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IBM C9050-042 Exam Syllabus Topics:

IBM Developing with IBM Enterprise PL/I Sample Questions:

1. Given the following code, what statement will produce an E level diagnostic?
%DCL ABC ENTRY;
%ABC: PROC (X,Y) STATEMENT;
DCL(X,Y) CHAR;
IF ^PARMSET(X) THEN NOTE('NO X ',0);
IF ^PARMSET(Y) THEN NOTE('NO Y ',8);
ANSWER('/* '!lX!!Yl!'C */'); %
END ABC;

A) ABC(,B);
B) ABC (A,B);
C) ABC(A,);
D) ABC Y(B)X(A);

2. The following code is supposed to find the earlier date between today's date and the given input date
which is input to the program in YYYYMMDD format. What happens when the program is executed?
DCL DATE JN CHAR (8);
DCL DAT E_MIN CHAR (8);
GET (DATE IN);
DATE_MIN = MIN(SUBSTR(DATETIME(), 1, 8), DATE IN);

A) The resulting value consists or a mixing or numbers and blanks.
B) The statement works as intended.
C) The resulting value consists of 8 blanks.
D) The program ends abnormally at runtime.

3. Given the following declarations, a list of 100 elements must be created so that the element created last
can be accessed as the first element of the list. A new element is always inserted in front of the element
created before. The variable NEXT in the last element should contain the value NULL. Which of the
following pieces of code implements this?
DCL 1 NODE BASED (ANCHOR).
2 NEXT POINTER,
2 DATA FIXED BIN(31);
DCL ANCHOR POINTER;
DCL P POINTER INIT (NULL());
DCL I FIXED BIN(31);

A) DO I = 1 TO 100; ALLOCATE NODE; NODE.DATA = I; NODE.NEXT = P; P = ANCHOR; END;
B) DO I = 1 TO 100; NODE.DATA = I; NODE.NEXT = P; ALLOCATE NODE; P = ANCHOR; END;
C) DO I = 1 TO 100; P = ANCHOR; ALLOCATE NODE; NOTE.DATA = I; NODE.NEXT = P; END;
D) I = 1 TO 100; ALLOCATE NODE; NODE.DATA = I; NODE.NEXT = P; ANCHOR = P; END;

4. What characterizes the relational model with respect to data storage?

A) Data is stored in variable length and format.
B) Data is stored in segments.
C) Data is stored in hierarchies.
D) Data is stored in tables.

5. Given the following code, which DECLARE should NOT be used for a reentrant program?
DCL B BIN FIXED(31,0) INIT (4); A = B;

A) DCL A BIN FIXED (31,0) INIT(0) STATIC;
B) DCL A BIN FIXED (31,0) INIT(0) AUTOMATIC;
C) DCL A BIN FIXED (31,0) INIT(0);
D) DCL A BIN FIXED (31,0);

Solutions: