IBM Mainframe Forum

An operand address that refers to storage is derived
from an intermediate value, which either is contained
in a register designated by an R field in the instruction
or is calculated from the sum of three binary
numbers: base address, index, and displacement.
The base address (B) is a 64-bit number contained in
a general register specified by the program in a fourbit
field, called the B field, in the instruction. Base
addresses can be used as a means of independently
addressing each program and data area. In arraytype
calculations, it can designate the location of an
array, and, in record-type processing, it can identify
the record. The base address provides for addressing
the entire storage. The base address may also be
used for indexing.
The index (X) is a 64-bit number contained in a general
register designated by the program in a four-bit
field, called the X field, in the instruction. It is
included only in the address specified by the RX-,
RXE-, RXF-, and RXY-format instructions. The RX-,
RXE-, RXF-, and RXY-format instructions permit double
indexing; that is, the index can be used to provide
the address of an element within an array.
The displacement (D) is a 12-bit or 20-bit number
contained in a field, called the D field, in the instruction.
A 12-bit displacement is unsigned and provides
for relative addressing of up to 4,095 bytes beyond
the location designated by the base address. A 20-bit
displacement is signed and provides for relative
addressing of up to 524,287 bytes beyond the baseaddress
location or of up to 524,288 bytes before it.
In array-type calculations, the displacement can be
used to specify one of many items associated with an
element. In the processing of records, the displacement
can be used to identify items within a record.
A 12-bit displacement is in bit positions 20-31 of
instructions of certain formats (see Figure 5-1 on
page 5-4). In instructions of some formats, a second
12-bit displacement also is in the instruction, in bit
positions 36-47.
A 20-bit displacement is in instructions of only the
RSY, RXY, or SIY format. In these instructions, the D
5-10 z/Architecture Principles of Operation
field consists of a DL (low) field in bit positions 20-31
and of a DH (high) field in bit positions 32-39. When
the long-displacement facility is installed, the numeric
value of the displacement is formed by appending the
contents of the DH field on the left of the contents of
the DL field. When the long-displacement facility is
not installed, the numeric value of the displacement
is formed by appending eight zero bits on the left of
the contents of the DL field, and the contents of the
DH field are ignored.
In forming the intermediate sum, the base address
and index are treated as 64-bit binary integers. A
12-bit displacement is treated as a 12-bit unsigned
binary integer, and 52 zero bits are appended on the
left. A 20-bit displacement is treated as a 20-bit
signed binary integer, and 44 bits equal to the sign bit
are appended on the left. The three are added as
64-bit binary numbers, ignoring overflow. The sum is
always 64 bits long and is used as an intermediate
value to form the generated address. The bits of the
intermediate value are numbered 0-63.
A zero in any of the B1, B2, X2, B3, or B4 fields indicates
the absence of the corresponding address
component. For the absent component, a zero is
used in forming the intermediate sum, regardless of
the contents of general register 0. A displacement of
zero has no special significance.

These users thanked the author Robert Sample for the post:

Andyo2424 (Thu Jul 11, 2019 5:07 am)

Andyo2424 wrote:Hi, everyone I'm currently taking an assembly course (HLASM), I just had one question about explicit addressing D(X,B), How do I know what register to use as an index register? how do I know what value to input as a displacement value? how do I know what register to use as a base register? I often see examples like
5,8(0,2) but I dont understand what that goes or what that does, is it like an array? thank you

Andyo2424 wrote:Hi, everyone I'm currently taking an assembly course (HLASM), I just had one question about explicit addressing D(X,B), How do I know what register to use as an index register? how do I know what value to input as a displacement value? how do I know what register to use as a base register? I often see examples like
5,8(0,2) but I dont understand what that goes or what that does, is it like an array? thank you

1. What register to use as an index register? You can use register 1 through 15 - the register that has the index value you need.
2. What value to use as a displacement? It depends on the data; the question cannot be answered here. This program fragment is an example.
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            USING IHADCB,5            ESATABLISH DCB DATA AREA ADDRESSING
            LA    5,ADCB              COMPUTE DCB ADDRESS
            MVC   RECFM,=CL8' '       INITIALIZE THE RECFM OUTPUT AREA
            LA    0,RTABN             LOAD ENTRIES IN RTAB
            LA    1,RTAB              COMPUTE ADDRESS OF RTAB
            LA    14,RECFM            COMPUTE ADDRESS OF RECFM OUTPUT AREA
   RECFM1   MVC   WKAREA,DCBRECFM     COPY BINARY RECFM TO A WORK AREA
            NC    WKAREA,0(1)         ISOLATE ATTRIBUTE BITS
            CLC   WKAREA,1(1)         ATTRIBUTE IN DCBRECFM?
            BNE   RECFM2              NO
            MVC   0(1,14),2(1)        COPY ATTRIBUTE TEXT TO OUTPUT
            LA    14,1(,14)           ADD 1 TO OUTPUT AREA POINTER
   RECFM2   LA    1,3(,1)             COMPUTE ADDRESS OF NEXT RTAB ENTRY
            BCT   0,RECFM1            CHECK IT
            DROP 5                    KILL DCB DATA AREA ADDRESSING
            ...
   RECFM    DC    CL5' '              RECFM TEXT
   WKAREA   DC    XL1'0'              WORK AREA
   ADCB     DCB   DSORG=PS,MACRF=PM,RECFM=VBA,LRECL=125  DEFINE A DCB
   *  RTAB - TABLE TO TRANSLATE DCBRECFM ATTRIBUTES TO TEXT
   *  OFFSET  PURPOSE
   *     0    ATTRIBUTE MASK
   *     1    ATTRIBUTE
   *     2    ATTRIBUTE TEXT
   RTAB     DC    AL1(DCBRECL,DCBRECF,C'F')
            DC    AL1(DCBRECL,DCBRECV,C'V')
            DC    AL1(DCBRECL,DCBRECU,C'U')
            DC    AL1(DCBRECBR,DCBRECBR,C'B')
            DC    AL1(DCBRECSB,DCBRECSB,C'S')
            DC    AL1(DCBRECTO,DCBRECTO,C'T')
            DC    AL1(DCBRECCC,DCBRECCA,C'A')
            DC    AL1(DCBRECCC,DCBRECCM,C'M')
   RTABN    EQU   (*-RTAB)/3
            ...
            DCBD  DEVD=DA,DSORG=PS
    
   
   The comments before the definition of the first RTAB entry describe the displacements.
   
   NC WKAREA,0(1)
   
   Displacement 0 in 0(1) can be readily deduced from the table. The displacement 0 in 0(1,14), where register 14 is an address in
   
   MVC 0(1,14),2(1)
   
   is pretty obvious. Displacement 2 in 2(1) is from the table.
3. What register to use as a base register? Any register from 1 to 15 that is not being used for some other purpose. In this fragment, registers 1, 5 and 14 are being used as base registers. An unspecified register in the fragment is also being used as a base register to address symbols such as WKAREA, RTAB, RECFM and other symbols.
4. Is it like an array? Yes and no. Perhaps like a C language (I presume you know C!) structure -
   
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   struct rtab
    {
     unsigned char rtmask, rtattrb, rtattrtxt;
    } *reg;

1. What is the reason for TABx EQU *-x? The index is a true value: day of year, for example is 1 through 366, so the first entry for day of year, 1, for example, has to be at TAB1 + 1.
2. TAB1 is awfully big, isn't it? Yes. However, we want to show the use of an index register, so we need tables. The most common example that does not use indexing uses a different, but smaller table, but uses a loop. This code is faster and does not use a loop.
3. Why use leap year day of year? Roughly 3 years in 4 are common years. Yes, but, as has been noted TAB1 is rather large. The equivalent table for a common year is almost as large. Better to have one large table than 2 large tables!