Addressing range seek

The computer is a 32-bit word length, its storage capacity is 256KB, the word addressed by its addressing range is 64 K

Word addressed by the addressing range is: 0 ~ 64KB.
Calculation step: 256KB = 256 bytes * 1024 * 8,
Computer word by word by 32-bit addressing. The word is 32 bits, the range of (256 * 1024 * 8) / 32 = 64KB
Further, if 32-bit computer word length by half-word addressing. The number of bits of 16-bit half-word, in the range of (2,561,024 * 8) / 16 = 128KB
Refers to the storage capacity of the memory can accommodate binary information, with memory stored in the address register MAR is addressed to the number of memory word represents the product of the number of bits.
1B = 8bit (bits)
Therefore addressing range = 256KB / 4B = 64K

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Origin www.cnblogs.com/claudia529/p/11073779.html