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IBM C6030-041 Exam Syllabus Topics:
| Section | Objectives |
|---|---|
| Topic 1: Input/Output and File Handling | - File attribute and format - READ and WRITE file operations |
| Topic 2: PL/I Data Types and Declarations | - Pointer and based variables - Variable declaration and data representation |
| Topic 3: Control Structures and Statements | - IF, SELECT and other control flow - Procedure and scope control |
| Topic 4: Program Structure and Storage | - Static, automatic and controlled storage - FETCH/RELEASE behavior |
| Topic 5: Expressions and Operations | - Arithmetic and string operations - Condition handling and conversions |
IBM Programming with IBM Enterprise PL/I Sample Questions:
CORRECT TEXT
Given the following code, what can be said about the scope of the variables in procedure P?
P: PROCEDURE;
B: BEGIN;
DCL K FIXED BIN (15);
END B;
D: DO;
DCL S CHAR (10);
END D;
END P;
- A. Variable K is known in the entire procedure.
- B. Variable K is known in group
- C. Variable S is not known in block B.
- D. Variable S is known in the entire procedure.
Correct Answer: D 🗳️
CORRECT TEXT
Given the following code, what value will be output?
TEST: PROC OPTIONS(MAIN);
DCL
P POINTER,
N1 FIXED BIN(31),
1 A BASED(P),
2 A1 FIXED BIN(3i),
2 A2 FIXED BIN(3i),
2 A3 FIXED BIN(3i),
2 A4( N1 REFER(A3)) CHAR(10) VAR;
N1 = 5;
ALLOC A;
PUT SKIP LIST( STG(A)); END;
- A. 24
- B. 22
- C. 62
- D. 72
Correct Answer: D 🗳️
CORRECT TEXT
Which is the most appropriate code to turn all of the bits in A ON?
DCL A BIT(8);
- A. A = 11111111B;
- B. A = 255;
- C. A = -1;
- D. A = '11111111'B;
Correct Answer: D 🗳️
CORRECT TEXT
What will be printed to SYSPRINT, if anything, after executing the following code?
DCL A PlC '9999'.
A =' 123';
PUT SKIP LIST('VALUE OF A lS:' !! A);
- A. No value will be displayed because the assignment statement will cause a conversion error.
- B. VALUE OF A IS: 123
- C. VALUE OF A lS: 123
- D. VALUE OF A lS :0123
Correct Answer: C 🗳️
CORRECT TEXT
What is the value of B after executing the following code?
DCL A CHAR(10) VAR;
DCL B BIN FIXED(31) INIT(0);
DCL C CHAR(5) INIT('ABCD');
A = C;
B = LENGTH(A);
- A. 10
- B. 5
- C. 7
- D. 4
Correct Answer: B 🗳️
