.NOS1 LSR A \ Divide A by 2, and also clear the C flag, as bit 0 of \ A is always zero (as A is a multiple of 8) ADC #3 \ Set Y = A + 3, so Y now points to the last byte of TAY \ four within the block of four-byte values LDX #7 \ We want to copy four bytes, spread out into an 8-byte \ block, so set a counter in Y to cover 8 bytes .NOL1 LDA #0 \ Set the X-th byte of XX16 to 0 STA XX16,X DEX \ Decrement the destination byte pointer LDA SFX,Y \ Set the X-th byte of XX16 to the value from SFX+Y STA XX16,X DEY \ Decrement the source byte pointer again DEX \ Decrement the destination byte pointer again BPL NOL1 \ Loop back for the next source byte RTS \ Return from the subroutineName: NOS1 [Show more] Type: Subroutine Category: Sound Summary: Prepare a sound blockContext: See this subroutine in context in the source code References: This subroutine is called as follows: * EXNO calls NOS1 * NOISE calls NOS1
Copy four sound bytes from SFX into XX16, interspersing them with null bytes, with Y indicating the sound number to copy (from the values in the sound table at SFX). So, for example, if we call this routine with A = 40 (long, low beep), the following bytes will be set in XX16 to XX16+7: &13 &00 &F4 &00 &0C &00 &08 &00 This block will be passed to OSWORD 7 to make the sound, which expects the four sound attributes as 16-bit big-endian values - in other words, with the low byte first. So the above block would pass the values &0013, &00F4, &000C and &0008 to the SOUND statement when used with OSWORD 7, or: SOUND &13, &F4, &0C, &08 as the high bytes are always zero.
Arguments: A The sound number to copy from SFX to XX16, which is always a multiple of 8
[X]
Label NOL1 is local to this routine
[X]
Variable SFX (category: Sound)
Sound data