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IBM C9050-041 Exam Syllabus Topics:
| Section | Objectives |
|---|---|
| Topic 1: Control Structures and Statements | - Procedure and scope control - IF, SELECT and other control flow |
| Topic 2: PL/I Data Types and Declarations | - Variable declaration and data representation - Pointer and based variables |
| Topic 3: Program Structure and Storage | - FETCH/RELEASE behavior - Static, automatic and controlled storage |
| Topic 4: Expressions and Operations | - Arithmetic and string operations - Condition handling and conversions |
| Topic 5: Input/Output and File Handling | - READ and WRITE file operations - File attribute and format |
IBM Programming with IBM Enterprise PL/I Sample Questions:
CORRECT TEXT
What is the result, if any, of executing the following code?
DCL A BIT(1) INIT('0'B);
DCL B BIT(1) INIT('0'B);
DCL C BIT(1) INIT('1'B);
A = B = C;
- A. The value of A is unpredictable.
- B. There is no result because the syntax is wrong.
- C. The value of A is'1'S.
- D. The value of A is 'D'S.
Correct Answer: D 🗳️
CORRECT TEXT
What happens, if anything, after end of rile has been reached in the following code?
DCL INF FILE RECORD INPUT;
DCL INFIELD CHAR(100) BASED(P);
DCL P PTR;
DCL EOF BIT(1) INIT('0'B);
ON ENDFILE(INF) BEGIN;
EOF = '16;
ALLOC INFIELD;
INFIELD = 'EOF REACHED';
END;
OPEN FILE(INF);
READ FILE(INF) SET(P);
DO WHILE(^EOF);
READ FILE(INF) SET(P);
END;
- A. Runtime error because there
SE statement. - B. There will be an infinite
loop. - C. INFIELD will have a value 'EOF REACHED' and the program ends.
- D. Nothing because end of file will never be reached.
Correct Answer: C 🗳️
CORRECT TEXT
What is the value of B after executing the following code?
DCL A CHAR(10) INIT('12A4BABCAB'); DCL B BIN FIXED(31) INIT(0); B = INDEX(A,'AB');
- A. 2
- B. 9
- C. 6
- D. 3
Correct Answer: C 🗳️
CORRECT TEXT
What is the most appropriate declaration for the variable X?
X= 1,123E+4;
- A. DCL X CHAR(4);
- B. DCL X DEC FLOAT(16);
- C. DCL X DEC FIXED(15,3);
- D. DCL X PlC '9999';
Correct Answer: B 🗳️
CORRECT TEXT
Which is the impact, if any, of LIKE in the following code?
DCL
1 XY
2 A CHAR(4),
2 B BIN FIXED(31);
DCL 1 YZ LIKE XY;
- A. YZ is based on the
structure XY. - B. YZ is exactly the same structure as XY but with its own storage.
- C. There is no impact, because LIKE is a syntax error.
- D. XY.A is always the same as YZ.
Correct Answer: B 🗳️
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