Answer
Jun 05, 2012 - 07:18 PM
Within ISO 8583, a bitmap is a field or subfield within a message which indicates which other data elements or data element subfields may be present elsewhere in a message.
A message will contain at least one bitmap, called the Primary Bitmap which indicates which of Data Elements 1 to 64 are present. A secondary bitmap may also be present, generally as data element one and indicates which of data elements 65 to 128 are present. Similarly, a tertiary, or third, bitmap can be used to indicate the presence or absence of fields 129 to 192, although these data elements are rarely used.
The bitmap may be transmitted as 8 bytes of binary data, or as 16 hexadecimal characters 0-9, A-F in the ASCII or EBCDIC character sets.
A field is present only when the specific bit in the bitmap is true. For example, byte '82x is binary '1000 0010' which means fields 1 and 7 are present in the message and fields 2, 3, 4, 5, 6, and 8 are not present.
Examples
Bitmap Defines presence of
4210001102C04804 Fields 2, 7, 12, 28, 32, 39, 41, 42, 50, 53, 62
7234054128C28805 Fields 2, 3, 4, 7, 11, 12, 14, 22, 24, 26, 32, 35, 37, 41, 42, 47, 49, 53, 62, 64
8000000000000001 Fields 1, 64
0000000000000003
(secondary bitmap) Fields 127, 128
Explanation of Bitmap (8 BYTE Primary Bitmap = 64 Bit) field 4210001102C04804
BYTE1 : 01000010 = 42x (counting from the left, the second and seventh bits are 1, indicating that fields 2 and 7 are present)
BYTE2 : 00010000 = 10x (field 12 is present)
BYTE3 : 00000000 = 00x (no fields present)
BYTE4 : 00010001 = 11x (fields 28 and 32 are present)
BYTE5 : 00000010 = 02x (field 39 is present)
BYTE6 : 11000000 = C0x (fields 41 and 42 are present)
BYTE7 : 01001000 = 48x (fields 50 and 53 are present)
BYTE8 : 00000100 = 04x (field 62 is present)
0________10________20________30________40________50________60__64
1234567890123456789012345678901234567890123456789012345678901234 n-th bit
0100001000010000000000000001000100000010110000000100100000000100 bit map
Fields present in the above variable length message record:
2-7-12-28-32-39-41-42-50-53-62
A message will contain at least one bitmap, called the Primary Bitmap which indicates which of Data Elements 1 to 64 are present. A secondary bitmap may also be present, generally as data element one and indicates which of data elements 65 to 128 are present. Similarly, a tertiary, or third, bitmap can be used to indicate the presence or absence of fields 129 to 192, although these data elements are rarely used.
The bitmap may be transmitted as 8 bytes of binary data, or as 16 hexadecimal characters 0-9, A-F in the ASCII or EBCDIC character sets.
A field is present only when the specific bit in the bitmap is true. For example, byte '82x is binary '1000 0010' which means fields 1 and 7 are present in the message and fields 2, 3, 4, 5, 6, and 8 are not present.
Examples
Bitmap Defines presence of
4210001102C04804 Fields 2, 7, 12, 28, 32, 39, 41, 42, 50, 53, 62
7234054128C28805 Fields 2, 3, 4, 7, 11, 12, 14, 22, 24, 26, 32, 35, 37, 41, 42, 47, 49, 53, 62, 64
8000000000000001 Fields 1, 64
0000000000000003
(secondary bitmap) Fields 127, 128
Explanation of Bitmap (8 BYTE Primary Bitmap = 64 Bit) field 4210001102C04804
BYTE1 : 01000010 = 42x (counting from the left, the second and seventh bits are 1, indicating that fields 2 and 7 are present)
BYTE2 : 00010000 = 10x (field 12 is present)
BYTE3 : 00000000 = 00x (no fields present)
BYTE4 : 00010001 = 11x (fields 28 and 32 are present)
BYTE5 : 00000010 = 02x (field 39 is present)
BYTE6 : 11000000 = C0x (fields 41 and 42 are present)
BYTE7 : 01001000 = 48x (fields 50 and 53 are present)
BYTE8 : 00000100 = 04x (field 62 is present)
0________10________20________30________40________50________60__64
1234567890123456789012345678901234567890123456789012345678901234 n-th bit
0100001000010000000000000001000100000010110000000100100000000100 bit map
Fields present in the above variable length message record:
2-7-12-28-32-39-41-42-50-53-62
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