Questions about the number of digits in the address line and the number of memory cells

1.Above: The storage unit is the basic unit for the CPU to access the memory.

8-bit binary is used as a memory unit, which is a byte. -- It is also said that the size of the memory unit is constant, that is, one byte.

Above: Storage word length: A storage unit stores a string of binary ** (storage words), the number of bits of this string of binary ** is called the storage word length, and the storage word length can be 8 bits, 16 bits, 32 bits, etc.

Combined with the definition of the memory unit, the memory word length is 8 bits

Above: Memory word: refers to a binary ** combination stored in a storage unit.

2.The address line determines one memory cell at a time, and all the possible combinations of the values on the address line determine the number of memory cells. So,Number of memory units = 2 Number of address lines.

3.Storage capacity refers to the amount of binary information that can be contained in memory, and is expressed as the product of the number of addresses in the memory address register mar and the number of stored word digits. (Storage capacity = address number of address register mar "number of memory cells" * number of stored words (storage word length)) - calculated is the total number of binary ** bits (not bytes) that can be stored

Refers to the basic unit in which a computer is addressed.

The length of the cable generally depends on the length of the cable.

If there are 32 address lines, then the storage word length is 32 bits.

The number of storage cells is called the storage capacity.

will transform from one form to another, or from an object.

Summary. If we take words as a unit, 4m 32 bits of memory is 4m words (4m 32 bits = 4m 4bytes = 4m bytes), since a word is 4 bytes, so the memory has a total of 2 24 word units, that is, 2 24 words, in bytes, there are 2 24 4 bytes = 2 26 bytes. Correspondingly, the range of addressing by word is 2 to the power of 2, which means that the range of words is 0 2 22, and in bytes, it is 0 2 24.

4m 32-bit memory, what is the number of address lines of this memory.

Dear hello your question I have received typing takes time rough wide please be patient and wait for a Chang blind, 4m 32 bit storage stool empty storage, what is the number of memory address lines.

The memory address line has a total of 36 bits, of which 2 bits are hand in hand control bits, 4 bits control the dust line address, 5 bits control column address, and 25 bits control memory cell number.

Can you write a little bit of detail?

This is not <> in detail

Whatever you want to ask, I can.

His answer is that the range of word addressing is 2 of 22 times, and the range of byte addressing is 2 of 24 The address line is 24 bits A23 A0

That's it, how did it come about.

4m 32-bit memory, the range of addressing by type is 2 of 22 times, and the range of addressing by word masking is 24 of 24 The address line is 24 bits A23 A0, just this data, how did it come from?

Is that so? That's right.

If the word is a single round bit, 4m 32 bits of memory is 4m words (4m 32 bits = 4m 4 bytes = 4m bytes), since a word is 4 bytes, so the memory has a total of 2 24 word units, that is, 2 24 words, in bytes, then there is regret vertical collapse 2 24 4 bytes = 2 26 bytes. Correspondingly, the range of addressing by word is 2 to the 22nd power, that is to say, the fan fiber of the word is 0 2 22, and by bytes, it is 0 2 24.

Summary. Hello, for example, 32-bit address (32 binary bits represent an address), there is a total of 2 to the power of 32 combinations, that is, it can correspond to 2 to the power of 32 addresses, each address can store one byte of data (8bit), that is to say, the addressable size is basically the size of RAM, where the 32 corresponding memory is 4GB. Hope mine is helpful to you.

Hello, for example, 32-bit address (32 binary bits represent an address of lead), there are a total of 2 to the 32nd power combination, that is, it can correspond to 2 to the 32nd power of the address, each address can store one byte of data (8bit), that is to say, the addressable Universiade is basically the size of RAM, here 32 corresponding to the memory of Huai quietly is 4GB. Hope mine is helpful to you.

Hello, e.g. MAR 36 bits, MDR 64 bits, addressed by bytes. The book says that the storage capacity is 2 36*64 bits, please:

1.Is the number of addresses also 36 bits (2 to the 36th power)?

2.If so, then an address should be 1b by byte addressing, and 2 36*1b is not equal to 2 36*64 bits.

3.If not, what is the number of addresses?

Thank you! I'll take a look at it later.

The storage capacity is 2 36 64 bits, but eight bits is a byte, so the source buries us with this number of bits to calculate that the hail swatting ant should be divided by 2 3 is the number of bits of the address.