can we declare s9(9)v9(9) in cobol ? if yes how many bytes
it will occupy ?(urgent plz answer it)
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Answer Posted / vikash
we can declare.It will take total bytes based on formula=(n/2+1).
Thus, 9+9=18
Here n=18,
Thus number of bytes=18/2+1=10 bytes
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I have program P1 which calls file F1 which has 100 records and following structure 001 .................. 002 .................. 003 .................. 098 .................... 099 ................... 100 .................... Now I want to read these files and write these records in file F2 in following manner. 001 ...... 051 ..... 002 ...... 052 ..... 003 ...... 053 ..... .......... ....... .......... ....... .......... ....... 048 ........ 098 ...... 049 .......... 099 ....... 050 .... 100 ......
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