C9050-042 無料問題集「IBM Developing with IBM Enterprise PL/I」
Requirement Copy a dataset of record length 100 to another dataset.
If the following code does not fulfill the requirement above, which is the most likely reason?
DCL DDIN FILE RECORD INPUT;
DCL DDOUT FILE RECORD OUTPUT;
DCL INSTRUC CHAR(100);
DCL EOF_IN BIT(1) INIT('0'B);
ON ENDFILE(DDIN) EOF_IN = '1'B;
READ FILE(DDIN) INTO(INSTRUC);
DO WHlLE(^EOF_IN);
WRITE FILE(DDOUT) FROM(INSTRUC);
READ FILE(DDIN) INTO(INSTRUC);
WRITE FILE(DDOUT) FROM(INSTRUC);
END;
If the following code does not fulfill the requirement above, which is the most likely reason?
DCL DDIN FILE RECORD INPUT;
DCL DDOUT FILE RECORD OUTPUT;
DCL INSTRUC CHAR(100);
DCL EOF_IN BIT(1) INIT('0'B);
ON ENDFILE(DDIN) EOF_IN = '1'B;
READ FILE(DDIN) INTO(INSTRUC);
DO WHlLE(^EOF_IN);
WRITE FILE(DDOUT) FROM(INSTRUC);
READ FILE(DDIN) INTO(INSTRUC);
WRITE FILE(DDOUT) FROM(INSTRUC);
END;
正解:B
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Given the following pseudocode inside a loop, where should the COMMIT action be placed assuming that
there is always a one-to-many relationship between header and detail tables?
1 Find header row 2 IF found 3 Update header row 4 Find detail row 5 IF found THEN 6 Update detail
rows 7 ELSE 8 ENDIF 9 ENDIF
there is always a one-to-many relationship between header and detail tables?
1 Find header row 2 IF found 3 Update header row 4 Find detail row 5 IF found THEN 6 Update detail
rows 7 ELSE 8 ENDIF 9 ENDIF
正解:C
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A lead developer reviewing code from one of the programmers found the following code. What should the
programmer be told about BY NAME assignments?
DCLl STR1, 3FB15 FIXED BIN (15), 3 CH10 CHAR (10), 3 NAME CHAR (25). 3 ADDR CHAR (30), 3
FB31 FIXED BIN(31);
DCL 1 STR2, 3 FB31 FIXED BIN (31), 3 FB31B FIXED BIN (31), 3 NAME CHAR (20), 3 ADDR CHAR
(30), 3 CH20 CHAR (20);
STR2 = STR1, BY NAME;
programmer be told about BY NAME assignments?
DCLl STR1, 3FB15 FIXED BIN (15), 3 CH10 CHAR (10), 3 NAME CHAR (25). 3 ADDR CHAR (30), 3
FB31 FIXED BIN(31);
DCL 1 STR2, 3 FB31 FIXED BIN (31), 3 FB31B FIXED BIN (31), 3 NAME CHAR (20), 3 ADDR CHAR
(30), 3 CH20 CHAR (20);
STR2 = STR1, BY NAME;
正解:A
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