It depends on how wide the data buses are on each chip, and how they're connected. If they're one byte wide, you could need over 256 million addresses, one for each byte. if they're wider, and connected to show an even wider combined data bus, it could be much less; around 32 million.
The program counter in the processor holds the address of the next instruction needed from main memory. The program counter copies its contents into the memory address register. The memory address register then sends the address along the address bus to main memory and the contents of the memory location specified by the address are sent along the data bus to the memory buffer register. The contents of the memory buffer register are then copied to the current instruction register where they are decoded and executed.
The program counter in the processor holds the address of the next instruction needed from main memory. The program counter copies its contents into the memory address register. The memory address register then sends the address along the address bus to main memory and the contents of the memory location specified by the address are sent along the data bus to the memory buffer register. The contents of the memory buffer register are then copied to the current instruction register where they are decoded and executed.
The program counter in the processor holds the address of the next instruction needed from main memory. The program counter copies its contents into the memory address register. The memory address register then sends the address along the address bus to main memory and the contents of the memory location specified by the address are sent along the data bus to the memory buffer register. The contents of the memory buffer register are then copied to the current instruction register where they are decoded and executed.
The program counter in the processor holds the address of the next instruction needed from main memory. The program counter copies its contents into the memory address register. The memory address register then sends the address along the address bus to main memory and the contents of the memory location specified by the address are sent along the data bus to the memory buffer register. The contents of the memory buffer register are then copied to the current instruction register where they are decoded and executed.
The number of address lines needed to access N-KB is given by log2N Then the number of address lines needed to access 256KB of main memory will be log2256000=18 address lines.
Direct address instructions specify the memory location of the operand directly within the instruction itself, requiring only one memory reference to fetch the operand. In contrast, indirect address instructions specify a memory location that contains the address of the operand, necessitating two memory references: one to retrieve the address and another to fetch the operand itself. Therefore, direct addressing is more efficient in terms of memory access.
In an instruction cycle with indirect addressing, the CPU fetches the instruction, decodes it to determine the memory address of the operand stored in a register, fetches the operand from the memory location pointed to by the register, and executes the instruction using the operand. Finally, the CPU stores the result back in memory if needed. This extra step of fetching the operand based on the indirect memory address adds complexity to the instruction cycle.
You need 30 address lines to access 1G of memory. 230 = 1,073,741,824. log2 (1,073,741,824) = 30.
The memory address space is 64 MB, which means 226. However, each word is 4 bytes, which means that you have 224 words. This means you need log2 224 or 24 bits, to address each word.
That depends on the memory architecture of the system.if the memory chips are byte wide and not used to create a multibyte bus, 11 address bits are needed.if the memory chips are 32 bits wide, 9 address bits are needed (with the CPU internally selecting which of the 4 bytes it will use).it the memory chips are 64 bits wide, 8 address bits are needed (with the CPU internally selecting which of the 8 bytes it will use.if the memory chips are 4 bits wide, 12 address bits will be needed and the CPU must perform 2 memory cycles per byte that it needs. (yes, I have seen a computer that worked this way!)etc.
Yes and no. All memory location from 0H to 0FFFFH are addressable, but some of them are needed for the program, interrupt vectors, and the stack, so you would need to pay attention to where things are located in memory to design an appropriate program. In addition, if your system is using memory mapped I/O, some locations will be reserved.
In a 256K x 16 memory system, the memory has 256K (256 * 1024 = 262,144) addressable locations and each location holds 16 bits of data. To calculate the number of address lines needed, we find the base-2 logarithm of 256K, which is 18 (since 2^18 = 262,144). For the data lines, since each location holds 16 bits, 16 data lines are required. Thus, the system requires 18 address lines and 16 data lines.