What is the no. of bytes occupied for a variable if it’s defined as PIC S9(7) COMP-3 and PIC S9(10) COMP
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2 and 3
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4 and 5
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7 and 6
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none of the above
PIC S9(7) COMP-3 (packed decimal) uses (7/2)+1 = 4 bytes (rounded up). PIC S9(10) COMP (binary) typically uses 4 bytes (fullword). So the answer is 4 and 4, which is 'none of the above'. COMP-3 stores 2 digits per byte plus sign byte; COMP uses binary representation.
To determine the number of bytes occupied for a variable in COBOL, you need to consider the data type and the storage format.
In COBOL, the PIC clause is used to define the data type and format of a variable. The S9 format is used for signed numeric variables, and the COMP-3 format is used for packed decimal variables.
For the given variable definitions:
PIC S9(7) COMP-3:
- The S9(7) indicates a signed numeric variable with a maximum of 7 digits.
- The COMP-3 format represents packed decimal, which stores each digit using half a byte (nibble).
- For a packed decimal variable, each digit requires 4 bits (half a byte) of storage.
- The number of digits in the variable (7) multiplied by the storage required per digit (4 bits) gives us the total storage required.
- Therefore, the total storage required for this variable is 7 * 4 bits = 28 bits, which is equivalent to 4 bytes.
PIC S9(10) COMP:
- The S9(10) indicates a signed numeric variable with a maximum of 10 digits.
- The COMP format represents binary storage, which uses full bytes to store the data.
- For a binary variable, each byte can store 8 bits.
- The number of digits in the variable (10) multiplied by the storage required per digit (1 byte) gives us the total storage required.
- Therefore, the total storage required for this variable is 10 * 1 byte = 10 bytes.
Based on the above explanations, the correct answer is D) none of the above, as the given options do not match the correct number of bytes occupied for the variables.