#255

HOMEBREW Digest Index (part 1)

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Published in 
 · 4 months ago
... D HABERMAN") July 1989 Digest Index ("MR. DAVID HABERMAN") Color computing (Pete Soper) Beer judge exam, Houston, Texas (drutx!homer) HOMEBREW Digest #255 (Sat 16 September 1989) Re: Color computing (Mike Fertsch) Lager ferment temp (M Nevar) Re: Wort Chiller (Bob Clark - Sun Engineering) Re: Homebrew Digest Indices (Greg Wageman) Hunter Energy Monitor (Crawford.WBST129) HOMEBREW Digest #256 (Sun 17 September 1989) restarting fermentation (Dick Dunn) indices homebrew digest indices (Wayne Hamilton) Re: [index for homebrew digest: useful?] (Dr. T. Andrews) HOMEBREW Digest #257 (Mon 18 September 1989) Re: Lager ferment temp (M Nevar) (John D. Polst ...

Reviews

eZine's profile picture
Published in 
 · 10 months ago
... b (25Mhz 68030+68882) ... #425 (yes, a genuine '030-with-MMU!) B5000-40/4Mb (40Mhz 68EC030) ....... #850 50Mhz 68882 for B5000-40 ........... #255 Extra 4Mb of RAM for B5000-40 ...... #212 (far cheaper elsewhere) CONTACT Solid State Leisure Ltd, 80 Finedon Road, Irthlingborough, Northants NN9 5TZ, England. Tel +44 933 650677 If you have any questions, feel free to e-mail me on the net. SSL are on the point of releasing some new products (including an '040 board), so my price list above is probably incomplete. Barry McConnell bmccnnll@unix1.tcd.ie KCS Power PC Board v4.3 IBM Emulator + 512Kb RAM expansion (undercarriage slot on A500 or ...

The midnight raid issue 8

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Published in 
 · 1 year ago
... dec). Finally a long is made up of 8 digits and is between 00 00 00 00 and FF FF FF FF (0 to 4294967295 dec). With other words: byte: $00 - $FF #0 - #255 word: $00 00 - $FF FF #0 - #65535 long: $00 00 00 00 - $FF FF FF FF #0 - #4294967295 As you can see a byte takes up one fourth of the memory a long does. This principle will be discussed further in the chapter dealing with HexEditors. I think it might be a good thing for you to learn how many digits a byte, a word and a long has. I will use these expressions later on. I chose to use these expressions (and not including floats and doubles) because I feel that even an experienced person can get f ...

Issue # 27 DTACK GROUNDED Newsletter - January 1984

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Published in 
 · 1 year ago
... ode. Consider the following hypothetical program line: 100 MAT SEARCH A$()=HEX(A900A2FF) TO ADR$(),N You may recognize the hex string as "LDA #0, LDX #255". That program line will search array A$() for all occurences of the four-byte substring, and return the addresses [with respect to the first byte of A$()] of every match to array ADR$(), plus the total number of matches to N. If the number of matches is larger than the number of elements in ADR$(), this will be identified in N. A question: if you program in BASIC, how have you managed to get along without such capability up to now ? Page 19, Column 2 (The preceding example is not quite identic ...

Issue # 25 DTACK GROUNDED Newsletter - November 1983

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 · 1 year ago
Issue # 25 DTACK GROUNDED Newsletter - November 1983
... 88 has 5 (five) dedicated interrupt pointers (#0 - #4), 27 reserved interrupt pointers (#5 - #31), and 224 (user) available interrupt pointers (#32 - #255). Page 2-4 of the IBM manual reveals that MS-DOS uses 25 of those 27 reserved interrupt vectors. Specifically, MS-DOS uses #5 and #8 thru #31 inclusive. Now turn to page 5-6 of the Intel manual. Table 5-2 reveals that the 80186/8 uses #5 as a vector for the array bounds exception, and also vectors #6 thru 15 and #18 and #19. Which means that MS-DOS programs will NOT, in general, run on the 80186/8. We are told that the better the program, the more likely that those MS- DOS vectors are b ...

Issue # 14 DTACK GROUNDED Newsletter - October 1982

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Published in 
 · 1 year ago
Issue # 14 DTACK GROUNDED Newsletter - October 1982
... m number from 1 to some limit such as 16. We call this 'N'. Then the host calculates N successive random values, each value ranging from 0 to #255. Thus, we have a string of random length composed of random values. We send the limit of memory (determined by the operator once when starting the program), N, and the random string to the 68000. Page 14, Column 2 The 68000 copies the string repeatedly end-to-end in the remaining available memory until the limit of memory is reached. Then comes a second pass during which those copies are compared back against the original string. In general, there will be no errors, so the 68000 reports the num ...

Issue # 13 DTACK GROUNDED Newsletter - September 1982

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Published in 
 · 1 year ago
Issue # 13 DTACK GROUNDED Newsletter - September 1982
... DC.L $EE6B2800 328 ; 329 ; PRINT A STRING FROM MEMORY 330 ; 00234E 3A7C 1932 331 PRSTRM MOVE.W #STR,A5 ;POINT A5 AT STR 332 ; 002352 7E FF 333 MOVEQ #255,D7 002354 61 08 334 BSR OUTCHR ;XMIT PRINT COMMAND 335 ; 336 ; HERE IS THE MAIN LOOP TO OUTPUT THE STRING 337 ; 002356 1E1D 338 OUTSTR MOVE.B (A5)+,D7 ;GET CHAR 002358 67 0A 339 BEQ OUTCH ;END IF NULL 340 ; 00235A 61 08 341 BSR OUTCH ;OUTPUT THE CHAR 00235C 60 F8 342 BRA OUTSTR ;CONTINUE 343 ; 344 ; SUBROUTINE; OUTPUT A CHAR TO HOST 345 ; 00235E 7C 06 346 OUTCHR MOVEQ #6,D6 ;SET UP D6 AND A0 002360 307C 0FFA 347 MOVE.W #STATUS,A0 ;FOR OUTPUT 348 ; 002364 0D10 349 OUTCH BTST D6,(A0) ;HOST READY? ...

BIT Manipulation

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Published in 
 · 1 year ago
... t around this and keep the other bits exactly the same but also changing bit 4 to equal 1 (Just assume all BITS in 532565 are 0): 10 LDA 53265 20 AND #255 30 ORA #16 40 STA 53265 Ok, firstly we get the value from 53265, we then AND this value with 255. The result of the AND is OR'ed with 16 to give the final result which is stored back at location 53265. 10 - Value 0000 0000 (We said assume that all bits in 53265 = 0) 20 - AND 1111 1111 The Result = 0000 0000 30 - ORA 0001 0000 The Result = 0001 0000 40 Store this back at 53265 As you can see bit 4 is a 1, but all the others are kept at 0. OK Here's a very useful tip for you. To invert a ...

More on using Sprites

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Published in 
 · 1 year ago
... own movement we will produce a bounce effect. Lets get straight into the coding: 10 LDA #147 20 JSR $FFD2 30 LDA #00 40 STA 53280 50 STA 53281 60 LDA #255 70 STA 53269 80 STA 53276 90 LDA #02 100 STA 53285 110 LDA #03 120 STA 53286 130 LDX #00 140 COLOUR LDA #04 150 STA 53287,X 160 INX 170 CPX #08 180 BNE COLOUR 190 LDA #64 200 LDX #00 210 BACK STA 2040,X 220 INX 230 CPX #08 240 BEQ DONE 250 LDA 2040,X 260 TAY 270 INY 280 TYA 290 JMP BACK 300 DONE CLC 310 LDA #30 320 LDX #00 330 XPOS STA 53248,X 340 ADC #30 350 INX 360 INX 370 CPX #16 380 BNE XPOS 390 MOVE LDX #00 400 DATA LDA TABLE,X 410 CMP #255 420 BNE NOTFIN 430 JMP MOVE 440 LDY #00 450 YPOS ...

Using Sprites

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Published in 
 · 1 year ago
... up X & Y position of Sprite Move it Down Move it Up So lets write the code: 10 LDA #147 20 JSR $FFD2 30 LDA #00 40 STA 53280 50 STA 53281 60 LDA #255 70 STA 53269 80 STA 53276 90 LDA #02 100 STA 53285 110 LDA #03 120 STA 53286 130 LDX #00 140 COLOUR LDA #04 150 STA 53287,X 160 INX 170 CPX #08 180 BNE COLOUR 190 LDX #64 200 STX 2040 210 INX 220 STX 2041 230 INX 240 STX 2042 250 INX 260 STX 2043 270 INX 280 STX 2044 290 INX 300 STX 2045 310 INX 320 STX 2046 330 INX 340 STX 2047 350 LDY # 70 360 STY 53249 370 STY 53251 380 STY 53253 390 STY 53255 400 STY 53257 410 STY 53259 420 STY 53261 430 STY 53263 440 LDX #30 450 STX 53248 460 LDX #60 470 S ...
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