A machine is 32-bit long and its storage capacity is 64 mb. If it is encoded by byte, what is its addressing range?

Problem:

1. A computer is 32 characters long and its storage capacity is 16 Mb.Double-CharacterAddressing, what is its addressing range?

2. A host has 32 characters and the storage capacity is 64 MB.BytesAddressing. What is its addressing range?

Answer:

My method is to convert all the elements into one-bit binary units for calculation. Calculate the total capacity first. For example, in the first question, in 16 MB, a B is 8 bits, that is, 8 Basic Units, 16 m = 2 ^ 24 bits = 2 ^ 24 a basic unit. Therefore, the basic unit is 2 ^ 24*8.

A word is n characters long, that is, a word is composed of N basic units. The number of basic units contained in a word is used as an address unit, which corresponds to an address. Likewise, double-word addressing is the number of basic units contained in two words as an address unit. Because one byte (1B) is always eight bits, byte addressing is always eight basic units as one address unit. The addressing range is the total number of such addresses.

In the first question, the length of a word is 32 characters. For address-based editing, an address unit has 32 basic units. For address-based editing, there are 64 address units, the size is 8 bytes and the total size is 2 ^ 24*8. Therefore, the number of address addresses by word is 2 ^ 24*8/32, the number of double words is 2 ^ 24*8/64, and the byte value is 2 ^ 24*8/8. Therefore, the answer to the first question is 2 ^ 21 =2 m.

Similarly, the answer to the second question is 2 ^ 26*8/8 = 2 ^ 26 =64 m.