search for:
never claim + (alwaysten)
assertion violations + (if within scope of claim)
acceptance cycles + (fairness disabled)
invalid end states - (disabled by -E flag)
State-vector 36 byte, depth reached 57, errors: 0
475 states, stored
162 states, matched
637 transitions (= stored+matched)
0 atomic steps
hash conflicts: 0 (resolved)
Stats on memory usage (in Megabytes):
0.024 equivalent memory usage for states (stored*(State-vector + overhead))
0.291 actual memory usage for states
64.000 memory used for hash table (-w24)
0.343 memory used for DFS stack (-m10000)
64.539 total actual memory usage
unreached in proctype P
(0 of 13 states)
unreached in claim alwaysten
_spin_nvr.tmp:8, state 10, "-end-"
(1 of 10 states)
pan: elapsed time 0.001 seconds
No errors found -- did you verify all claims?
This is because your LTL is translated into a claim with an assert statement. See the following automaton.
So, without checking for assertion violations, no error can be found.
(A possible explanation of different behaviors: previous versions of Spin might translate this differently, perhaps using accept instead of assert.)
Given a NxN matrix and a (row,column) position, what is a method to select a different position in a random (or pseudo-random) order, trying to avoid collisions as much as possible?
For example: consider a 5x5 matrix and start from (1,2)
0 0 0 0 0
0 0 X 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
I'm looking for a method like
(x,y) hash (x,y);
to jump to a different position in the matrix, avoiding collisions as much as possible
(do not care how to return two different values, it doesn't matter, just think of an array).
Of course, I can simply use
row = rand()%N;
column = rand()%N;
but it's not that good to avoid collisions.
I thought I could apply twice a simple hash method for both row and column and use the results as new coordinates, but I'm not sure this is a good solution.
Any ideas?
Can you determine the order of the walk before you start iterating? If your matrices are large, this approach isn't space-efficient, but it is straightforward and collision-free. I would do something like:
Generate an array of all of the coordinates. Remove the starting position from the list.
Shuffle the list (there's sample code for a Fisher-Yates shuffle here)
Use the shuffled list for your walk order.
Edit 2 & 3: A modular approach: Given s array elements, choose a prime p of form 2+3*n, p>s. For i=1 to p, use cells (iii)%p when that value is in range 1...s-1. (For row-length r, cell #c subscripts are c%r, c/r.)
Effectively, this method uses H(i) = (iii) mod p as a hash function. The reference shows that as i ranges from 1 to p, H(i) takes on each of the values from 0 to p-1, exactly one time each.
For example, with s=25 and p=29 or 47, this uses cells in following order:
p=29: 1 8 6 9 13 24 19 4 14 17 22 18 11 7 12 3 15 10 5 16 20 23 2 21 0
p=47: 1 8 17 14 24 13 15 18 7 4 10 2 6 21 3 22 9 12 11 23 5 19 16 20 0
according to bc code like
s=25;p=29;for(i=1;i<=p;++i){t=(i^3)%p; if(t<s){print " ",t}}
The text above shows the suggestion I made in Edit 2 of my answer. The text below shows my first answer.
Edit 0: (This is the suggestion to which Seamus's comment applied): A simple method to go through a vector in a "random appearing" way is to repeatedly add d (d>1) to an index. This will access all elements if d and s are coprime (where s=vector length). Note, my example below is in terms of a vector; you could do the same thing independently on the other axis of your matrix, with a different delta for it, except a problem mentioned below would occur. Note, "coprime" means that gcd(d,s)=1. If s is variable, you'd need gcd() code.
Example: Say s is 10. gcd(s,x) is 1 for x in {1,3,7,9} and is not 1 for x in {2,4,5,6,8,10}. Suppose we choose d=7, and start with i=0. i will take on values 0, 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, which modulo 10 is 0, 7, 4, 1, 8, 5, 2, 9, 6, 3, 0.
Edit 1 & 3: Unfortunately this will have a problem in the two-axis case; for example, if you use d=7 for x axis, and e=3 for y-axis, while the first 21 hits will be distinct, it will then continue repeating the same 21 hits. To address this, treat the whole matrix as a vector, use d with gcd(d,s)=1, and convert cell numbers to subscripts as above.
If you just want to iterate through the matrix, what is wrong with row++; if (row == N) {row = 0; column++}?
If you iterate through the row and the column independently, and each cycles back to the beginning after N steps, then the (row, column) pair will interate through only N of the N^2 cells of the matrix.
If you want to iterate through all of the cells of the matrix in pseudo-random order, you could look at questions here on random permutations.
This is a companion answer to address a question about my previous answer: How to find an appropriate prime p >= s (where s = the number of matrix elements) to use in the hash function H(i) = (i*i*i) mod p.
We need to find a prime of form 3n+2, where n is any odd integer such that 3*n+2 >= s. Note that n odd gives 3n+2 = 3(2k+1)+2 = 6k+5 where k need not be odd. In the example code below, p = 5+6*(s/6); initializes p to be a number of form 6k+5, and p += 6; maintains p in this form.
The code below shows that half-a-dozen lines of code are enough for the calculation. Timings are shown after the code, which is reasonably fast: 12 us at s=half a million, 200 us at s=half a billion, where us denotes microseconds.
// timing how long to find primes of form 2+3*n by division
// jiw 20 Sep 2011
#include <stdlib.h>
#include <stdio.h>
#include <sys/time.h>
double ttime(double base) {
struct timeval tod;
gettimeofday(&tod, NULL);
return tod.tv_sec + tod.tv_usec/1e6 - base;
}
int main(int argc, char *argv[]) {
int d, s, p, par=0;
double t0=ttime(0);
++par; s=5000; if (argc > par) s = atoi(argv[par]);
p = 5+6*(s/6);
while (1) {
for (d=3; d*d<p; d+=2)
if (p%d==0) break;
if (d*d >= p) break;
p += 6;
}
printf ("p = %d after %.6f seconds\n", p, ttime(t0));
return 0;
}
Timing results on 2.5GHz Athlon 5200+:
qili ~/px > for i in 0 00 000 0000 00000 000000; do ./divide-timing 500$i; done
p = 5003 after 0.000008 seconds
p = 50021 after 0.000010 seconds
p = 500009 after 0.000012 seconds
p = 5000081 after 0.000031 seconds
p = 50000021 after 0.000072 seconds
p = 500000003 after 0.000200 seconds
qili ~/px > factor 5003 50021 500009 5000081 50000021 500000003
5003: 5003
50021: 50021
500009: 500009
5000081: 5000081
50000021: 50000021
500000003: 500000003
Update 1 Of course, timing is not determinate (ie, can vary substantially depending on the value of s, other processes on machine, etc); for example:
qili ~/px > time for i in 000 004 010 058 070 094 100 118 184; do ./divide-timing 500000$i; done
p = 500000003 after 0.000201 seconds
p = 500000009 after 0.000201 seconds
p = 500000057 after 0.000235 seconds
p = 500000069 after 0.000394 seconds
p = 500000093 after 0.000200 seconds
p = 500000099 after 0.000201 seconds
p = 500000117 after 0.000201 seconds
p = 500000183 after 0.000211 seconds
p = 500000201 after 0.000223 seconds
real 0m0.011s
user 0m0.002s
sys 0m0.004s
Consider using a double hash function to get a better distribution inside the matrix,
but given that you cannot avoid colisions, what I suggest is to use an array of sentinels
and mark the positions you visit, this way you are sure you get to visit a cell once.