Chapter 6 Applied Instructions
6-479
API Instruction code Operand
Function
2113 $MIDR P
S
1
, S
2
, D
Retrieving a part of the string
Device
X Y M S T C HC D L SM SR E PR K 16#
“$” DF
2
16-bit instruction (7-13 steps)
Symbol:
S
1
：
String Word
S
2
：
Part of the string which is
retrieved
Word
D
：
Device in which the characters
retrieved are stored
Word
Explanation:
1. Suppose the values in S
2
and S
2
+1 are n and m respectively. The m characters starting from
the n
th
character in the string in S
1
are retrieved, and stored in D.
b15
b0
b7
b8
16#00
~
~
D
+1
D
...........
b15
b0
b7
b8
~
First ASCII code
Second ASCII code
Third ASCII code
Fourth ASCII code
Last ASCII code
n ASCII code from the last
th
n+1 ASCII code from the last
th
n+m-1 ASCII code from the last
th
n+m-3 ASCII code from the last
th
n+m-2 ASCII code from the last
th
16#00
S1
+1
S1
.....................
n+2 ASCII code from the last
th
n+3 ASCII code from the last
th
n+m ASCII code from the last
th
n+m-1 ASCII code from the last
th
~
n+1 ASCII code from the last
th
n ASCII code from the last
th
n+3 ASCII code from the last
th
n+2 ASCII code from the last
th
2. If the data in S
1
is ABCDEFGHIJK, the value in S
2
is 3, and the value in S
2
+1 is 7, the seven
characters starting from the third characters in the string are retrieved from the left. The
conversion result is as follows.
b15
b0
b7 b8
0016#
S1
16#41(A) 16#42(B)
43(C)16# 44(D)16#
46(F)16#
45(E)16#
47(G)16#
48(H)16#
4A(J)16#
49(I)16#
4B(K)16#
b15
b0
b7 b8
D
16#45(E)
46(F)16#
47(G)16# 48(H)16#
49(I)16#
Third character
0016#
+1
S1
+2
S1
+3
S1
+4
S1
+5 S1
S2
Seventh character
43(C)16#
44(D)16#
D+3
D+1
D+2
3. If the value in S
2
+1 is 0, the instruction is not executed.
4. If the value in S
2
+1 is -1, the characters in S
1
starting from the character indicated by the value
in S
2
to the last character in S
1
are retrieved.
5. If the data in S
1
is ABCDEFGHIJK, the value in S
2
is 5, and the value in S
2
+1 is -1, the
conversion result is as follows.
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