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IBM C9050-042 Exam Syllabus Topics:

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

IBM Developing with IBM Enterprise PL/I Sample Questions:

1. Which compiler option causes the compiler to flag any IF, WHILE, UNTIL and WHEN clauses that do not
have the attributes BIT(1) NONVARYING?

A) RULES(NOLAXWHEN)
B) RULES(NOLAXUNTIL)
C) RULES(NOLAXWHILE)
D) RULES(NOLAXIF)


2. Prerequisite:
A sorted input dataset with record length 100 contains at least one record for each of the values '1', '2', '3'
in the first byte. The applied sort criteria is 1,100,ch,a.
Requirements:
1 .) All records with '1' in the first byte must be ignored.
2 .) All records with '2' in the first byte must be written to the output dataset.
3 .) If there is a '3' in the first byte, the read iteration must be left.
4 .) The program must not abend or loop infinitely.
If the code below does not fulfill the specifications provided above, which of the following is the most likely
reason?
DCL DDIN FILE RECORD INPUT;
DCL DDOUT FILE RECORD OUTPUT;
DCL 1 INSTRUC,
3 A CHAR(1),
3 * CHAR(99);
DCL EOF_IN BIT(1) INIT('0'B);
DCL (Z1,Z2,Z3,ZO) BIN FIXED(31) INIT(0);
ON ENDFILE(DDIN) EOF_IN = '1'B;
READ FILE(DDIN) INTO (INSTRUC);
LOOP: DO WHILE (^EOF_IN);
SELECT(INSTRUC.A);
WHEN('1') DO;
Z1 +-= Z1;
ITERATE LOOP;
END;
WHEN('3') DO;
Z3 = Z3+1;
LEAVE LOOP;
END;
WHEN('2') DO;
Z2 = Z2+1;
WRITE FILE(DDOUT) FROM(INSTRUC);
END;
OTHER DO;
ZO = ZO+1;
PUT SKIP LIST(INSTRUC.A);
END;
END;/*select*/
READ FILE(DDIN) INTO(INSTRUC);
END ;/*loop*/

A) The code fulfills the requirement.
B) The code does not fulfill the requirement because not all records with '2' in the first byte will be written to the output dataset.
C) The code does not fulfill the requirement because the last record with '2' in the first byte will be written twice to the output dataset.
D) The code does not fulfill the requirement because the program will loop infinitely.


3. How much storage in bytes will be allocated for the following declaration?
DCL 1 S(10) ALIGNED,
2 A BIN FIXED(15),
2 B DEC FIXED(3),
2 C CHAP(3);

A) 80
B) 69
C) 79
D) 70


4. Given the following code, which procedure call would cause the "Pointer corrupted!" message to be output
provided that the PL/I compile time option CHECK(STORAGE) has been specified?
DCLX FIXED BIN(31) BASED (P);
DCL Y CHAR (5) BASED (P);
DCL P POINTER;
SUB: PROCEDURE (P);
DCL P POINTER;
if CHECKSTG(P) THEN PUT ('Pointer 0k!');
ELSE PUT ('Pointer corrupted!');
END;

A) P = ALLOCATE (100);
Y = 'ABCDE';
CALL SUB (ADDR(X));
B) ALLOCATEY;
Y = 'ABCDE';
CALL SUB (ADDR(X));
C) ALLOCATEX;
Y = 'ABCD';
CALL SUB (P);
D) ALLOCATEY;
Y = 'ABCDE';
CALL SUB (P);


5. Given the lollowing declarations, what statement will raise STRINGSIZE condition if enabled?
DCLA_STR CHAR (100) VARYING;
DCLB_STR CHAR(10) STATIC;
DCL C_STR CHAR (100);

A) C_STR = A_STR;
B) C_STR = B_STR;
C) A_STR = B_STR;
D) SUBSTR(C_STR, 92) = B_STR;


Solutions:

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

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