U UNIVAC 1218 - Bitsavers

different manner. The differences are illustrated by the example shown below
for a 3-word input transfer on channel 02. Assume that the same memory area
(addresses 10000, 10001, and 10002) is to be used for buffer storage. The
input buffer control word locations for channel 02 are 00064 and 00065. The
coding required to initiate the input buffer is as follows:
Address n-l - previous instruction.
Address n - 501102 (I/O instruction).
Address n+l - 010002 (buffer terminal address).
Address n+l - 010000 (buffer initial address).
Address n+3 - Next instruction.
When the computer executes the I/O instruction 501102, channel 02 is activated
and the buffer limits are transferred "from addresses n+l and n+2 to the input
buffer control word locations for channel 02 (addresses 00064 and 00065). (Note
that in this case the buffer terminal address is specified as the address of
the last buffer wordo) The I/O section then assumes control of the input operation
and the program proceeds to the instruction located at address n+3. The
manner in which the I/O section executes the input buffer is illustrated by the
sequence of actions shown below.
I/O Action State of Buffer Control Words
Honor 10 request
Read and compare control words
Increment (000065) by one
Set first word, (010000), on data lines
Set 10 acknowledge line
Honor 10 request
Read and compare control words
Increment (000065) by one
Set second word, (010001), on data lines
Set 10 acknowledge line
Honor 10 request
Read and compare control words
Clear active flip-flop for channel 02
Increment (000065) by one
Set third word, (010002), on data lines
Set 10 acknowledge line
Terminate buffer
Note that the differences are as follows:
(000064) = 010002
(000065) = 010000
010001 --+ 000065
(000064) = 010002
(000065) = 010001
010002 ----.000065
(000064) = 010002
(000065) = 010002
010003 ---+ 000065
(000064) = 010002
(000065) = 010003
1) The terminal address control word must initially be equal to the address of
the last buffer word rather than one greater than the address of the last
buffer word (one less for decrementing buffer).
2) The active flip-flop is cleared immediately after equality of buffer control
words but the sequence continues in order to complete the transfer
of the last buffer word.
3) After termination the current address control word is one greater than the
terminal address control word (one less for decrementing buffer).
T ,.. , ....
.l-v-.l0