64 Bits Max Value

[Kemal Allan]

Abit (binary digit) is the smallest unit of data that a computer can process and store. A bit is always in one of two physical states, similar to an on/off light switch. The state is represented by a single binary value, usually a 0 or 1. However, the state might also be represented by yes/no, on/off or true/false. Bits are stored in memory through the use of capacitors that hold electrical charges. The charge determines the state of each bit, which, in turn, determines the bit's value.

Although a computer might be able to test and manipulate data at the bit level, most systems process and store data in bytes. A byte is a sequence of eight bits that are treated as a single unit. References to a computer's memory and storage are always in terms of bytes. For example, a storage device might be able to store 1 terabyte (TB) of data, which is equal to 1,000,000 megabytes (MB). To bring this into perspective, 1 MB equals 1 million bytes, or 8 million bits. That means a 1 TB drive can store 8 trillion bits of data.

Each bit in a byte is assigned a specific value, which is referred to as the place value. A byte's place values are used to determine the meaning of the byte as a whole, based on the individual bits. In other words, the byte values indicate what character is associated with that byte.

The place values are used in conjunction with the bit values to arrive at the byte's overall meaning. To calculate this value, the place values associated with each 1 bit are added together. This total corresponds to a character in the applicable character set. A single byte can support up to 256 unique characters, starting with the 00000000 byte and ending with the 11111111 byte. The various combinations of bit patterns provide a range of 0 to 255, which means that each byte can support up to 256 unique bit patterns.

For example, the uppercase "S" in the American Standard Code for Information Interchange (ASCII) character set is assigned the decimal value of 83, which is equivalent to the binary value of 01010011. This figure shows the letter "S" byte and the corresponding place values.

The "S" byte includes four 1 bits and four 0 bits. When added together, the place values associated with 1 bits total 83, which corresponds to the decimal value assigned to the ASCII uppercase "S" character. The place values associated with the 0 bits are not added into the byte total.

Because a single byte supports only 256 unique characters, some character sets use multiple bytes per character. For example, Unicode Transformation Format character sets use between 1 and 4 bytes per character, depending on the specific character and character set. Despite these differences, however, all character sets rely on the convention of 8 bits per byte, with each bit in either a 1 or 0 state.

The term octet is sometimes used instead of byte, and the term nibble is occasionally used when referring to a 4-bit unit, although it's not as common as it once was. In addition, the term word is often used to describe two or more consecutive bytes. A word is usually 16, 32 or 64 bits long.

it means the combination of all the bits represents one value like 010 represents 2 and 101 represents 5. the expression for the same is n bits can represent 2n values because at the heart 1 bit can represent two values.

Since here we are talking about bits then definetly it means "0/1" which is the representation of "False/True" or "OFF/ON". According to Wiki - A bit is the basic unit of information in computing and digital communications. A bit can have only one of two values, and may therefore be physically implemented with a two-state device. The most common representation of these values are 0 and 1.

Note: The following information is providedin part by the Extreme Science and Engineering Discovery Environment(XSEDE), a National Science Foundation (NSF) project that provides researcherswith advanced digital resources and services that facilitatescientific discovery. For more, see the XSEDE website.

Because bits are so small, you rarely work with information one bit ata time. Bits are usually assembled into a group of eight to form abyte. A byte contains enough information to store asingle ASCII character, like "h".

Many hard drive manufacturers use a decimal number system to defineamounts of storage space. As a result, 1 MB is defined as one millionbytes, 1 GB is defined as one billion bytes, and so on. Since yourcomputer uses a binary system as mentioned above, you may notice adiscrepancy between your hard drive's published capacity and thecapacity acknowledged by your computer. For example, a hard drivethat is said to contain 10 GB of storage space using a decimal systemis actually capable of storing 10,000,000,000 bytes. However, in abinary system, 10 GB is 10,737,418,240 bytes. As a result, instead ofacknowledging 10 GB, your computer will acknowledge 9.31 GB. This is not a malfunction but a matter of differentdefinitions.

Note: The names and abbreviations for numbers ofbytes are easily confused with the notations for bits. Theabbreviations for numbers of bits use a lower-case "b" instead of anupper-case "B". Since one byte is made up of eight bits, thisdifference can be significant. For example, if a broadband Internetconnection is advertised with a download speed of3.0 Mbps, its speed is 3.0 megabitsper second, or 0.375 megabytes per second (whichwould be abbreviated as 0.375 MBps). Bits and bit rates(bits over time, as in bits per second [bps]) are most commonly usedto describe connection speeds, so pay particular attention whencomparing Internet connection providers and services.

This document was developed with support from National Science Foundation (NSF) grants 1053575 and 1548562. Any opinions, findings, conclusions, or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the NSF.

I tracked down an extremely nasty bug hiding behind this little gem. I am aware that per the C++ spec, signed overflows are undefined behavior, but only when the overflow occurs when the value is extended to bit-width sizeof(int). As I understand it, incrementing a char shouldn't ever be undefined behavior as long as sizeof(char) Although getting impossible results for undefined behaviour is a valid consequence, there is actually no undefined behaviour in your code. What's happening is that the compiler thinks the behaviour is undefined, and optimises accordingly.

If c is defined as int8_t, and int8_t promotes to int, then c-- is supposed to perform the subtraction c - 1 in int arithmetic and convert the result back to int8_t. The subtraction in int does not overflow, and converting out-of-range integral values to another integral type is valid. If the destination type is signed, the result is implementation-defined, but it must be a valid value for the destination type. (And if the destination type is unsigned, the result is well-defined, but that does not apply here.)

A compiler can have bugs which are other than nonconformances to the standard, because there are other requirements. A compiler should be compatible with other versions of itself. It may also be expected to be compatible in some ways with other compilers, and also to conform to some beliefs about behavior that are held by the majority of its user base.

In this case, it appears to be a conformance bug. The expression c-- should manipulate c in a way similar to c = c - 1. Here, the value of c on the right is promoted to type int, and then the subtraction takes place. Since c is in the range of int8_t, this subtraction will not overflow, but it may produce a value which is out of the range of int8_t. When this value is assigned, a conversion takes place back to the type int8_t so the result fits back into c. In the out-of-range case, the conversion has an implementation-defined value. But a value out of the range of int8_t is not a valid implementation-defined value. An implementation cannot "define" that an 8 bit type suddenly holds 9 or more bits. For the value to be implementation-defined means that something in the range of int8_t is produced, and the program continues. The C standard thereby allows for behaviors such as saturation arithmetic (common on DSP's) or wrap-around (mainstream architectures).

The compiler is using a wider underlying machine type when manipulating values of small integer types like int8_t or char. When arithmetic is performed, results which are out of range of the small integer type can be captured reliably in this wider type. To preserve the externally visible behavior that the variable is an 8 bit type, the wider result has to be truncated into the 8 bit range. Explicit code is required to do that since the machine storage locations (registers) are wider than 8 bits and happy with the larger values. Here, the compiler neglected to normalize the value and simply passed it to printf as is. The conversion specifier %i in printf has no idea that the argument originally came from int8_t calculations; it is just working with an int argument.

Freaky ey? I don't know much about what the compiler does to expressions like i++ or i--. It's likely promoting the return value to an int and passing it. That's the only logical conclusion I can come up with because you ARE in fact getting values that cannot fit into 8-bits.

I guess that the underlying hardware is still using a 32-bit register to hold that int8_t. Since the specification does not impose a behaviour for overflow, the implementation does not check for overflow and allows larger values to be stored as well.

What you see there is the result of compiler optimizations combined with you telling printf to print a 32bit number and then pushing a (supposedly 8bit) number onto the stack, which is really pointer sized, because this is how the push opcode in x86 works.

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