Basic Knowledge |
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1.1 | Explain the format of a basic assembler statement. |
1.2 | What is the difference between DS and DC? |
1.3 | What does each of the following storage allocations mean?
FLDA DS CL8 FLDA DS XL4 FLDA DS PL3 FLDA DS F FLDA DS H FLDA DS D |
1.4 | How many general purpose registers are available for use by the application? How are they labeled? How many bytes in length is each one? |
1.5 | How many decimal values can be represented in one byte of storage? How many bits in each byte? What are the possible values of each bit? What is the actual decimal value of each bit in a byte? |
1.6 | For a sequential file (QSAM access method), what is the DCB? Which macros are used to read and write to the sequential file? If you are working with variable length records, the first 4 bytes of each record is the RDW. What is that and what is the format of those 4 bytes? |
1.7 | What is a DSECT? |
Coding Questions |
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The following is from a test sample program. These storage descriptions will be used for the remaining questions.
FLDA DC CL5ABCDE FLDB DC CL3123 FLDC DS CL100 FLDD DC PL225 FLDE DS CL3 FLDF DS F REC1 DSECT DATA DS CL100In the next 6 questions, Id like you to tell me what the instruction does and what is in the identified fields after execution of the instruction. |
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2.1 |
MVC FLDB,FLDA |
2.2 |
MVC FLDB,FLDA+2 |
2.3 |
MVC FLDB(1),FLDA |
2.4 |
MVI FLDB,CQ |
2.5 |
MVI FLDC,X40 MVC FLDC+1(99),FLDC |
2.6 |
UNPK FLDE,FLDDThe following 2 questions are based on your answer. What does the high order 4 bits of the low order byte mean? What additional instruction would you code to make the field print correctly? |
2.7 | Assume FLDF contains the address of data defined by the REC1 DSECT. Also assume registers 5 thru 9 are available for use. How would you make the data within the REC1 DSECT addressable? |
2.8 | What is the maximum length of data that may be moved with an MVC instruction? What op-code would you use if more bytes must be moved? |
2.9 | The following code will assemble correctly with no errors. However, there are 2 errors present. One is a logical error (the program will not execute as intended) and the second error will cause the program to abend when it executes. The BAL instruction itself is correct. The 2 errors are found in the other 8 statements. What are those 2 errors?
BAL R14,SUBR . . DS F SUBR EQU * ST R14,SUBR-4 CLC FLDD,=C25 BE SUBREND MVC FLDE,FLDB SUBREND L R14,SUBR-4 B R14 |
Version 1.2
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