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IBM C6030-041 Exam Syllabus Topics:

SectionObjectives
Topic 1: File and I/O Handling- Input/Output statements and file access
Topic 2: PL/I Fundamentals- Basic PL/I syntax and structure
- Data types and declarations
Topic 3: Control Structures- Conditional and iterative constructs
Topic 4: Advanced PL/I Topics- Pointers, bit manipulation and system constructs
Topic 5: Procedures and Functions- Procedure declaration and invocation

IBM Programming with IBM Enterprise PL/I Sample Questions:

CORRECT TEXT
What is the result, if any, of executing the following code?
DCL A CHAR(2) INIT(' ');
DCL B BIT(2) INIT('11'B);
A = ^B;

  • A. The value of A is unpredictable.
  • B. The value of A is'11'.
  • C. There is no result, because the syntax is wrong.
  • D. The value of A is '00'.
Reveal Solution  Discussion  0

Correct Answer: D  🗳️

CORRECT TEXT
Given the following code, what declaration of I will cause an infinite loop under default condition enablement?
DO I = 1 TO 99;

  • A. DCL I FIXED BIN (7);
  • B. DCL I PIC'99';
  • C. DCL I FLOAT;
  • D. DCL I FIXED BIN (3,0);
Reveal Solution  Discussion  0

Correct Answer: B  🗳️

CORRECT TEXT
Given the following declaration, the compiler will issue a warning message that says "INITIAL list for the array A contains only one item.".
Which INIT statement would correctly eliminate this message? DCL A(10) CHAR(5) INIT( '');

  • A. INIT( COPY(' '.10))
  • B. INIT( (10)(' '))
  • C. INIT(LOW(10))
  • D. INIT((10)' ')
Reveal Solution  Discussion  0

Correct Answer: B  🗳️

CORRECT TEXT
Given the following declarations, which code is always a valid way to test if the value of B is 7?
DCL P POINTER NIT (NULL());
DCL B FIXED BIN (31) BASED (P);

  • A. IF B ^ = 7 THEN; ELSE
    PUT ('OK');
  • B. IF P ^= NULL() ! B = 7
    THEN PUT ('OK');
  • C. IF B = 7
    THEN PUT ('OK');
  • D. IF ^ = NULL() THEN IF B = 7 THEN
    PUT ('OK');
Reveal Solution  Discussion  0

Correct Answer: D  🗳️

CORRECT TEXT
Given the following code, what is the value of X after execution of the assignment?
DCL X CHAR (5) lNlT ('ABCDE');
DCL Y CHAR (3) lNlT ('123');
X = Y

  • A. 'ABCDE'
  • B. '123' followed by blanks
  • C. '123' followed by 'DE'
  • D. '123' followed by low values
Reveal Solution  Discussion  0

Correct Answer: B  🗳️

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