Maximum main memory possible for 32 bit instruction word and 7 bit opcode

Question

If we have a system with 32 bit instruction words from which 7 bits are dedicated for the opcodes; is the maximum main memory possible for such a system, 2^32 words or is it 2^25 words?

Answer 1

Maximum addressable memory depends on the address register's size. If your system is designed so that 7 bits are used for opcode, therefore 25 bits are left for addressing at your instruction register and the address register is 25 bits long, then yes the total addressable memory is 2^25 words.

Take this basic computer example from Morris Mano's Computer System Architecture:

The 15th bit is I bit, which represents the addressing mode (direct/indirect).

There are 3 bits for opcode. The address register in the introduced basic computer architecture is 12 bits. You see, the system is designed that way. Check your address register's size and you'll find your answer.

Answer 2

Generally memory pointers will be residing in some register rather than in the opcode itself, so if you have a 32 bit instruction word, out of which 7 bits are opcode, out of the remaining 25bits some bits can be assigned to register combinations used by the opcode (Like src/destination etc).

So for an instruction like

MOV MEM(R1) MEM(R3)

depending on your machine's architecture, R1 and R3 can be 32 bits long or 64 bits long, or however long the designed made it to be. The memory addressing shouldn't be limited to any amount by the instruction bit-width except when designed that way.