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Instructions are stored in memory. When they are to be executed they are brought into instruction register. Now it is divided as Mode, opcode, operand (log (memory size) is divided in three parts) . In direct addressing, operand represent the effective memory address. If bits in mode + opcode + operand (log mem size) > bits required to address a memory then how such a big instruction is stored in memory whose operand itself has those many bits?

in CO and Architecture by Loyal (7.2k points) | 153 views
+3
It's a disadvantage of direct addressing mode.
Suppose operand field consists of 16 bits then addresses $ from\ 0 \ to \ 2^{16}-1$ are only addressable using this mode that's why we go for other AM like register addressing, indirect addressing etc.
0
@soumya that means we consider memory is divided into blocks and when we try to store instruction in particular block of memory it cannot hold those many bits. So we can address fewer words in memory that actually present.

 

correct me if m wrong.
+2
Not just the block address but the complete address is present in operand field.
And yes, we can address fewer words than actually present USING direct addressing mode provided that lesser bits are available than the required.

Suppose MM address or virtual address actually consists of 24 bits but you have 16 bits for operand and you want the content of location say $2^{22}$.
It won't be possible using direct addressing. What will you do then?
If you want to go for register indirect AM then you will store this 24-bit address in a 32-bit register and then give the name of that register in your instruction.
+1
Thanx for the detailed explanation. :)

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