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Programming with IBM Enterprise PL/I

Question No: 1 – (Topic 1)

What is the most appropriate declaration for the variable X?

X = 1.123E 4;

  1. DCL X CHAR(4);

  2. DCL X PIC #39;9999#39;;

  3. DCL X DEC FLOAT(16);

  4. DCL X DEC FIXED(15,3);

Answer: C

Question No: 2 – (Topic 1)

Which of the following describes when a program is NOT FETCHABLE?

  1. When the main program need not be recompiled to reflect the changes made in the called program

  2. When the called program will be loaded from the library at execution time

  3. When the called program is part of the main program in the load module

  4. When two main programs referring to the called program at the same time cannot have different versions of the called program

Answer: C

Question No: 3 – (Topic 1)

What code needs to be executed, if any, before the variable A can be successfully accessed?

DCL X PTR;

DCL B CHAR(100) INIT(#39; #39;); DCL A CHAR(100) BASED(X);

  1. A can be accessed without any further action.

  2. X = ADDR(B);

  3. A = B;

  4. X = NULL();

Answer: B

Question No: 4 – (Topic 1)

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(#39;ABCD#39;);

A = C;

B = LENGTH(A);

  1. 10

  2. 7

  3. 5

  4. 4

Answer: C

Question No: 5 – (Topic 1)

Given the following code, what will be output?

MP: PROC OPTIONS(MAIN); DCL A CHAR(1) INIT(#39;A#39;);

DCL B CHAR(1) INIT(#39;B#39;);

DCL C CHAR(1) STATIC INIT(#39;C#39;); 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 = #39;1#39;;

B = #39;2#39;;

C = #39;3#39;; END SR1;

END;

  1. 1B3

  2. 1BC

C. 123

D. 12C

Answer: B

Question No: 6 – (Topic 1)

What happens to the STATIC variables in the program U, if anything, after executing the following code?

FETCH U; CALL U; RELEASE U; FETCH U;

  1. STATIC variables cannot be used in program U.

  2. STATIC variables will have the values from the last time U was called.

  3. STATIC variables will have their INITIAL values.

  4. Nothing because there is a syntax error.

Answer: C

Question No: 7 – (Topic 1)

What is the result of executing the following code?

DCL A CHARACTER (4) INIT(#39;10.5#39;); DCL B DEC FIXED(7,1) INIT(10.5);

B = A B;

  1. CONVERSION is raised.

  2. ERROR is raised.

  3. No condition is raised and the value of B is 21.

  4. No condition is raised and the value of B is 20.5

Answer: D

Question No: 8 – (Topic 1)

What code must be added after EOF = #39;1#39;B, if any, to print #39;EOF REACHED#39;?

DCL INF FILE RECORD INPUT;

DCL INFIELD CHAR(100) BASED(P); DCL P PTR;

DCL EOF BIT(1) INIT(#39;0#39;B);

ON ENDFILE(INF) BEGIN; EOF = #39;1#39;B;

PUT SKIP LIST(INFIELD); END;

OPEN FILE(INF);

READ FILE(INF) SET(P);

DO WHILE(^EOF); READ FILE(INF) SET(P); END;

  1. ALLOC INFIELD; INFIELD = #39;EOF REACHED#39;;

  2. INFIELD = #39;EOF REACHED#39;;

  3. It cannot be printed, as it is not sure if INFIELD contains the last record

  4. There is a syntax error.

Answer: A

Question No: 9 – (Topic 1)

Given the following piece of code, how many times is the loop executed?

DCL I FIXED BIN (31); I = 20;

DO UNTIL (I = 0); PUT (I);

I = I – 3; END;

  1. less than 6 times

  2. 6

  3. 7

  4. more than 7 times

Answer: D

Question No: 10 – (Topic 1)

Which is the impact, if any, of LIKE in the following code?

DCL 1 XY,

2 A CHAR(4),

2 B BIN FIXED(31);

DCL 1 YZ LIKE XY;

  1. XY.A is always the same as YZ.A.

  2. YZ is exactly the same structure as XY but with its own storage.

  3. YZ is based on the structure XY.

  4. There is no impact, because LIKE is a syntax error.

Answer: B

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