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IBM Programming with IBM Enterprise PL/I Sample Questions:

1. CORRECT TEXT
Given the following piece of code, what will happen when the code is executed if STRINGSIZE is not enabled?
DCL X CHAR (5) INIT ('ABCDE');
DCL Y CHAR (3) INIT (");
Y = X;

A) Y receives the value 'ABC' and the next two bytes following Y's storage are overridden by 'DE'.
B) Y receives the value 'ABC' and 'DE' is not assigned to any storage area.
C) The assignment is not executed and program execution continues after the assignment.
D) The assignment is not executed and program execution abends.

2. CORRECT TEXT
What should be done, if anything, when the following compiler message appears?
Multiple closure of blocks, one extra END statement assumed.

A) Take out the extra END statement.
B) Nothing needs to be done.
C) PUT an extra END statement at the end of the program.
D) Find the missing END statement and add it at the right place.

3. CORRECT TEXT
What would be printed, if anything, to SYSPRINT after executing the following code?
DCL A BIN FIXED(15) INIT(1000);
DCL B PlC '99999' INIT(2000);
B = A + B;
PUT SKIP LIST('THE VALUE OF B IS:' !! B);

A) THE VALUE OF B IS :03000
B) THE VALUE OF B IS:3000
C) Nothing will be printed because CONVERSION would be raised.
D) THE VALUE OF B IS :3000

4. CORRECT TEXT
If the physical dataset referred to by DDIN has a record length of 200 and a RECFM of F, what happens after executing the following code?
DCL DDIN FILE RECORD INPUT;
DCL P PTR;
DCL 1 INSTR BASED(P),
2 A CHAR(100),
2 B CHAR(100);
ALLOCATE INSTR;
OPEN FILE(DDIN);
READ FILE(DDIN) INTO(INSTR);

A) READ INTO cannot be used on a BASED structure.
B) One record
will be read into the structure INSTR.
C) One record will be read into the buffer and the pointer will be set accordingly.
D) Program will abend because P has not been properly initialized.

5. CORRECT TEXT
Which of the following methods of initializing the variable X is most efficient because it results in
the fewest compiler instructions?
DCL A PlC '999V.99';
DCL A CHAR(6)INIT( '000.00');
DCL B DEC FIXED(5,2) INIT( 0);
DCL C BIT(8)INIT( '00000000'B);
DCL D FLOAT(6)INIT( 0E0);

A) X=B;
B) X=D;
C) X=C;
D) X=A;

Solutions: