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In an 11-bit computer instruction format, the size of address field is 4-bits. The computer uses expanding OP code technique and has 5 two-address instructions and 32 one-address instructions. The number of zero-address instructions it can support is ________

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No. of possible instruction encoding = $2^{11} = 2048$

No. of encoding taken by two-address instructions = $5 \times 2^4 \times 2^4 = 1280$

No. of encoding taken by one-address instructions = $32 \times 2^4 = 512$

So, no. of possible zero-address instructions = $2048 - (1280 + 512) = 256$
answered by Veteran (332k points)
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what is the logic behind the calculation of no. of 2 address and 3 address instructions ?

@arjun sir @habibkhan
Out of 11 bits, 3 bits are for opcode in 2-Address. So 2^3=8 instructions(2-Address) are possible.

But 5 are being used. So 3 are free.

We use these 3 for 1-address instruction by using 4 bits of 1 of the addresses i.e. 3*16=48

Now out of 48 1-address instructions, 32 are being used. So we are left with 16.

Again use these 16 for 0-address instruction by using 4 bits of address i.e. 16*16=256
answered ago by Junior (559 points)