Question

The computer is a PDP-8. The program counter (PC) is a 400. The contents of the various memory locations are follows:

1050 320 3620

Show what will be stored in the AC in each case if the

content of location 400 is:

a)    1152

b)    1352

c)    1552

d)    1752

Answer 1

(a) For the instruction 1152 in octal the equivalent binary instruction is 001 001 101 010.

First bit on the left most side is bit #0. The right­ most bit is bit #11. Bit #4 is 0. This means the ad­dress is on Page 0. Hence, the last 7 bits give the address = 152. Bit #5 is also 0. So, it is direct ad­dressing. Hence, location 152 contains the operand. Thus, the AC gets the value of location 152, which is 1050.

(b) For the instruction 1352 in octal the equivalent binary instruction is 001 011 101 010. Bit #4 is 1. This means that the address is on the same page as the in­struction. But, the instruction is in location 400, i.e., on Page 2. Hence, the address is also on Page 2. Hence, the address is given by the first five bits of the PC concatenated with the last seven bits of the instruction. The program counter contains 0400, i.e. 000 100 000 000 and the instruction is 001 011 101010. Hence, the add­ress is: 00010 1101 010, i.e., 0552. The contents of location 552 are 320. So, 320 gets stored in the AC.

(c) For the instruction 1552 in octal the equivalent binary instruction is 001 101 101 010. Bit #4 is 0. So, address is on Page 0. Thus, the ad­dress is: 00000 1101 010

= 000 001 101 010

= 0 152

The Bit #3 is 1. This means it is Indirect Addressing. Hence, the location 0152 on Page 0 contains not the operand itself, but another address where the operand can be found. The content 1050 of location 152 is therefore an add­ress. So, the content of location 1050 or 5 is the operand which goes to the AC.

(d) For the instruction 1752 in octal the equivalent binary instruction is 001 111 101 010. The #4 bit is 1. So, it indicates the same page as the instruction at 400, i.e. at 000 100 000 000. Hence, the address is: 00010 1101 010

000 101 101 010 = 0552.

Now, the #3 bit is 1. So, it means indirect address­ing, and the content of 552 is an address of the op­erand. The content of 552 is: 320.
Hence, 320 is the address of the operand. Now, the content of 320 is 3620. Hence, 3620 is the operand which goes to the AC.
So, the AC contains the following:

(a) 1050

(b) 320

(c) 5

(d) 3620

Hope this helps.