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Elite on the BBC Micro

Text: TT26 [Electron version]

Name: TT26 [Show more] Type: Subroutine Category: Text Summary: Print a character at the text cursor by poking into screen memory Deep dive: Drawing text
Context: See this subroutine in context in the source code Variations: See code variations for this subroutine in the different versions References: This subroutine is called as follows: * BPRNT calls TT26 * S% (Part 1 of 2) calls TT26 * TT160 calls TT26 * TT161 calls TT26 * TT16a calls TT26 * TT214 calls TT26 * TT25 calls TT26 * TT42 calls TT26 * TT74 calls TT26 * cmn calls TT26 * gnum calls TT26 * DIALS (Part 3 of 4) calls entry point rT9 * BULB calls entry point RREN

Print a character at the text cursor (XC, YC), do a beep, print a newline, or delete left (backspace). WRCHV is set to point here by the loading process. Arguments: A The character to be printed. Can be one of the following: * 7 (beep) * 10-13 (line feeds and carriage returns) * 32-95 (ASCII capital letters, numbers and punctuation) * 127 (delete the character to the left of the text cursor and move the cursor to the left) XC Contains the text column to print at (the x-coordinate) YC Contains the line number to print on (the y-coordinate) Returns: A A is preserved X X is preserved Y Y is preserved C flag The C flag is cleared Other entry points: RR3+1 Contains an RTS RREN Prints the character definition pointed to by P(2 1) at the screen address pointed to by (A SC). Used by the BULB routine rT9 Contains an RTS
.TT26 STA K3 \ Store the A, X and Y registers, so we can restore STY YSAV2 \ them at the end (so they don't get changed by this STX XSAV2 \ routine) LDY QQ17 \ Load the QQ17 flag, which contains the text printing \ flags CPY #255 \ If QQ17 = 255 then printing is disabled, so jump to BEQ RR4 \ RR4, which doesn't print anything, it just restores \ the registers and returns from the subroutine CMP #7 \ If this is a beep character (A = 7), jump to R5, BEQ R5 \ which will emit the beep, restore the registers and \ return from the subroutine CMP #32 \ If this is an ASCII character (A >= 32), jump to RR1 BCS RR1 \ below, which will print the character, restore the \ registers and return from the subroutine CMP #10 \ If this is control code 10 (line feed) then jump to BEQ RRX1 \ RRX1, which will move down a line, restore the \ registers and return from the subroutine LDX #1 \ If we get here, then this is control code 11-13, of STX XC \ which only 13 is used. This code prints a newline, \ which we can achieve by moving the text cursor \ to the start of the line (carriage return) and down \ one line (line feed). These two lines do the first \ bit by setting XC = 1, and we then fall through into \ the line feed routine that's used by control code 10 .RRX1 INC YC \ Print a line feed, simply by incrementing the row \ number (y-coordinate) of the text cursor, which is \ stored in YC BNE RR4 \ Jump to RR4 to restore the registers and return from \ the subroutine (this BNE is effectively a JMP as Y \ will never be zero) .RR1 \ If we get here, then the character to print is an \ ASCII character in the range 32-95. The quickest way \ to display text on-screen is to poke the character \ pixel by pixel, directly into screen memory, so \ that's what the rest of this routine does \ \ The first step, then, is to get hold of the bitmap \ definition for the character we want to draw on the \ screen (i.e. we need the pixel shape of this \ character). The MOS ROM contains bitmap definitions \ of the system's ASCII characters, starting from &C000 \ for space (ASCII 32) and ending with the £ symbol \ (ASCII 126) \ \ There are definitions for 32 characters in each of the \ three pages of MOS memory, as each definition takes up \ 8 bytes (8 rows of 8 pixels) and 32 * 8 = 256 bytes = \ 1 page. So: \ \ ASCII 32-63 are defined in &C000-&C0FF (page 0) \ ASCII 64-95 are defined in &C100-&C1FF (page 1) \ ASCII 96-126 are defined in &C200-&C2F0 (page 2) \ \ The following code reads the relevant character \ bitmap from the above locations in ROM and pokes \ those values into the correct position in screen \ memory, thus printing the character on-screen \ \ It's a long way from 10 PRINT "Hello world!":GOTO 10 TAY \ Copy the character number from A to Y, as we are \ about to pull A apart to work out where this \ character definition lives in memory \ Now we want to set X to point to the relevant page \ number for this character - i.e. &C0, &C1 or &C2. \ The following logic is easier to follow if we look \ at the three character number ranges in binary: \ \ Bit # 76543210 \ \ 32 = %00100000 Page 0 of bitmap definitions \ 63 = %00111111 \ \ 64 = %01000000 Page 1 of bitmap definitions \ 95 = %01011111 \ \ 96 = %01100000 Page 2 of bitmap definitions \ 125 = %01111101 \ \ We'll refer to this below LDX #&BF \ Set X to point to the first font page in ROM minus 1, \ which is &C0 - 1, or &BF ASL A \ If bit 6 of the character is clear (A is 32-63) ASL A \ then skip the following instruction BCC P%+4 LDX #&C1 \ A is 64-126, so set X to point to page &C1 ASL A \ If bit 5 of the character is clear (A is 64-95) BCC P%+3 \ then skip the following instruction INX \ Increment X \ \ By this point, we started with X = &BF, and then \ we did the following: \ \ If A = 32-63: skip then INX so X = &C0 \ If A = 64-95: X = &C1 then skip so X = &C1 \ If A = 96-126: X = &C1 then INX so X = &C2 \ \ In other words, X points to the relevant page. But \ what about the value of A? That gets shifted to the \ left three times during the above code, which \ multiplies the number by 8 but also drops bits 7, 6 \ and 5 in the process. Look at the above binary \ figures and you can see that if we cleared bits 5-7, \ then that would change 32-53 to 0-31... but it would \ do exactly the same to 64-95 and 96-125. And because \ we also multiply this figure by 8, A now points to \ the start of the character's definition within its \ page (because there are 8 bytes per character \ definition) \ \ Or, to put it another way, X contains the high byte \ (the page) of the address of the definition that we \ want, while A contains the low byte (the offset into \ the page) of the address STA P+1 \ Store the address of this character's definition in STX P+2 \ P(2 1) LDA #128 \ Set SC = 128 for use in the calculation below STA SC LDA YC \ If YC < 24 then we are in the top part of the screen, CMP #24 \ so skip the following two instructions BCC P%+8 JSR TTX66 \ We are off the bottom of the screen, so we don't want \ to print anything, so first clear the screen and draw \ a white border JMP RR4 \ Jump to RR4 to restore the registers and return from \ the subroutine \ The text row is on-screen, so now to calculate the \ screen address we need to write to, as follows: \ \ SC = &5800 + (char row * 256) + (char row * 64) + 32 \ \ See the deep dive on "Drawing pixels in the Electron \ version" for details LSR A \ Set (A SC) = (A SC) / 4 ROR SC \ = (4 * ((char row * 64) + 32)) / 4 LSR A \ = char row * 64 + 32 ROR SC ADC YC \ Set SC(1 0) = (A SC) + (YC 0) + &5800 ADC #&58 \ = (char row * 64 + 32) STA SC+1 \ + char row * 256 \ + &5800 \ \ which is what we want, so SC(1 0) contains the address \ of the first visible pixel on the character row we \ want LDA XC \ Fetch XC, the x-coordinate (column) of the text cursor \ into A ASL A \ Multiply A by 8, and add to SC. As each character is ASL A \ 8 pixels wide, this gives us the screen address of the ASL A \ character block where we want to print this character ADC SC STA SC BCC P%+4 \ If the addition of the low byte overflowed, increment INC SC+1 \ the high byte CPY #127 \ If the character number (which is in Y) <> 127, then BNE RR2 \ skip to RR2 to print that character, otherwise this is \ the delete character, so continue on DEC XC \ We want to delete the character to the left of the \ text cursor and move the cursor back one, so let's \ do that by decrementing YC. Note that this doesn't \ have anything to do with the actual deletion below, \ we're just updating the cursor so it's in the right \ position following the deletion DEC SC+1 \ Decrement the high byte of the screen address to point \ to the address of the current character, minus one \ page LDY #&F8 \ Set Y = &F8, so the following call to ZES2 will count \ Y upwards from &F8 to &FF JSR ZES2 \ Call ZES2, which zero-fills from address SC(1 0) + Y \ to SC(1 0) + &FF. SC(1 0) points to the address of the \ current character, minus one page, and adding &FF to \ this would point to the cursor, so adding &F8 points \ to the character before the cursor, which is the one \ we want to delete. So this call zero-fills the \ character to the left of the cursor, which erases it \ from the screen BEQ RR4 \ We are done deleting, so restore the registers and \ return from the subroutine (this BNE is effectively \ a JMP as ZES2 always returns with the Z flag set) .RR2 \ Now to actually print the character INC XC \ Once we print the character, we want to move the text \ cursor to the right, so we do this by incrementing \ XC. Note that this doesn't have anything to do \ with the actual printing below, we're just updating \ the cursor so it's in the right position following \ the print EQUB &2C \ Skip the next instruction by turning it into \ &2C &85 &08, or BIT &0885, which does nothing apart \ from affect the flags. We skip the instruction as we \ already set the value of SC+1 above .RR3 \ A contains the value of YC - the screen row where we \ want to print this character - so now we need to \ convert this into a screen address, so we can poke \ the character data to the right place in screen \ memory .RREN STA SC+1 \ Store the page number of the destination screen \ location in SC+1, so SC now points to the full screen \ location where this character should go LDY #7 \ We want to print the 8 bytes of character data to the \ screen (one byte per row), so set up a counter in Y \ to count these bytes .RRL1 LDA (P+1),Y \ The character definition is at P(2 1) - we set this up \ above - so load the Y-th byte from P(2 1), which will \ contain the bitmap for the Y-th row of the character EOR (SC),Y \ If we EOR this value with the existing screen \ contents, then it's reversible (so reprinting the \ same character in the same place will revert the \ screen to what it looked like before we printed \ anything); this means that printing a white pixel on \ onto a white background results in a black pixel, but \ that's a small price to pay for easily erasable text STA (SC),Y \ Store the Y-th byte at the screen address for this \ character location DEY \ Decrement the loop counter BPL RRL1 \ Loop back for the next byte to print to the screen .RR4 LDY YSAV2 \ We're done printing, so restore the values of the LDX XSAV2 \ A, X and Y registers that we saved above and clear LDA K3 \ the C flag, so everything is back to how it was CLC .rT9 RTS \ Return from the subroutine .R5 JSR BEEP \ Call the BEEP subroutine to make a short, high beep JMP RR4 \ Jump to RR4 to restore the registers and return from \ the subroutine using a tail call