peek (was Re: CHDK 1.4 release + bug reports)

Caefix

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peek (was Re: CHDK 1.4 release + bug reports)

« on: 27 / November / 2019, 12:58:58 »

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Troubleshouting is an old bavarian tradition. We call it "Jodeln".

Here I have a simple Lua ask for value[addr]:

Code: [Select]

--[[--
@title peekdemo
@chdk_version 1.4
@param    p peek Address /4*4
 @range   p 4 999996
 @default p 1000
@subtitle :...,....1....,....2....,....3....,....4..
@param    u Logging
 @default u 1
 @range   u 0 1
@param    l Log:
 @default l -23
 --]]--
 print_screen(l)
p=p/4*4 -- avoid nil loop 
print(string.format("%08x at 0x%08x",peek(p,4),p ))
print(string.format("%04x:%04x %08x ",peek(p-2,2),peek(p,2),peek(p,4)))
print(string.format("%02x %02x:%02x %02x at %d",peek(p-3,1),peek(p-2,1),peek(p-1,1),peek(p,1),p))

 print_screen(0)
wait_click(0)

The answer I expect is repeating only 4 char values, but only the peek(addr,1) line fits.
9002a011 at 0x00000490
7684:a011 9002a011
95 84:76 11 at 1168

that´s my current private ubasic factor peek, returning int´s 4 chars in expected order.

Code: [Select]

  } elseif x=6 exec {
	?"Line1---------",x,bug:=% p,=%peek p,4.
? =%peek p-2,2;":";=%peek p,2,=%peek p,4
? =%peek p-3,1,=%peek p-2,1;":";=%peek p-1,1,=%peek p,1
  break

0x0490 0x95847611
0x9584:0x7611 0x95847611
0x95 0x84:0x76 0x11

edit: upper big endian values outdated, peek polarity switched below.

Code: [Select]

  case TOKENIZER_PEEK:
    accept(TOKENIZER_PEEK);
	long x,a=expr(); char s=4;
	if (tokenizer_token()==TOKENIZER_COMMA) {
		accept(TOKENIZER_COMMA); s=expr()%5; }
    if (a & (s-1)) {ended=1; break;}
    if (s==4) r=(unsigned int) *(unsigned int *)(a); else
    if (s==2) r=(unsigned short) *(unsigned short *)(a); else
    if (s==1) r=(unsigned char) *(unsigned char *)(a);
  break;

	/*
	memset(&r, (char)(*(char *)a),1); 
    if (s>1) { x=0;
	memset(&x, (char)(*(char *)(a+1)),1); // for big-endian (a-1) ...
		x<<=8; r+=x;
    if (s>3) {	x=0; 
	memset(&x, (char)(*(char *)(a+2)),1);
		x<<=16; r+=x; x=0;
	memset(&x, (char)(*(char *)(a+3)),1);
		x<<=24; r+=x;	} 
	}
	break; */

« Last Edit: 29 / November / 2019, 15:49:12 by Caefix »

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reyalp

14110

Re: peek (was Re: CHDK 1.4 release + bug reports)

« Reply #1 on: 27 / November / 2019, 13:38:53 »

Code: [Select]

print(string.format("%04x:%04x %08x ",peek(p-2,2),peek(p,2),peek(p,4)))

p-2 is a different address. If you want to look at the same word as the first version, you need to use peek(p,2) and peek(p+2,2), order depending on your endian preferences.

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Caefix

947
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Re: peek (was Re: CHDK 1.4 release + bug reports)

« Reply #2 on: 27 / November / 2019, 14:49:50 »

I expected that line printing 'left short' : 'right short' 'same int'.

edit: int following 95 86:76 11 > e0 02 90 da (p+1,p+2..)

« Last Edit: 27 / November / 2019, 15:03:26 by Caefix »

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reyalp

14110

Re: peek (was Re: CHDK 1.4 release + bug reports)

« Reply #3 on: 27 / November / 2019, 16:20:53 »

Quote from: Caefix on 27 / November / 2019, 14:49:50

I expected that line printing 'left short' : 'right short' 'same int'.

edit: int following 95 86:76 11 > e0 02 90 da (p+1,p+2..)

I don't understand.

If p is 100 in your example, then

Code: [Select]

print(string.format("%04x:%04x %08x ",peek(p-2,2),peek(p,2),peek(p,4)))

prints the halfword starting at 98, the halfword starting at 100, and the word starting at 100
The halfword at 98 is a totally different value, not part of the word at 100.
The halfword at 100 is lower two bytes of the word at 100, which on a little endian system like this will be the least significant values, appearing on the right when you print the whole word.
Example (chdkptp, because I'm lazy)

Code: [Select]

con 1> =p=0xff810000 return string.format('0x%08x',peek(p))
2:return:'0xea000001'
con 3> =p=0xff810000 return string.format('0x%04x0x%04x',peek(p,2),peek(p+2,2))
4:return:'0x0001 0xea00'
con 4> =p=0xff810000 return string.format('0x%02x 0x%02x 0x%02x 0x%02x',peek(p,1),peek(p+1,1),peek(p+2,1),peek(p+3,1))
5:return:'0x01 0x00 0x00 0xea'

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Caefix

947
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Re: peek (was Re: CHDK 1.4 release + bug reports)

« Reply #4 on: 28 / November / 2019, 11:09:29 »

We are talking about int value | eulav tni ( ( (

90.02:a0.11 at 0x00000490
7684:a011 9002a011
95 84:76 11 at 1168
95 86:76 < 11 > e0 02 90 (da) | 90.02:a0.11

peeking for char at 0x00000490 =11 is the axis.
I expect there the lowest char of "my" int, Lua-peek takes it as the first char of a string.
"My" int/short is built reading backwards from 11, the other is reading forward and returns that mirrored.
(btw. it´s a 'funny' address in Sx710, something lives there.)

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reyalp

14110

Re: peek (was Re: CHDK 1.4 release + bug reports)

« Reply #5 on: 28 / November / 2019, 16:58:16 »

I still don't understand what you are trying to say, but I'm confident Lua peek works correctly. It literally just does return *(type *)(address)

That returns the byte/halfword/word starting at the given address in the cameras native little endian byte order. Nothing is "mirrored", that's just how it's stored in memory.

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Caefix

947
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Re: peek (was Re: CHDK 1.4 release + bug reports)

« Reply #6 on: 30 / November / 2019, 11:10:04 »

Thx @vergilius for that journey, gimmy a "

".

Now I know

, that little endian is not a font, and I will not distort the native bytes.

Peek is done.
My con-fusion is gone!

Oops,

?? that´s conf.using....

Where

is it gone?

? ( ( (

Beware of not-endian confusion outside!!

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Caefix

947
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Re: peek (was Re: CHDK 1.4 release + bug reports)

« Reply #7 on: 23 / April / 2022, 16:22:03 »

Code: [Select]

 
function printf(...)
  local str=string.format(...)
  print(str)
return  str
end

printf("0x%08x = %d",0xFE020004,0xFE020004)

Quote

=== CHDK Lua local ====================
Run Script: printf.lua
=======================================

Process started >>>
---------------------------------------
conf = D:\hostluaPortable\bin\local_conf.lua
=== START =============================
0xfe020004 = 4261543940
---------------------------------------
<<< Process finished. (Exit code 0x0)

================ READY ================

I thought to have an equivalent in

Code: [Select]

Ubasic.c: 
	if (x>0xFFFF) {sprintf(string, "0x%08x",(unsigned)x); 

printf.bas:

  ?"'";mem$(0xFE020004,q);"'":0xFE020004,,=%0xFE020004

Should be 'gaonisoy' on Sx270, but it outputs:

Quote

Ð¥Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾Ð¾
2147483647, 0x7fffffff

Code: [Select]

		case TOKENIZER_AND:
             sprintf(string, "?=&:0x%08x=>%s",0xFE020004,
             memstr((void*)0xFE020004,40)); break;

Quote

?=&:0xfe020004=>gaonisoyЯш„Р

... as expected, but how to grep from expr() ?

« Last Edit: 25 / April / 2022, 16:07:10 by Caefix »

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Caefix

947
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Re: peek (was Re: CHDK 1.4 release + bug reports)

« Reply #8 on: 26 / April / 2022, 16:06:36 »

...

atoi(ptr) has limited range, max 0x7fffffff.

Now I call after parsing "0x" hexval=

Code: [Select]

static int hex2u()
{ int c=0;
  register unsigned u=0; 
  while ((*nextptr>'/' && *nextptr<':')
        || strpbrk(nextptr,"abcdefABCDEF")==nextptr) {
    c++; if (c==9) return 0x80000000;
    u<<=4;
    if (*nextptr<64) u|=*nextptr-48;
    else u|=(*nextptr|32)-87;
    nextptr++;
  }
  return u;
}
long  tokenizer_value(void)  {return hexval ? hexval:decval} // !atoi(ptr);

// could ((avoid) atoi(ptr)) with :

Code: [Select]

    decval=0;
    for (i = 0; i <= (MAX_NUMLEN); ++i) {
      if(isdigit(ptr[i]))
        {t=decval<<3; decval<<=1; decval+=t+ptr[i]-48;}
      else { break; }
    }

So in hex_dump.log appears:

Quote

gaonisoyßø„Ð ð(ø JhAð`HI K™B<¿Pø+Aø+

and console prints
?bug,mem$(0xFE020004,q):0xfe020004,,=%0xFE020004:=&:=(0xFE020004).
61 gaonisoy??„?
-33423356, 0xfe020004
&:=(0xFE020004).

&& my confusion is gaon...

« Last Edit: 28 / April / 2022, 08:24:22 by Caefix »

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Caefix

947
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Re: peek (was Re: CHDK 1.4 release + bug reports)

« Reply #9 on: 28 / April / 2022, 12:12:02 »

Summary ...

Quote

static int str2u(const char *s)
{
int t, c=0, com=0, neg=0;
register unsigned u=0;
while (*s && *s<'!') s++;
if (!*s) return 0;
if (*s=='+') {s++;}
if (*s=='~') {s++; com=1;}
if (*s=='-') {s++; neg=1;}
while (*s && *s<'!') s++;
if (*s=='0' && (s[1]|32)=='x') { s+=2;
while (( *s>'/' && *s<':')
|| (((*s|32)>96 && (*s|32)<'g'))) {
if (c==9) break;
u<<=4;
u |= (*s<64) ? (*s-48)*s|32)-87;
c++; s++;}
}else
while (*s>'/' && *s<':') {
if (c==11) break;
t=u<<3; u<<=1;
u += t+*s-48;
c++; s++;}
if (neg) u=-u;
if (com) u=~u;
return u;
}

« Last Edit: 01 / May / 2022, 13:21:31 by Caefix »

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