It depends on which of several coding standards you use.
ANSI or ASCII uses one byte to define a character, as does BCDIC and EBCDIC.
Multi-byte character sets typically have a special character that is used to indicate that the following character is from a different character set than the base one. If the character u-umlaut cannot be represented in the standard set of characters, for instance, you could use two characters, one to say the following character is special, and then the special u0umlaut character. This coding standard requires somewhere between one and two bytes to encode a character.
The Unicode system is intended to support all possible characters, including Hebrew, Russian / Cyrillic, Greek, Arabic, and Chinese. As you can imagine, in order to support all these different characters, you need a lot of bits. The initial standard, U16, used two bytes per character; but this proved to be insufficient, so a new standard, U24 which uses three bytes per character, is also now available.
Eight.
There are 8 bits (not bytes) in a standard ascii character.
8 bits = 1 byte.
A character in ASCII format requires only one byte and a character in UNICODE requires 2 bytes.
15,383 Bytes
If you're referring to kilobyte, then it contains 1024 bytes and if the characters are the standard ASCII character set where 1 character is 1 byte, then a kilobyte would have 1024 characters.
An extended ASCII byte (like all bytes) contains 8 bits, or binary digits.
011000110110000101110100 is cat in Binary. That is 23 Bits, or just under 3 Bytes.
First, let's assume you are talking about modern computers that use the hexadecimal system. {Old systems use octal, and we won't go there in this answer.) The ASCII chart runs from hex 0 to hex 7F, so it takes two bytes to store one ASCII character. {see the Related Link for an ASCII chart} One kilobyte = 1000 bytes, thus 500 ASCII characters can normally by stored in one kilobyte. http://www.asciitable.com/
The total number of bytes allocated to the union will be the same number as would have been allocated if instead of the union was declared the largest member of the union. For example, if you declared: union myUnion { char c; int i; double d; } u;, then the space allocated to u will be the size of a double.
If you are using the ASCII system, the word "duck", as it has four letters, contains 4 bytes, or 32 bits.
One.
4
In the UTF 8 standard of representing text, which is the most commonly used has a varying amount of bytes to represent characters. The Latin alphabet and numbers as well as commonly used characters such as (but not limited to) <, >, -, /, \, $ , !, %, @, &, ^, (, ), and *. Characters after that, however, such as accented characters and different language scripts are usually represented as 2 bytes. The most a character can use is 4, I think (Can someone verify? I can't seem to find the answer).
128