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IBM C6030-041 Exam : Programming with IBM Enterprise PL/I

C6030-041 actual test
  • Exam Code: C6030-041
  • Exam Name: Programming with IBM Enterprise PL/I
  • Updated: Aug 29, 2026
  • Q & A: 146 Questions and Answers
  • PDF Demo
  • PC Test Engine
  • Online Test Engine
  • Total Price: $49.99  

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IBM C6030-041 Exam Syllabus Topics:

SectionWeightObjectives
Topic 1: PL/I Language Fundamentals25%- Scope and visibility
- Expressions and operators
- Data types and attributes
  • 1. Initialization and constants
    • 2. Storage classes and alignment
      • 3. Scalar and aggregate declarations
        Topic 2: Input/Output and File Handling15%- Environment options and z/OS integration
        - File declarations and attributes
        - Record and stream I/O
        Topic 3: Control Structures and Program Flow20%- Conditional statements
        - Procedures, functions, and entry points
        - Iteration and looping
        - Parameter passing mechanisms
        Topic 4: Data Structures, Pointers, and Memory Management20%- Controlled and automatic variables
        - Pointer and based variables
        - Dynamic storage allocation
        - Structures, unions, and arrays
        Topic 5: Compilation, Preprocessing, and Performance10%- Optimization and best practices
        - Compiler options and directives
        - Macro and SQL preprocessors
        Topic 6: Error Handling and Conditions10%- ON units and condition handling
        - Error recovery and program termination

        IBM Programming with IBM Enterprise PL/I Sample Questions:

        Question 1

        Given the following program, which subroutine improperly uses pointers to address memory and could lead to data corruption or an exception?

        A. SUB2
        B. SUB4
        C. PROC OPTIONS(MAIN);
        DCL
        1 X,
        2 X1 CHAR(76),
        2 X2 PIC'9999';
        DCL Y CHAR(20);
        CALL SUB1( ADDR(X));
        CALL SUB2( ADDR(X));
        CALL SUB3( ADDR(X));
        CALL SUB4( ADDR(X));
        SUB 1: PROC( P);
        DCL P POINTER;
        DCL
        1 XBASED(P),
        2 X1 CHAR(76),
        2 X2 PIC'9999'
        X = ";
        END;
        SUB2: PROC( P);
        DCL P POINTER;
        DCL
        1 X BASED(P),
        2 X 1 PIC'(76)X'
        2 X 2 PIC'9999'
        X = ";
        END;
        SUB3: PROC( P);
        DCL P POINTER;
        DCL X CHAR(60) BASED(P);
        X = ";
        END;
        SUB4: PROC( P );
        DCL P POINTER;
        DCLX CHAR(100) BASED(P);
        X = ";
        END; END;
        D. SUB1
        E. SUB3


        Question 2

        CORRECT TEXT
        Under default condition enablement, what is the result of executing the following code?
        DCL A CHAR(5);
        A = '123ABC';

        A. A will have a value '23ABC'.
        B. STRINOSIZE would be raised.
        C. A will have a value '123AB'.
        D. CONVERSION would be raised.


        Question 3

        CORRECT TEXT
        Given the following code, what will be output?
        MP: PROC OPTIONS(MAIN);
        DCL A CHAR(1) INIT('A');
        DCL B CHAR(1) INIT('B');
        DCL C CHAR(1) STATIC INIT('C');
        CALL SR1(A);
        PUT SKIP LIST(A!!B!!C);
        SR1: PROC(A);
        DCL A CHAR(1);
        DCL B CHAR(1);
        DCL C CHAR(1);
        A = '1'
        B = '2'
        C = '3';
        END SR1;
        END;

        A. 1B3
        B. 123
        C. 12C
        D. 1BC


        Question 4

        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. 62
        C. 72
        D. 22


        Question 5

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


        Solutions:

        Question 1
        Answer: E
        Question 2
        Answer: C
        Question 3
        Answer: D
        Question 4
        Answer: C
        Question 5
        Answer: C

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