In ASCII code, each letter, number or punctuation mark takes one byte, or 8 bits. That gives you 256 discrete combinations.
Two letters take 2 bytes, or 16 bits.
It differs slightly depending on what platform your computer uses or what language you are using. for the Java programming language, which is platform independent, it is 64-bits.
It requires 6 bits to address 64 words. It does not matter what the word size is.
HI I am Ahtarva,The addressibility is how many bits does that particular processor or micro-controller's architecture use to specify the address of a memory location in the memory. For example if someone say that addressibility is 8 bit then your memory address contains 8 bits and at maximum you have 2^8 different memory locations (or say memory addresses in your device). Here 2^8 is called Address space.
That depends on the memory architecture of the system.if the memory chips are byte wide and not used to create a multibyte bus, 11 address bits are needed.if the memory chips are 32 bits wide, 9 address bits are needed (with the CPU internally selecting which of the 4 bytes it will use).it the memory chips are 64 bits wide, 8 address bits are needed (with the CPU internally selecting which of the 8 bytes it will use.if the memory chips are 4 bits wide, 12 address bits will be needed and the CPU must perform 2 memory cycles per byte that it needs. (yes, I have seen a computer that worked this way!)etc.
To store the hexadecimal number FF, we need to convert it to binary first. FF in hexadecimal is equivalent to 1111 1111 in binary, which requires 8 bits to represent. Each hexadecimal digit corresponds to 4 bits in binary, so two hexadecimal digits (FF) require 8 bits to store.
integer data type consumes memory of 4 bytes or 32 bits
a character/byte as defined in the C programming language is one byte (8 bits). A string can be as short as one byte and can be as long as the physical memory limits can contain it.
The memory in the computer is stored in the form of bits and bytes
From a technical standpoint of view, computer memory stores multiple bits of information. Each bit is an atomic boolean entity (it can be in one of two states, True or False, 0 or 1, etc). Computer memory typically stores large numbers of bits, commonly but not necessarily arranged in groups of bits. Eith bits (an octet) are commonly known as a byte, but other grouping, typically in powers of two, is common.Semantically, computer memory can store anything for which there is a digital description. It can store numbers, or the works of Shakespeare, or you digital holiday photos. It cannot store items such as a perfume, or a smell (such as the perfume's frangrance), but it can store the perfume's recipie along with a chemical formula for the fragrance's aromatics.
In an 8085 system, the memory word size required is 8 bits. This means that each memory location can store 8 bits or one byte of data. The 8085 processor accesses memory locations using these 8-bit memory addresses to read or write data during program execution. The memory word size of 8 bits allows the 8085 system to handle data in small, manageable chunks efficiently.
You need 20 bits of address bus to address 1 Mb of memory.
It differs slightly depending on what platform your computer uses or what language you are using. for the Java programming language, which is platform independent, it is 64-bits.
15 bits
40 bits or 5 byrtes
4
1024 bytes is 8192 bits.
8 bits equals to one character.