This code appears in the following versions (click to see it in the source code):
Code variations between these versions are shown below.
Name: TT26 Type: Subroutine Category: Text Summary: Print a character at the text cursor, with support for verified text in extended tokens Deep dive: Extended text tokens
Arguments: A The character to print
Returns: X X is preserved C flag The C flag is cleared
Other entry points: DASC DASC does exactly the same as TT26 and prints a character at the text cursor, with support for verified text in extended tokens
rT9 Contains an RTS
.DASC .TT26 STX SC \ Store X in SC, so we can retrieve it below LDX #%11111111 \ Set DTW8 = %11111111, to disable the effect of {19} if STX DTW8 \ it was set (as {19} capitalises one character only) CMP #'.' \ If the character in A is a word terminator: BEQ DA8 \ CMP #':' \ * Full stop BEQ DA8 \ * Colon CMP #10 \ * Line feed BEQ DA8 \ * Carriage return CMP #12 \ * Space BEQ DA8 \ CMP #' ' \ then skip the following instruction BEQ DA8 INX \ Increment X to 0, so DTW2 gets set to %00000000 below .DA8 STX DTW2 \ Store X in DTW2, so DTW2 is now: \ \ * %00000000 if this character is a word terminator \ \ * %11111111 if it isn't \ \ so DTW2 indicates whether or not we are currently \ printing a word LDX SC \ Retrieve the original value of X from SC BIT DTW4 \ If bit 7 of DTW4 is set then we are currently printing BMI P%+5 \ justified text, so skip the next instruction JMP CHPR \ Bit 7 of DTW4 is clear, so jump down to CHPR to print \ this character, as we are not printing justified text \ If we get here then we are printing justified text, so \ we need to buffer the text until we reach the end of \ the paragraph, so we can then pad it out with spaces
The enhanced versions use this routine for in-flight text, where carriage returns are buffered.
This variation is blank in the Disc (docked) version.
Tap on a block to expand it, and tap it again to revert.
CMP #12 \ If the character in A is a carriage return, then we BEQ DA1 \ have reached the end of the paragraph, so jump down to \ DA1 to print out the contents of the buffer, \ justifying it as we go \ If we get here then we need to buffer this character \ in the line buffer at BUF LDX DTW5 \ DTW5 contains the current size of the buffer, so this STA BUF,X \ stores the character in A at BUF + DTW5, the next free \ space in the buffer LDX SC \ Retrieve the original value of X from SC so we can \ preserve it through this subroutine call INC DTW5 \ Increment the size of the BUF buffer that is stored in \ DTW5 CLC \ Clear the C flag RTS \ Return from the subroutine .DA1 \ If we get here then we are justifying text and we have \ reached the end of the paragraph, so we need to print \ out the contents of the buffer, justifying it as we go TXA \ Store X and Y on the stack PHA TYA PHA .DA5 LDX DTW5 \ Set X = DTW5, which contains the size of the buffer BEQ DA6+3 \ If X = 0 then the buffer is empty, so jump down to \ DA6+3 to print a newline CPX #(LL+1) \ If X < LL+1, i.e. X <= LL, then the buffer contains BCC DA6 \ fewer than LL characters, which is less than a line \ length, so jump down to DA6 to print the contents of \ BUF followed by a newline, as we don't justify the \ last line of the paragraph \ Otherwise X > LL, so the buffer does not fit into one \ line, and we therefore need to justify the text, which \ we do one line at a time LSR SC+1 \ Shift SC+1 to the right, which clears bit 7 of SC+1, \ so we pass through the following comparison on the \ first iteration of the loop and set SC+1 to %01000000 .DA11 LDA SC+1 \ If bit 7 of SC+1 is set, skip the following two BMI P%+6 \ instructions LDA #%01000000 \ Set SC+1 = %01000000 STA SC+1 LDY #(LL-1) \ Set Y = line length, so we can loop backwards from the \ end of the first line in the buffer using Y as the \ loop counter .DAL1 LDA BUF+LL \ If the LL-th byte in BUF is a space, jump down to DA2 CMP #' ' \ to print out the first line from the buffer, as it BEQ DA2 \ fits the line width exactly (i.e. it's justified) \ We now want to find the last space character in the \ first line in the buffer, so we loop through the line \ using Y as a counter .DAL2 DEY \ Decrement the loop counter in Y BMI DA11 \ If Y <= 0, loop back to DA11, as we have now looped BEQ DA11 \ through the whole line LDA BUF,Y \ If the Y-th byte in BUF is not a space, loop back up CMP #' ' \ to DAL2 to check the next character BNE DAL2 \ Y now points to a space character in the line buffer ASL SC+1 \ Shift SC+1 to the left BMI DAL2 \ If bit 7 of SC+1 is set, jump to DAL2 to find the next \ space character \ We now want to insert a space into the line buffer at \ position Y, which we do by shifting every character \ after position Y along by 1, and then inserting the \ space STY SC \ Store Y in SC, so we want to insert the space at \ position SC LDY DTW5 \ Fetch the buffer size from DTW5 into Y, to act as a \ loop counter for moving the line buffer along by 1 .DAL6 LDA BUF,Y \ Copy the Y-th character from BUF into the Y+1-th STA BUF+1,Y \ position DEY \ Decrement the loop counter in Y CPY SC \ Loop back to shift the next character along, until we BCS DAL6 \ have moved the SC-th character (i.e. Y < SC) INC DTW5 \ Increment the buffer size in DTW5
\LDA #' ' \ This instruction is commented out in the original \ source, as it has no effect because A already contains \ ASCII " ". This is because the last character that is \ tested in the above loop is at position SC, which we \ know contains a space, so we know A contains a space \ character when the loop finishes
\ We've now shifted the line to the right by 1 from
\ position SC onwards, so SC and SC+1 both contain
\ spaces, and Y is now SC-1 as we did a DEY just before
\ the end of the loop - in other words, we have inserted
\ a space at position SC, and Y points to the character
\ before the newly inserted space
\ We now want to move the pointer Y left to find the
\ next space in the line buffer, before looping back to
\ check whether we are done, and if not, insert another
\ space
.DAL3
CMP BUF,Y \ If the character at position Y is not a space, jump to
BNE DAL1 \ DAL1 to see whether we have now justified the line
DEY \ Decrement the loop counter in Y
BPL DAL3 \ Loop back to check the next character to the left,
\ until we have found a space
BMI DA11 \ Jump back to DA11 (this BMI is effectively a JMP as
\ we already passed through a BPL to get here)
.DA2
\ This subroutine prints out a full line of characters
\ from the start of the line buffer in BUF, followed by
\ a newline. It then removes that line from the buffer,
\ shuffling the rest of the buffer contents down
LDX #LL \ Call DAS1 to print out the first LL characters from
JSR DAS1 \ the line buffer in BUF
LDA #12 \ Print a newline
JSR CHPR
TAX \ Copy the new value of DTW5 into X BEQ DA6+3 \ If DTW5 = 0 then jump down to DA6+3 to print a newline \ as the buffer is now empty \ If we get here then we have printed our line but there \ is more in the buffer, so we now want to remove the \ line we just printed from the start of BUF LDY #0 \ Set Y = 0 to count through the characters in BUF INX \ Increment X, so it now contains the number of \ characters in the buffer (as DTW5 is a zero-based \ pointer and is therefore equal to the number of \ characters minus 1) .DAL4 LDA BUF+LL+1,Y \ Copy the Y-th character from BUF+LL to BUF STA BUF,Y INY \ Increment the character pointer DEX \ Decrement the character count BNE DAL4 \ Loop back to copy the next character until we have \ shuffled down the whole buffer BEQ DA5 \ Jump back to DA5 (this BEQ is effectively a JMP as we \ have already passed through the BNE above) .DAS1 \ This subroutine prints out X characters from BUF, \ returning with X = 0 LDY #0 \ Set Y = 0 to point to the first character in BUF .DAL5 LDA BUF,Y \ Print the Y-th character in BUF using CHPR, which also JSR CHPR \ clears the C flag for when we return from the \ subroutine below INY \ Increment Y to point to the next character DEX \ Decrement the loop counter BNE DAL5 \ Loop back for the next character until we have printed \ X characters from BUF
RTS \ Return from the subroutine .DA6 JSR DAS1 \ Call DAS1 to print X characters from BUF, returning \ with X = 0 STX DTW5 \ Set the buffer size in DTW5 to 0, as the buffer is now \ empty PLA \ Restore Y and X from the stack TAY PLA TAX LDA #12 \ Set A = 12, so when we skip BELL and fall through into \ CHPR, we print character 12, which is a newline .DA7 EQUB &2C \ Skip the next instruction by turning it into \ &2C &A9 &07, or BIT &07A9, which does nothing apart \ from affect the flags \ Fall through into CHPR (skipping BELL) to print the \ character and return with the C flag cleared