Instantly download C9050-042 updated real questions
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Updated: Jun 02, 2026
No. of Questions: 140 Questions & Answers with Testing Engine
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IBM Developing with IBM Enterprise PL/I Sample Questions:
1. Which of the following structures will NOT contain padding bytes if the PL/I default for alignment is
applied?
A) DCL 1 A, 2 B FLOAT DEC (16), 2 F FLOAT DEC (16), 2 E FIXED BIN (31), 2 C FIXED DEC (5,2), 2 D
CHAR (3);
B) DCL 1 A, 2 E FIXED BIN (31), 2 C FIXED DEC (5,2), 2 B FLOAT DEC (16), 2 F FLOAT DEC (16), 2 D
CHAR (3);
C) DCL 1 A, 2 B FLOAT DEC (16), 2 C FLOAT DEC (5,2), 2 D CHAR (3), 2 E FIXED BIN (31), 2 F FLOAT
DEC (16);
D) DCL 1 A, 2 B FLOAT DEC (16), 2 F FLOAT DEC (16), 2 C FIXED DEC (5,2), 2 E FIXED BIN (31), 2 D
CHAR (3);
2. Given the following declaration, how many bytes will be occupied by this structure under the default
alignment rules?
DCL 1 A DIM (10),
2 B FIXED BIN(31),
2 C CHAR (1)VAR;
A) 77
B) 79
C) 80
D) 70
3. Given the following code, which statement will produce an E level preprocessor message?
%DCL ABC ENTRY;
%ABC: PROC (X,Y) STATEMENT RETURNS(CHAR); DCL(XY) CHAR;
IF ^PARMSET(X) THEN NOTE('NO X ',0);
IF ^PARMSET(Y) THEN NOTE('NO Y '8');
RETURN('/* '!!X!!YI!'C */');
%END ABC;
A) ABC Y(B)X(A);
B) ABC (A,B);
C) ABC (A,);
D) ABC(,B);
4. A programmer has submitted the following declaration for review. What feedback should be provided to
the programmer?
DCL 1 A,
2 B DIM (1000) FIXED BIN (31) INIT (0),
2 C DIM (1000) FIXED BIN(15) INIT (0);
A) The code is good as written.
B) Discuss with the programmer how many elements of the arrays need to be initialized.
C) A is incorrectly initialized and the code must be changed.
D) The declaration of A should be changed because the current declaration contains padding bytes.
5. Requirement:
If the value of the numeric variable I is 1 it needs to be changed to 2 and vice versa. In all other cases it
must remain unchanged. Which of the following solutions meets the requirement and does not require
essential structural modifications when the requirement is changed to the following: If the value of the
numeric variable I is 512 it needs to be changed to 731 and if the value is 814 it needs to be changed to 5.
In all other cases it must be set to 111.
A) lF I = 1 ! 1 = 2
THEN I = 3 - I;
B) SELECT (I);
WHEN(1) I = 2;
WHEN(2) I = 1;
OTHER;
END;
C) IF I = 1 THEN I = 2;
IF I = 2 THEN I = 1;
D) DCL ONETWO(2) BIN FIXED(15) INIT(2,1);
IF I = 1! I = 2
THEN I = ONETWO(I);
Solutions:
| Question # 1 Answer: A | Question # 2 Answer: B | Question # 3 Answer: C | Question # 4 Answer: B | Question # 5 Answer: B |
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