Programming instructions

Since bit 0 of a word is used for the sign of a number, the largest positive
number that can be represented is 217-l. If, in l’s complement addition,
the addends are of like sign and the sign of the sum is different, overflow
is said to have occurred and the Link is set to 1.
2’S COMPLEMENT ARITHMETIC
In 2’s complement addition (see tad instruction), a carry out of the high-
order bit is not added into the low order position. Instead, if a carry occurs,
the Link is complemented. The signs of the addends and sum are not
examined. Two’s complement addition is used primarily in multiple
precision arithmetic.
All memory reference instructions require an Execute cycle (see Control
States above) to transfer data between Core Memory and the MB and
execute the instruction. When indirect addressing is specified, an extra
cycle is required to determine the effective address. The jmp instruction,
while it requires an address, does not require an operand; an Execute
cycle is thus not needed, and the instruction is performed in only one cycle.
MEMORY REFERENCE INSTRUCTIONS
Explanation of Special Terms
C(A)
contents of A aL exclusive OR
A => B A replaces B V inclusive OR
Yl-4
bits 1 - 4 of Y A AND
Y,
a given bit in Y ii l’s complement of A
MNEMONIC %iz- TIME
SYMBOL (BITS O-3)
(w=>
OPERATION
~ ~~.~
lac Y 20 16
Load AC. The C(Y) are loaded into the AC.
The previous C(AC) are lost.
C(Y) => C(AC).
dac Y
04 16
Deposit AC. The C(AC) are deposited in
the memory cell at location Y. The previous
C(Y) are lost; the C(AC) are unchanged.
C(AC) => C(Y).
dzm Y
add Y
14
30
16
Deposit Zero in Memory. Zero is deposited
in memory cell Y. The original C(Y) are
lost. The AC is unaffected by this operation.
0 => C(Y).
16
Add (l’s complement). The C(Y) are adde
to the C(AC) in l’s complement arithmeti .
The result is left in the AC and the origin I
C(AC) are lost. The C(Y) are unchange
t
.
The Link is set to 1 on overflow. (See text .
C(Y) + C(AC) => C(AC).
18