The 8086 can address 1,114,080 bytes. (One Mb + 64Kb - 16) That does not count I/O space, it only counts memory space.
In the 8086 microprocessor architecture, each segment can contain 64 kilobytes (KB) of data. Since 1 KB is equal to 1024 bytes, this means each segment can hold 65,536 bytes. The four segments typically used are the code segment, data segment, stack segment, and extra segment, allowing the processor to manage different types of information efficiently within its 1 MB addressable memory space.
6 bytes
6 bytes
Of the 128-byte internal RAM of the 8051, only 16 bytes are bit-addressable. The rest must be accessed in byte format. The bit-addressable RAM locations are 20H to 2FH.
The bit addressable memory in 8051 is compose from 210 bits: - bit address space: 20H - 2FH bytes RAM = 00H - 7FH bits address; - SFR registers; The following addresses are NOT bit addressable, only 1-byte addressable: - 32 bytes RAM from 00H to 1FH (R0 - R7 registers in all four banks); - 80 bytes RAM general user from 30H to 7FH.
The 16 bytes (128 bits) at internal RAM locations 0x20-0x2F are bit-addressable.
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The 8086/8088 has 20 address lines. It can access 220, or 1MB, or 1,048,576 bytes of memory.
The 8086/8088 has 20 address lines. It can access 220, or 1MB, or 1,048,576 bytes of memory.
There are 256 different interrupt vectors in the 8086/8088. Each vector is a far CS:IP address, which is four bytes. That makes the interrupt vector table 1,024 bytes.
Hey ! in bit addressable , one can reach the bits (PORT 1.0,PORT1.1,PORT1.2,..etc..) of the ports in the programming while byte addressable only allows the bytes(PORT 1) to be addressed. therefore,there is difference in programming instructions too.
In the 8086 microprocessor, the maximum length of an instruction is 6 bytes. This includes the opcode, any necessary prefixes, and operands. The architecture allows for complex addressing modes, which can contribute to the instruction length, but the limit remains at 6 bytes for any single instruction.