Bit Manipulation & Editing
Ch. 10
Page 240
Boolean Instructions
Bits in the sending field are used to modify bits in the receiving field – one byte at a time
Instruction
SI
SS
RX
RR
OR
OI
OC
O
OR
AND
NI
NC
N
NR
EXCLUSIVE OR
XI
XC
X
XR
OR Logic
If one bit or both bits are “on”,
then the resulting bit is “on”
AND Logic
Both bits must be “on” in order for the
resulting bit to be “on”
X-OR Logic
If one bit or the other bit is
“on”, but not both, then the
resulting bit is “on”.
OR Family
O
R1,D2(X2,B2)
OI D1(B1),I2
OC
D1(L,B1),D2(B2)
OR
R1,R2
The OR of the 1st and 2nd operands
are placed at the first operand location
Example:
OI AREA+10,X’5C'
Before: AREA+10
Mask
After: AREA+10
0011 0101
0101 1100
0111 1101
Notice the result in the example to the left:
Begin with the left-hand bit in the result field – the result bit if
“0” because neither of the target fields (AREA+10 and MASK)
is a “1”. In the case of the second bit from the left, the result
is “1” because the bit in the same position in MASK is “1”
even though it is a “0” in AREA+10. The third bit from the left
is the same situation. Notice the fourth bit from the left. The
result is “1” because both target bits are “1”. Simply apply the
same rules to the second half of the byte – the result is “0”
only if both of the target bits are “0”.
This is the instruction you have been using to change the flag
bit in the DCB to accept the data as EBCDIC rather than
ASCII. You have also, probably, used this same instruction to
make your sign-byte printable following a CVD and UNPK
instruction for an output field.
AND Family
N
R1,D2(X2,B2)
NI D1(B1),I2
NC
D1(L,B1),D2(B2)
NR
R1,R2
The AND of the 1st and 2nd operands
are placed at the first operand location
Example:
NI AREA+10,X’5C'
Before: AREA+10
Mask
After: AREA+10
0011 0101
0101 1100
0001 0100
This time the target fields are being AND’ed – both
operand fields must contain “1” bits in order for the result
to be a “1” in the same bit position. Notice the only bits
that are “1” in the result field are because that is the only
part of the target fields that both bits are “1” – all others
are “0”. This instruction is typically used to turn selected
bits to “0”. In your mask field, set a bit to “1” for those bits
you do not want to change, but set the bits to “0” if you
wish to turn the target bit to “0”.
X-OR Family
X
R1,D2(X2,B2)
XI
D1(B1),I2
XC
D1(L,B1),D2(B2)
XR
R1,R2
The X-OR of the 1st and 2nd operands
are placed at the first operand location
Example:
XI
AREA+10,X’5C'
Before: AREA+10
Mask
After: AREA+10
0011 0101
0101 1100
0110 1001
This is probably the trickiest of the Boolean sets of
instructions. Set the result to a “1” if either the target or the
mask is a “1” bit – but not both are “1” bits (that would then be
the same as AND). Map the result field back to the two input
fields to see that in each case, only one of the two bits being
compared cause the result to be “1”. Use this instruction
when OR nor AND will do the right thing for you.
Test Under Mask - TM
•
•
•
•
Check the status of bits in a byte
Storage-Immediate format only
Sets Condition Code based on results of the test
Indicate in the mask the bits in the byte to be tested by setting
those bits to 1’s
Practically every program uses a switch or some indicator to indicate some event has occurred. TM instruction
provides a very efficient means to test that switch to see if your event has occurred yet. Do you remember the
Parcheesi exercise back near the beginning of class? In it, you set a switch (using OI perhaps) to indicate that
you landed on a Blue Square and are due to lose your next roll of the die. On entry to your Parcheesi Player,
you could test that switch using TM. The smallest amount of memory that you can allocate is a byte, so let’s
assume that you used the high-order byte for your switch. The example using Parcheesi for testing the LOSE
TURN follows. Of course, the purpose of TM is to set the condition code so you can actually check on the
status of the comparison.
FLAGBYTE,X’C3’
TM
CC set:
If tested
bits are:
CC = 0
CC = 1
CC = 3
0's
mixed
1's
FLAGBYTE
1111 1011
Mask
1100 0011
Result:
CC=3
All tested bits are 1’s
Remember Parcheesi ?
The flowchart that you created lies to the far
left – at least the part of the flowchart that
relates to our current discussion. The
Assembler code that is relevant is under the
flowchart symbol labeled RESET SWITCH.
Early in the turn of PLAYER1, the test needs
to be made to see if a Blue Square was
landed on the last turn and if so, this turn
needs to be bypassed. The TM instruction
satisfies the test of the Blue Square (the
switch is the high-order bit of the BLUESW
byte. The define is done in the last
displayed instruction at the bottom. The
switch is set once it has been determined
that the square landed on is blue. This can
be easily accomplished with the OI
instruction, as shown near the bottom. Note
also that the switch is tested prior to rolling
the die. Finally, where the switch is tested,
then the branch to SWITCHON is taken, you
want to remember to shut the switch off.
This could be a good use of the Exclusive
Or instruction:
XI
BLUESW,X’80’
Translation Instructions
• Translate – TR
• Translate and Test – TRT
• Execute - EX
Translate one bit pattern
into a different bit pattern
– up to 256 unique characters per byte
TR and TRT essentially work the same way, but they end up a little differently. Read on! The EX
instruction is not really a Translation instruction, but it does most often get used with the TR/TRT
instructions.
p.328
Translate - TR
2 Operands:
1.
2.
TR
pattern to be translated
table of replacement characters
EBCDIC,ASCIITAB
We will create an example to change EBCDIC digits (F0-F9)
into ASCII digits (30-39). EBCDIC operand is the digits to be
translated & ASCIITAB is the replacement digits.
0
1
7
8
9
B
C
D
E
F
0_
40
40
40 40
40 40
40 40
40
40
40 40
40
40
40
40
1_
40
40
40 40
40
40
40 40
40
40
40 40
40
40
40
40
2_
40
40
40 40
40
40
40 40
40
40
40 40
40
40
40
40
3_
40
40
40 40
40
40
40 40
40
40
40 40
40
40
40
40
4_
40
40
40 40
40
40
40 40
40
40
40 40
40
40
40
40
5_
40
40
40 40
40
40
40 40
40
40
40 40
40
40
40
40
6_
40
40
40 40
40
40
40 40
40
40
40 40
40
40
40
40
7_
40
40
40 40
40
40
40 40
40
40
40 40
40
40
40
40
8_
40
40
40 40
40
40
40 40
40
40
40 40
40
40
40
40
9_
40
40
40 40
40 40
40 40
40
40
40 40
40
40
40
40
A_ 40
40
40 40
40
40
40 40
40
40
40 40
40
40
40
40
B_ 40
40
40 40
40
40
40 40
40
40
40 40
40
40
40
40
C_ 40
40
40 40
40
40
40 40
40
40
40 40
40
40
40
40
D_ 40
40
40 40
40
40
40 40
40
40
40 40
40
40
40
40
E_
40
40 40
40
40
40 40
40
40
40 40
40
40
40
40
F_ 30 31 32 33 34 35 36 37 38 39 40 40
40
40
40
40
40
2
3
4
5
6
A
Take notice of the table shown on the previous slide. Nearly all the blocks in the table contain the EBCDIC
character for a blank – X’40’. These are the EBCDIC character patters to be ignored for translation. While the
characters around the table – across the top and down the left side are not part of the table in memory – these
characters are just showing us the 256 positions in memory which is occupied by the table. The only characters
other than X’40’ are used in the bytes representing F0-F9 – the EBCDIC characters for the digits 0-9. That’s
where the characters to be translated are placed in the table. The values that are replacing the X’40’ are those of
the ASCII characters that represent the digits 0-9. The idea should now be quite obvious to you – when a byte
that contains the EBCDIC character X’F0’ it will be translated into ASCII character X’30’, X’F1’ will be translated to
ASCII X’31’ and so on. Any other characters will remain untouched (un-translated) which is what the X’40’ means
to accomplish. The hardest part (not really hard, just tedious or cumbersome) is building and defining the table in
memory.
Recall that each character, as it appears in memory, is made up of two hex digits, up to 256 unique characters. In
the table on the previous slide, the first hex character is represented down the left column and the second hex
character is represented across the table (in a row).
Translate - TR
TR
EBCDIC,ASCIITAB
where EBCDIC contains:
which in hex is:
1984
F1 F9 F8 F4
From left-to-right, each EBCDIC character is used as an index into
ASCIITAB. The resulting character is the character at the indexedto location in memory.
Translate - TR
TR
EBCDIC,ASCIITAB
Any other bit configuration would give back a blank character – X’40’
Defining ASCIITAB ?
ASCIITAB
DS
DC
DC
DC
0CL256
16X’40’
16X’40’
16X’40’
DC
DC
DC
DC
16X’40’
*E0-EF
X’303132333435’
X’36373839’
12X’40’
*00-0F
*10-1F
*20-2F
*F0-F5
*F6-F9
*FA-FF
Translate-and-Test - TRT
• Similar to TR in the way it works
– 1st operand acts as an index into 2nd operand
– 2nd operand is a table
• No replacement of characters – just checks that table byte is
X’00’
– If X’00’, process the next byte
– If not X’00’, stop processing TRT and place address of the
byte in GPR 1, place the byte from the table into low-order
byte of GPR 2
– You had better not be using GPRs 1 & 2 for anything
So What Are Uses of TRT?
• When looking at a variable-length field such as a NAME, Find
the blank following a name shorter than the fixed-character
length in an input area
• Find the first significant digit in a numeric field
• Locate a comma separator
• Find any character that has special meaning
TRT Table
• X’00’ – character with no special meaning
• Non-X’00’ – special meaning
– X’40’ represents characters without interest
– X’00’ represents alpha or numeric characters
– Other fields identify some special characters such as a
blank (40) is identified with X’04’ and from there, the
special characters are identified in increments of 4 bytes
(04, 08, 0C, etc.)
As you may have guessed by now, this is another of those instructions that takes significantly more time to
think through, design, and build all the support areas, and the instruction itself is very easy to code.
0
1
2
3
0_
40
40
40 40
1_
40
40
2_
40
3_
4
5
6
7
8
9
40 40
40 40
40
40 40
40
40
40 40
40
40 40
40
40
40
40
40 40
40
4_
04
40
40 40
40
5_
14
40
6_
24
7_
A
B
C
D
E
F
40
40 40
40
40
40
40
40
40
40 40
40
40
40
40
40 40
40
40
40 40
40
40
40
40
40
40 40
40
40
40 40
40
40
40
40
40
40 40
40
40
40 08
40 0C 10
40
40 40
40 40
40 40
40
40
40 18 1C 20
40
40
28
40 40
40
40
40 40
40
40
40 2C 40
40
40
40
40
40
40 40
40
40
40 40
40
40
40 30
34
38 3C 40
8_
40
40
40 40
40
40
40 40
40
40
40 40
40
40
40
40
9_
40
40
40 40
40
40
40 40
40
40
40 40
40
40
40
40
A_ 40
40
40 40
40
40
40 40
40
40
40 40
40
40
40
40
B_ 40
40
40 40
40
40
40 40
40
40
40 40
40
40
40
40
C_ 40
00
00 00
00
00
00 00
00
00
40 40
40
40
40
40
D_ 40
00
00 00
00
00
00 00
00
00
40 40
40
40
40
40
E_
40
40
00 00
00
00
00 00
00
00
40 40
40
40
40
40
F_
00
00
00 00
00 00
00 00
00
00
40 40
40
40
40
40
Review closely the table on the previous slide:
X’00’ in the table represent the alphanumeric characters in EBCDIC – letters and numbers
X’40’ is used to indicate invalid characters – those that have no interest
Some special characters are of interest and are identified with other than X’00’ or X’40’ –
for example a blank (40) is identified with a X’04’ and the period (4B) is identified with X’08’.
Notice that the special characters are identified with symbols that are in increments of 4 bytes
– 04, 08, 0C, 10, etc.
The table to the right outlines the characters that are “special” for the TRT instruction.
• Assume the table is located beginning at location X‘2000’ – so
the letter upper-case A is the translate byte at X’20C1’
• A branch table is located at X’3004’ with 16 branch instructions
– one for each of the special characters- the first of which is for
a blank
• The characters to be translated are at location X’CA80’ which
contains:
(CA80-CA94)
UNPK PROUT(9),WORD(5)
E4D5D7D2 40D7D9D6 E4E34DF9 5D6BE6D6 D9C44DF5 5D
• Translate-and-Test instruction is:
LA
TRT
R2,BRTABLE
UNPKINST(21),TRTTABLE
• 1st character to translate is U – X’E4’
• Translation characters are X’00’ at Table+E4 – take no special
action
• Translate the rest of the op-code until translation reaches the
blank (b)
• A blank character is X’40’ which takes us to table+40 which is
non-X’00’
– Addr of the source byte is placed in GPR 1
GPR 1
00 00 00 00 00 00 CA 84
– Table entry (X’04’) is placed in low order byte of GPR 2
(which contained X’3000’)
GPR 2
00 00 00 00 00 00 30 04
– CC is set to 1 (non-zero byte found)
BC
4,0(2)
*blank found
This is complicated! At table+40 (address in GPR1) is found X’04” so the X”04” is placed in the low-order byte of GPR2.
Prior to this, GPR2 contained X”3000” – the address of the Branch Table which you constructed and can be seen on the
next slide. With the value found by TRT (X”04”), the condition code is set to 1, so the BR instruction above branches on
condition (cond. Code 4 = Cond. Code set to 1) to location 0000 in memory, indexed by GPR2 which contains the address
of the Branch Table entry for “Found a Blank” which was X”04” in the Translate Table. Phew!
Condition Code Set
Mask Bit Value
0
8
1
4
2
2
3
1
TRT instruction set the Condition Code to 1 to indicate non-zero found when the “blank”
was translated. With Cond.Code 1, the code to test in the Branch-on-Condition is X”4”.
Branch Table beginning at Loc.
X’3000’
Memory
Location
3000
3004
3008
300C
3010
3014
3018
301C
3020
3024
3028
302C
3030
3034
3038
303C
Contents of
Branch Table
BC 4,0(2)
BLANKSP
BC 4,0(2)
PERIOD
BC 4,0(2)
LEFT PAREN
BC 4,0(2)
PLUS SIGN
BC 4,0(2)
AMPERSAND
BC 4,0(2)
DOLLAR SIGN
BC 4,0(2)
ASTERISK
BC 4,0(2)
RIGHT PAREN
BC 4,0(2)
DASH
BC 4,0(2)
FORW’D SLASH
BC 4,0(2)
COMMA
BC 4,0(2)
POUND SIGN
BC 4,0(2)
AT SIGN
BC 4,0(2)
APOSTROPHE
BC 4,0(2)
EQUAL SIGN
I think you get the idea…
Execute - EX
• Causes just one target instruction to be executed out of sequence
without actually branching to that instruction.
• After OR’ing bits 8-15 (2nd byte) of the 2nd operand, which could be
the length field of SS-type instructions without changing the byte
in the instruction (OR’s from low-order byte of 1st operand)
• Most often used with TRT
• RX format:
EX
R1,D2(X2,B2)
As we all learned possibly on the very first day of class, we don’t mix instructions and definitions in the same part of the
program. Normally instructions come first, then the definitions. But there are circumstances where that rule may be
disregarded and the use of the EX instruction is one of those times. As we also know, with the EX instruction, we can target
some other instruction that is out of sequence, execute that target instruction, then return to the instruction following the EX
instruction. In this case, it makes sense to place that target instruction out of the way of normal sequential instruction
execution. Place the target instruction along with the definitions of memory.
The ability of the EX instruction to modify the second byte of a target instruction gives you the power, depending on the
instruction format, to change a register reference, or an immediate operand, or an operand length. Review each instruction
format for what is located at the second byte location, therefore subject to change:
FORM
AT
OPERANDS
RR
R1,R2
RS
R1,R3,D2(B2)
RX
R1,D2(X2,B2)
SI
D1(B 1),I2
SS(1)
D1(L,B1),D2(B2)
SS(2)
D1(L1,B1),D2(L2,B2)
In the chart to the left, the underlined operand in each
instruction format shows the possibilities for the values
that can be altered by using the EX instruction.
Operand 1 designates a register containing a value used
to modify the second byte of the target instruction. However,
if register 0 is specified, no modification is done.
Operand 2 refers to the target instruction, defined elsewhere in
memory, to be executed by EX. Any instruction except another
EX may be target to EX.
EX then Ors bits 8-15 (byte 2) of the target instruction with bits
24-31 (the rightmost byte) of the operand 1 register. The target
instructions, however, is not actually changed in memory.
After executing the target instruction, the program resumes either with the instruction following EX or, if the target instruction is a branch,
where the branch is directed.
Examples of Using EX
The following RR-format example inserts X”14” into GPR5 and Ors byte 2 of ADDREG with the X”14”. This process
causes the second byte of ADDREG to reference GPR 1 and 4 temporarily. ADDREG executes out of line as “add”
the contents of GPR4 to the contents of GPR1.
EXRR
ADDREG
IC
EX
--DS
AR
5,=X’14’
5,ADDREG
insert X’14’ into R5
Add R4 to R1
0H
0,0
target instruction
The RS-format example Ors the second byte of STOREGS with X’DF’. This process causes byte 2 of STOREGS to
reference R13 and R15. STOREGS executes out of line as “Store the contents of registers 13,14, and 15 in
SAVEREGS.
SAVEREGS
STOREGS
IC
EX
--DS
DS
STM
6,X’DF’
6,STOREGS
insert X’DF’ into R6
store regs 13-15
3F
0H
0,0,SAVEREGS
3 fullwords
boundary alignment to even
RS target instruction
Examples of Using EX
This RX-format example Ors the second byte of STORCH with X’20’. This process causes byte 2 of STORCH to reference
base reg 2 and index reg 0. STORCH executes as “store the rightmost 8 bits of R2 in SAVEBYTE.
SAVEBYTE
STORCH
IC
EX
--DS
DS
STC
7,=X’20’
7,STORCH
put X’20’ into R7
store right byte of R7
C
0H
0,SAVEBYTE
one byte – character
force even address
RX Target instruction
This SI-format example Ors the second byte of MOVIMM with X’5B’ (a dollar sign), then moves the dollar sign to PRINT+20,
and the instruction can thus move any immediate value.
PRINT
MOVIMM
IC
EX
--DC
DS
MVI
8,=X’5B’
8,MOVIMM
place X’5B’ into R8
move $ to print area
CL133’ ‘
0H
PRINT+20,X’00’
print area blanked out
force boundary alignment
target instrution
Examples of Using EX
Using EX on SS-format instructions causes it to modify operand lengths. Remember that
when executing an SS-format instruction, the computer adds 1 to the length. The first
example uses MOVECHAR to move a specified number of asterisks (up to 3) to the print area
and could be used to print one, two, or three asterisks to denote the total level:
PRINT
MOVECHAR
IC
EX
--DC
DS
MVC
9,=X’02’
9, MOVECHAR
insert X’02’ into R9
move *** to print area
CL133’ ‘
print area definition
0H
PRINT+30(0),=C’CCC’ target instruction for EX
This example executes an AP operation is which byte 2 contains a length code for both operands:
FLD1
FLD2
ADDPK
IC
EX
--DS
DS
DS
AP
10,=X’32’
10,ADDPK
insert X’32’ into R10
Add FLD2 to FLD1
PL4
PL3
0H
FLD1(0),FLD2(0)
target of EX instruction
Last EX Example
003802
003804
003808
00380C
00380E
003810
003814
003818
00381A
00381E
003820
003824
1711
4190
4310
1881
0610
4190
4410
1A98
4310
0610
4190
4410
003828
003852
0038D7 00
0038D8
0038DE
4026
9000
9001
40D6
9000
9001
40DC
6
7
8
9
11
12
13
14
16
17
18
19
20
21
22
23
24
*
*
*
NAMADDR
PRLINE
26 M30MOVE
27 M40MOVE
XR
LA
IC
LR
BCTR
LA
EX
AR
IC
BCTR
LA
EX
------DS
DS
1,1
9,NAMADDR
1,0(0,9)
8,1
1,0
9,1(0,9)
1,M30MOVE
9,8
1,0(0,9)
1,0
9,1(0,9)
1,M40MOVE
Clear Reg 1
Addr 1st length indicator
length of NAME in Reg 1
save length in Reg 8
decrement Reg 1 by 1
increment for NAME field
execute M30MOVE
addr of NAME + length for 2nd indicator
length of addr in Reg 1
decrement reg 1 by 1
increment for address field
execute M40MOVE
CL42
CL133
variable name & address
print area
MVC
MVC
---
PRLINE+20(0),0(9)
PRLINE+50(0),0(9)
move NAME to Print
move ADDRESS to Print
If you follow this example carefully, there is a length byte in front of each variable-length field in the input that is read from a
device. The length is one byte and in binary format. There are two fields alternating: NAME and ADDRESS.
0A ADAM SMITH 0C 423 BASIN ST nn NEXT NAME nn NEXT ADDR
Last EX Example Continued
Since the name is 10 characters long, the preceding length indicator contains X’0A’. The address is 12 characters long,
and its preceding length indicator contains X’0C’. The entire record consists of 1 + 10 + 1 + 12 = 24 bytes. Other records
would vary in length accordingly. The maximum record length is assumed to be 42 bytes.
The purpose is to move the name and the address separately to the print area. IC extracts the length indicator, and the
STC inserts the length indicator (minus 1 byte) into the second byte, the length, of the MVC instruction. At execution time,
the computer adds 1 to the instruction length. At the end of execution, the print area looks like this:
Print area:
ADAM SMITH
423 BASIN ST
PRINT + 20
PRINT + 40
Editing
• Used to Edit (Beautify) numeric data to packed decimal (and it
UNPK’s it too)
• Insert $-sign
• Include commas in larger numbers
• Insert the decimal point in financial numbers
• Suppress leading zeros
p.82 in Ch.4
Some Edit Patterns
Char Meaning
20 Digit selector
21
40
4B
60
6B
Significance starter
Blank character
Decimal point
Minus sign
comma
Detailed Edit Example
1200
1203
02 57 42 6C
Source:
1000
Pattern:
100C
40 20 20 6B 20 21 20 4B 20 20 40 C3 D9
b d d ,
ED
d (
d .
d d b C R
PATTERN(13),SOURCE(4)
The SOURCE field is the value that you want to have beautified by inserting commas, if appropriate, dollar sign, and
decimal point. The PATTERN field tells EDIT how to go about accomplishing the beautification. The PATTERN is shown
in two places – the hex format in the light green box and the character interpretation in the white box below the green
box. The instruction executes like most others (right to left) and the PATTERN must be the receiving field. SOURCE is
unchanged after the instruction executes.
Before decimal data in the packed format can be used in a printed report, digits and signs must be converted to printable
characters. Moreover, punctuation marks, such as commas and decimal points, may have to be inserted in appropriate
places. The highly flexible EDIT instruction performs these functions in a single instruction execution.
The example on the previous slide shows step-by-step one way that the EDIT instruction can be used. The field to be
edited (the source) is four bytes long; it is edited against a pattern 13 bytes long. The following symbols are used:
SYMBOL
b (hex 40)
( (hex 21)
d (hex 20)
MEANING
blank character
significance started
digit selector
As the instruction executes, it looks at each byte from left to right. The table below shows how each byte is treated by
instruction execution.
PATTERN
b
d
d
,
d
(
d
.
d
d
b
C
R
DIGIT
0
2
5
7
4
2
6+
SIGNIFICANCE
INDICATOR
(before/after)
off/off
off/off
off/on(2)
on/on
on/on
on/on
on/on
on/on
on/on
on/off(3)
off/off
off/off
off/off
RULE
LOCATION
1000-100C
leave (1) bdd,d(d.ddbCR
fill
bbd,d(d.ddbCR
digit
bb2,d(d.ddbCR
leave
same
digit
bb2,5(d.ddbCR
digit
bb2,57d.ddbCR
digit
bb2,574.ddbCR
leave
same
digit
bb2,574.2dbCR
digit
bb2,574.26bCR
fill
same
fill
bb2,574.26bbR
fill
bb2,574.26bbb
In the table above there are 3 notes:
1. This character is the fill byte.
2. First nonzero decimal source digit turns on the significance indicator.
3. Plus sign in the four rightmost bits of the byte turns off the significance indicator.
And the Answer Is …
Pattern: 40 40 F2 6B F5 F7 F4 4B F2 F6 40 40 40
which prints as: 2,574.26
After ED executes, the condition code is set.
0
result is 0
1
result is negative
2
result is positive
3
not used
More Examples of ED on p. 83 in Your Textbook
•
•
•
•
Suppress Leading Zeroes
Significance Starting
Insert commas and decimal points
Negative values handling
EDMK
• Used for printing money values
• Same as Edit instruction except that the address of the 1st
significant digit is saved in GRP 1 when EDMK executes
• This makes it easier to insert a dollar-sign or other currency
symbol into the field to the left of the amount
p.238
EDMK Example
Source:
02 57 42 6C
Pattern:
40 20 20 6B 20 21 20 4B 20 20 40 C3 D9
EDMK
BCTR
MVI
PATTERN(13),SOURCE(4)
1,0
0(1),C’$’
Pattern: 40 5B F2 6B F5 F7 F4 4B F2 F6 40 40 40
Printed Output:
$2,574.26
EDMK Examples on p.239
•
•
•
•
•
•
•
Suppress leading zeroes
Significance Starting
Insert commas and decimal point
Non-blank fill character
Negative value handling
Editing dates
Suppressing fields
Exercise
• Due next week
• Optional / Extra Credit
• Bit Bash Assignment