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16KB, or 16384 bytes, can be addressed with 14 address lines. (214 = 16384)
The memory is to be designed so that 16-bitdata can be accessed in one .Two 64K X 8 SRAM chips have a capacity of 128KB.
A memory with a 16 bit address bus can address 216 or 65536 distinct items. If each item is 32 bits in size, then the item is 4 bytes. The size of this memory is then 262144 bytes. (256Kb)
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.
Max. memory address space= 216 X 2 bytes = 128 Kbytes
2 bytes or 16 bits
The 8086/8088 has 20 address lines. It can access 220, or 1MB, or 1,048,576 bytes of memory.
microprocessor can access 2^8 points which is 256 then we have 8 bit memory = 1 bytes then 1*256 =256 bytes
In most languages with a null reference, it is simply a memory address to a zero-length memory block. So the only memory it would occupy in these cases would be enough for a memory pointer: usually around 4 bytes.
2147483648 bytes
A memory address a, is said to be n-byte aligned when a is a multiple of n bytes (where n is a power of 2). In this context a byte is the smallest unit of memory access, i.e. each memory address specifies a different byte.
The 8086/8088 has 20 address lines. It can access 220, or 1MB, or 1,048,576 bytes of memory.