ARRAY EXAMPLE 3
  LOC  OBJECT CODE    ADDR1 ADDR2  STMT   SOURCE STATEMENT
000000                                2 P1       START
000000 05C0                           3          BALR  12,0
                            00002     4          USING *,12
                                      5 * OBJECTIVE LEARN HOW TO USE ONE-DIMENSIONAL WITH NON-CONSTANT
                                      6 * SUBSCRIPTS
                                      7 * A:  ARRAY[0..10] OF INTEGER
                                      8 *
                                      9 * A[I] := 1
000002 4110 C0AA            000AC    10          LA    1,A      R1 = ADDRESS OF A, USE LA TO LOAD ADDRESS
000006 5830 C0D6            000D8    11          L     3,I
00000A 1A33                          12          AR    3,3
00000C 1A33                          13          AR    3,3      R3 NOW = 4*I, (4 BYTES PER WORD)
00000E 1A13                          14          AR    1,3      R1 NOW = ADDRESS OF A(I)
000010 5820 C09A            0009C    15          L     2,F1
000014 5020 1000            00000    16          ST    2,0(0,1)
                                     17 * A [ I + 1 ] := 2
000018 4110 C0AA            000AC    18          LA    1,A      R1 = ADDRESS OF A, USE LA TO LOAD ADDRESS
00001C 5830 C0D6            000D8    19          L     3,I
000020 5A30 C09A            0009C    20          A     3,F1
000024 1A33                          21          AR    3,3
000026 1A33                          22          AR    3,3      R3 NOW = 4*(I+1)
000028 1A13                          23          AR    1,3      R1 NOW = ADDRESS OF A(I)
00002A 5820 C09E            000A0    24          L     2,F2
00002E 5020 1000            00000    25          ST    2,0(0,1)
                                     26 * A [ I + 3 ] := A [ I ] +  A [ I + 1 ]
000032 4110 C0AA            000AC    27          LA    1,A      R1 = ADDRESS OF A, USE LA TO LOAD ADDRESS
000036 5830 C0D6            000D8    28          L     3,I
00003A 1A33                          29          AR    3,3
00003C 1A33                          30          AR    3,3      R3 NOW = 4*I
00003E 1A13                          31          AR    1,3      R1 NOW = ADDRESS OF A(I)
000040 5840 1000            00000    32          L     4,0(0,1) R4 WILL ACCUMULATE THE SUM
000044 4110 C0AA            000AC    33          LA    1,A      R1 = ADDRESS OF A, USE LA TO LOAD ADDRESS
000048 5830 C0D6            000D8    34          L     3,I
00004C 5A30 C09A            0009C    35          A     3,F1
000050 1A33                          36          AR    3,3
000052 1A33                          37          AR    3,3      R3 NOW = 4*(I+1)
000054 1A13                          38          AR    1,3      R1 NOW = ADDRESS OF A(I+1)
000056 5A40 1000            00000    39          A     4,0(0,1) R4 WILL ACCUMULATE THE SUM
00005A 4110 C0AA            000AC    40          LA    1,A      R1 = ADDRESS OF A, USE LA TO LOAD ADDRESS
00005E 5830 C0D6            000D8    41          L     3,I
000062 5A30 C0A2            000A4    42          A     3,F3
000066 1A33                          43          AR    3,3
000068 1A33                          44          AR    3,3      R3 NOW = 4*(I+3)
00006A 1A13                          45          AR    1,3      R1 NOW = ADDRESS OF A(I+3)
00006C 5040 1000            00000    46          ST    4,0(0,1) PUT THE SUM IN A(I+3)
         ARRAY EXAMPLE 3
  LOC  OBJECT CODE    ADDR1 ADDR2  STMT   SOURCE STATEMENT
                                     48 *  I := A[A[I+3]]
000070 4110 C0AA            000AC    49          LA    1,A      R1 = ADDRESS OF A, USE LA TO LOAD ADDRESS
000074 5830 C0D6            000D8    50          L     3,I
000078 5A30 C0A2            000A4    51          A     3,F3
00007C 1A33                          52          AR    3,3
00007E 1A33                          53          AR    3,3      R3 NOW = 4*(I+3)
000080 1A13                          54          AR    1,3      R1 NOW = ADDRESS OF A(I+3)
000082 5830 1000            00000    55          L     3,0(0,1) R3 CONTAINS A(I+3)
000086 4110 C0AA            000AC    56          LA    1,A
00008A 1A33                          57          AR    3,3
00008C 1A33                          58          AR    3,3      R3 NOW = 4*A(I+3)
00008E 1A13                          59          AR    1,3      R1 NOW = ADDRESS OF A(A(I+3))
000090 5840 1000            00000    60          L     4,0(0,1)
000094 5040 C0D6            000D8    61          ST    4,I
000098 07FE                          62          BR    14
                                     63 *
00009A 0000
00009C 00000001                      64 F1       DC    F'1'
0000A0 00000002                      65 F2       DC    F'2'
0000A4 00000003                      66 F3       DC    F'3'
0000A8 00000004                      67 F4       DC    F'4'
                                     68 *
0000AC  0000000000000000             69  A       DC    11F'0' LAST SUBSCRIPT IS TEN
0000D8 00000003                      70 I        DC    F'3'
                                     71          END