Confused in understanding JLT *-19 in SIC/XE means

Question

This is a program from LELAND L beck's book on System Software (An introduction to system programming)

I am trying to understanding how MACROS work

this demonstrates the use of macro

My doubt is what does JEQ *-19 means and what does JEQ *-3 means

When i googles it says that it jumps 19 instructions back then where will it reach How can i figure 19 instructions and pin point the place where the control flow would land

Is there any way i can practise sic/xe programs ...any virtual machines or so that i can try it ..I would be much grateful to any suggestions...

     COPY START 0
     RDBUFF MACRO &INDEV,&BUFADR,&RECLTH

    CLEAR X
    CLEAR A
    CLEAR S
    +LDT #4096
    TD =X'&INDEV'
    JEQ *-3
    STCH &BUFADR,X
    TIXR T
    JLT *-19
    STX &RECLTH
    MEND



WRBUFF MACRO  &OUTDEV,&BUFADR,&RECLTH
     CLEAR X
     LDT  &RECLTH
     TD  =X'&OUTDEV'
     JEQ  *-3
     WD   =X'&OUTDEV'
     TIXR  T
     JLT  *-14
      MEND



...
MAIN PROGRAM
FIRST STL RETADR

CLOOP  RDBUFF F1,BUFFER,LENGTH  
       LDA LENGTH
       COMP  #0
       JEQ   ENDFIL
       WRBUFF 05,BUFFER,LENGTH
       J     CLOOP
ENDFIL WRBUFF 05,EOF,THREE
      J      @RETADR
 EOF    BYTE   C`EOF`
THREE   WORD   3
RETADR  RESW  1
LENGTH  RESW   1
BUFFER RESB   4096
     END   FIRST

Any help would be highly appreciated .Thanks and regards in advance

Answer 1

The character * evaluates the address of the current instruction and -3 subtracts three bytes (not instructions) from it.

    TD =X'&INDEV'
    JEQ *-3

In this case * will evaluate to the value of JEQ's address so *-3 will evaluate to TD's address (since TD instruction takes 3 bytes in memory).

Similarly, JLT *-19 will jump to CLEAR A (CLEARs and TIXR are in format 2, +LDT in format 4 and the others in format 3, which adds to 19).