IBM Developing with IBM Enterprise PL/I : C9050-042

C9050-042 real exams

Exam Code: C9050-042

Exam Name: Developing with IBM Enterprise PL/I

Updated: Oct 10, 2026

Q & A: 140 Questions and Answers

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IBM C9050-042 Exam Syllabus Topics:

SectionObjectives
Topic 1: Data Processing- Character and String Handling
  • 1. Pattern and conversion operations
  • 2. Substring processing
  • 3. Built-in string functions
- Arrays and Structures
  • 1. Structure definitions
  • 2. Array manipulation
  • 3. Based and controlled storage
Topic 2: File and I/O Processing- Input and Output Formatting
  • 1. GET and PUT statements
  • 2. Formatting data
- File Operations
  • 1. Sequential and direct access files
  • 2. OPEN, READ, WRITE, and CLOSE statements
  • 3. Record processing
Topic 3: PL/I Language Fundamentals- Data Types and Declarations
  • 1. Scalar and aggregate data
  • 2. Attributes and storage classes
  • 3. Initialization and constants
- Expressions and Operators
  • 1. String operations
  • 2. Arithmetic expressions
  • 3. Logical and relational operators
Topic 4: Program Structure and Control- Procedures and Blocks
  • 1. Parameter passing
  • 2. Main and subprocedures
  • 3. Scope and declarations
- Control Flow
  • 1. Iterative processing
  • 2. Conditional statements
  • 3. Branching and exception handling
Topic 5: IBM Enterprise PL/I Features- Compiler and Runtime Features
  • 1. Optimization and diagnostics
  • 2. Compiler options
  • 3. Debugging techniques
- Integration Technologies
  • 1. Interoperability with C and Java
  • 2. XML support
  • 3. JSON support

IBM Developing with IBM Enterprise PL/I Sample Questions:

Question #1

A dummy argument is NOT created when a routine is called in which of the following situations?

  • A. With a constant as parameter
  • B. With the attributes(s) of the arguments and parameters are not the same
  • C. Using SUBSTR as one of the arguments
  • D. Where the attributes(s) of the arguments and parameters are the same
Reveal Solution  Discussion  0

Correct Answer: D  🗳️

Question #2

What happens to STATIC variables declared in a procedure when the procedure is called recursively?

  • A. The values of STATIC variables from the previous invocation are saved and are available when the
    current invocation ends.
  • B. STATIC variables are allocated and initialized at each new invocation.
  • C. The values of STATIC variables from the previous invocation are lost.
  • D. STATIC variables are allocated only once and can, therefore, be used for communication between
    invocations.
Reveal Solution  Discussion  0

Correct Answer: D  🗳️

Question #3

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 C FLOAT DEC (5,2), 2 D CHAR (3), 2 E FIXED BIN (31), 2 F FLOAT
    DEC (16);
  • B. 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);
  • C. 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);
  • 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);
Reveal Solution  Discussion  0

Correct Answer: B  🗳️

Question #4

Requirement:
All the characters of the CHAR(3) variable X must be tested to be numeric. Which of the following
solutions meets the requirement and does not require essential structural modifications when the
requirement is changed to the following: The first character of the CHAR(3) variable X must be tested to
be uppercase alphabetic, while the two other characters must be tested to be numeric.

  • A. DCL NUM CHAR(10) VALUE('0l 23456789');
    IF VERIFY(X,NUM) = 0
    THEN ... /*NUMERIC*/
  • B. DCL ALPHA CHAR(26) VALUE('ABCDEFGHIJKLMNOPQRSTUVWXYZ');
    DCL NUM CHAR(10) VALUE('0123456789');
    IF TRANSLATE(X,(26)'A'!I(10)'9'ALPHA!!NUM) = '999'
    THEN ... ; /*NUMERIC*/
  • C. DCL Y PIC'999';
    DCL SWITCH BIT(1) INIT('1'B);
    ON CONVERSION BEGIN;
    SWITCH =
    ONSOURCE = '000';
    END;
    Y = X;
    IF SWITCH
    THEN ... ; /*NUMERIC*/
  • D. IFX >= '000' & X <= '999'
    THEN ... /*NUMERIC*/
Reveal Solution  Discussion  0

Correct Answer: B  🗳️

Question #5

Requirement:
The function LEAPYEAR evaluates a given 4-digit number and returns '1'B if it is a leap year, '0'B if it is
not. This function is supposed to work for the years 2004 to 2015.
Leap years occur every four years, except for years ending in 00 that are not divisible by 400. Which of
the following solutions meets the requirement and does NOT need to be changed if the requirement
changes to: The function is supposed to work for the years 1900 to 3000.

  • A. LEAPYEAR:PROC(YEAR) RETURNS(BIT(1));
    DCL YEAR PlC '9999';
    DCL (MOD,VERIFY) BUILTIN;
    SELECT;
    WHEN (VERIFY(YEAR '0123456769') ^= 0) RETURN('0'B); WHEN (MOD(YEAR,100) = 0)
    RETURN('0'B);
    WHEN (MOD(YEAR,400) = 0) RETURN('1'B);
    WHEN (MOD(YEAR,4) = 0) RETURN('1'B);
    OTHERRETURN('0'B);
    END;
    END LEAPYEAR;
  • B. LEAPYEAR: PROC(YEAR) RETURNS(BIT(1));
    DCL YEAR PlC '9999';
    DCL VERIFY BUILTIN;
    IFVERIFY(YEAR,0123456789)^= 0 THEN RETURN('0'B);
    SELECT(YEAR);
    WHEN (2004) RETURN('1'B);
    WHEN (2008) RETURN('1'B);
    WHEN (2012) RETURN('1'B);
    OTHER RETURN('0'B);
    END;
    END LEAPYEAR;
  • C. LEAPYEAR: PROC(YEAR) RETURNS(BIT(1));
    DCL YEAR PlC '9999';
    DCL (MOD,VERIFY) BUILTIN;
    SELECT;
    WHEN (VERIFY(YEAR,'0l23456789') ^= 0) RETURN('0'B);
    WHEN (MOD(YEAR,100) = 0)RETURN('0'B);
    WHEN (MOD(YEAR,4) = 0)RETURN('1'B);
    OTHERRETURN('0'B);
    END;
    END LEAPYEAR;
  • D. LEAPYEAR:PROC(YEAR) RETURNS(BIT(1));
    DCL YEAR PlC '9999';
    DCL (MOD,VERIFY) BUILTIN;
    SELECT;
    WHEN (VERIFY(YEAR '0123456789') ^= 0) RETURN('0'B);
    WHEN (MOD(YEAR,400) = 0) RETURN('l'B); WHEN (MOD(YEAR,100) = 0) RETURN('0'B);
    WHEN (MOD(YEAR,4) = 0) RETURN('1'B); OTHER RETURN('0'B);
    END;
    END LEAPYEAR;
Reveal Solution  Discussion  0

Correct Answer: D  🗳️

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