To answer this question, we need to understand the storage format of the PIC s9(6)v99 comp-3 data type.
The PIC s9(6)v99 comp-3 data type is a packed decimal representation. Packed decimal uses a specific encoding scheme to store decimal numbers in a compact manner.
In the given data type, s9(6)v99 represents a signed decimal number with 6 digits to the left of the decimal point and 2 digits to the right of the decimal point.
In the comp-3 format, each digit is stored in half a byte (4 bits), with the last half-byte reserved for the sign. The sign half-byte uses the following encoding scheme:
- Positive sign (+) is represented by the value 0x0C.
- Negative sign (-) is represented by the value 0x0D.
Therefore, each digit requires 4 bits of storage, and the sign requires an additional half-byte (4 bits).
For the given data type PIC s9(6)v99 comp-3, we have:
- 6 digits to the left of the decimal point, requiring 6 * 4 = 24 bits or 3 bytes.
- 2 digits to the right of the decimal point, requiring 2 * 4 = 8 bits or 1 byte.
- 1 half-byte for the sign.
So, the total storage required for the PIC s9(6)v99 comp-3 data type is 3 bytes + 1 byte + 0.5 byte = 4.5 bytes.
Therefore, the statement "If a field is defined as Pic s9(6)v99 comp-3, it uses 5 bytes of storage" is incorrect.
The correct answer is B) False.