I'm using mplabX 4.20, and xc8 compiler. I'm trying to understand which is the difference between uint8_t and unsigned char. Both of them have size from 0 till 255.
Both of can hold characters and numbers. But which is better to use, and for which case?
Example if i want to create a buffer for holding a string.
uint8_t buffer[20]="Hello World";
unsigned char buffer[20]="Hello World";
In most cases i need to hold characters. Which is the best practise for this action?
I'm using mplabX 4.20, and xc8 compiler. I'm trying to understand
which is the difference between uint8_t and unsigned char. Both of
them have size from 0 till 255. Both of can hold characters and
numbers. But which is better to use, and for which case?
unsigned char is the unsigned integer type corresponding to signed char. Its representation does not use any padding bits. Both of these occupy the same amount of storage as type char, which is at least 8 bits, but may be more. The macro CHAR_BIT tells you how many it comprises in your implementation. Every conforming C implementation provides all of these types.
uint8_t, if available, is an unsigned integer data type exactly 8 bits wide and with no padding bits. On an implementation having CHAR_BIT defined as 8, this is the same type as unsigned char. On such systems you may use the two types interchangeably wherever the declarations provided by stdint.h are in scope. On other systems, uint8_t will not be declared at all.
Example if i want to create a buffer for holding a string.
If you want to declare a buffer for holding a string then as a matter of style, you should use type char, not either of the other two:
char buffer[20] = "Hello World";
Although either of the other two, or signed char, can also be used for string data (provided in the case of uint8_t that the type is defined at all), type char is the conventional one to use for character data. Witness, for example, that that's the type in terms of which all the string.h functions are declared.
You should use uint8_t where and only where you need an integer type with exactly its properties: unsigned, 8 value bits, no padding bits.
You should use unsigned char where you want the smallest unsigned integer type available, but you don't care whether it is exactly 8 bits wide, or where you want to emphasize that it is the same size as a char -- the smallest discrete unit of storage available.
You should use signed char where you want the smallest signed integer type available but don't care about the exact size or representation.
You should use int8_t where you want a signed integer type with exactly 7 value bits, one sign bit, and no padding bits, expressed in two's complement representation.
You should remain mindful that uint8_t and int8_t are not guaranteed to be available from every C implementation, and that where they are available, their use requires inclusion of stdint.h. Furthermore, this header and these types were not part of C90 at all, so you should not use them if compatibility with legacy C implementations is important to you.
difference between uint8_t and unsigned char
If you're on a some exotic system where CHAR_BIT > 8, then uint8_t isn't going to be defined at all.
Otherwise (if CHAR_BIT == 8) there is no difference between unsigned char and uint8_t.
i need to hold characters
Then use plain char.
Functions that operate in strings usually have [const]char * parameters, and you won't be able to pass your unsigned char arrays to them.
Related
since C language using the char as integer internally(correspondent ASCII is stored). for internal calculation we can use signed and unsigned char.
other than that, any other use??
signed and unsigned char are first and foremost just small integers. Do you need to store a large quantity of small numbers (in the range [-127, +127]ยน or [0, 255])? You can use an array of signed or unsigned chars and save memory compared to pretty much any other type. That's what is done for e.g. images - a grayscale image is generally stored as an array of unsigned char (and an RGB image is generally stored as an array of 3 unsigned char components).
The second usage of char is for character strings, which you probably already saw; notice that char is a distinct type from both signed char and unsigned char, and its signedness is implementation defined. This is stupid and inconvenient in many situations - and leads to sad stuff such as the mandatory cast to unsigned char when calling functions of the toupper/isupper family.
Finally, char & co. are defined as the "underlying storage" of the C abstract machine. sizeof(char) == 1 by definition, and any type can be aliased through a (signed|unsigned)? char pointer to access its underlying bit representation.
Yes, -127; [-127, +127] is the minimum range allowed for signed char by the standard, as it still allows sign and magnitude representation; more realistic, on any real-world machine of this century it will be at least [-128, 127].
What is the need for signed and unsigned characters in C?
Is there some special reason for having a signed and unsigned char in C? Or was it simply added for completeness so that the compiler does not have to check the data type before adding signed/unsigned modifier?
I am not asking about signed and unsigned variables. My doubt is about the special cases where an unsigned character variable will not be sufficient such that you have to depend on a signed character variable.
A char can be either signed or unsigned depending on what is most efficient for the underlying hardware. The keywords signed and unsigned allow you to explicitly specify that you want something else.
A quote from the C99 rationale:
Three types of char are specified: signed, plain, and unsigned. A plain char may be represented as either signed or unsigned depending upon the implementation, as in prior practice. The type signed char was introduced in C89 to make available a one-byte signed integer type on those systems which implemented plain char as unsigned char. For reasons of symmetry, the keyword signed is allowed as part of the type name of other integer types.
Information #1: char in C is just a small int, which uses 8 bits.
Information #2: Difference between signed and unsigned, is that one bit in the representation is used as the sign bit for a signed variable.
Information #3: As a result of (#2), signed variables hold different ranges (-128 to 127, in char case) compared to unsigned (0 to 255 in char case).
Q-A #1: why do we need unsigned?
In most cases (for instance representing a pointer) we do not need signed variables. By convention all locations in the memory are exposed to the program as a contiguous array of unsigned addresses.
Q-A #2: why do we need signed?
Generally, to do signed arithmetic.
I assume you are using a char to hold numbers, not characters.
So:
signed char gives you at least the -128 to 127 range.
unsigned char gives you at least the 0 to 255 range.
A char is required by standard to be AT LEAST 8 bits, so that is the reason for my saying at least. It is possible for these values to be larger.
Anyway, to answer your question, having a char as unsigned frees the requirement for the first bit to be the 'sign' bit, thus allowing you to hold near double that of a signed char.
The thing you have to understand is that datatype "char" is actually just an integer, typically 8-bits wide. You can use it like any other inter datatype, assuming you respect the reduced value limits. There is no reason to limit "char" to characters.
On a 32/64-bit processor, there is typically no need to use such small integer fields, but on an 8-bit processor such as the 8051, 8-bit integers are no only much faster to process and use less (limited) memory.
In this answer and the attached comments, Pavel Minaev makes the following argument that, in C, the only types to which uint8_t can be typedef'd are char and unsigned char. I'm looking at this draft of the C standard.
The presence of uint8_t implies the presence of a corresponding type int8_t (7.18.1p1).
int8_t is 8 bits wide and has no padding bits (7.18.1.1p1).
Corresponding types have the same width (6.2.5p6), so uint8_t is also 8 bits wide.
unsigned char is CHAR_BIT bits wide (5.2.4.2.1p2 and 6.2.6.1p3).
CHAR_BIT is at least 8 (5.2.4.2.1p1).
CHAR_BIT is at most 8, because either uint8_t is unsigned char, or it's a non-unsigned char, non-bit-field type whose width is a multiple of CHAR_BIT (6.2.6.1p4).
Based on this argument, I agree that, if uint8_t exists, then both it and unsigned char have identical representations: 8 value bits and 0 padding bits. That doesn't seem to force them to be the same type (e.g., 6.2.5p14).
Is it allowed that uint8_t is typedef'd to an extended unsigned integer type (6.2.5p6) with the same representation as unsigned char? Certainly it must be typedef'd (7.18.1.1p2), and it cannot be any standard unsigned integer type other than unsigned char (or char if it happens to be unsigned). This hypothetical extended type would not be a character type (6.2.5p15) and thus would not qualify for aliased access to an object of an incompatible type (6.5p7), which strikes me as the reason a compiler writer would want to do such a thing.
If uint8_t exists, the no-padding requirement implies that CHAR_BIT is 8. However, there's no fundamental reason I can find why uint8_t could not be defined with an extended integer type. Moreover there is no guarantee that the representations are the same; for example, the bits could be interpreted in the opposite order.
While this seems silly and gratuitously unusual for uint8_t, it could make a lot of sense for int8_t. If a machine natively uses ones complement or sign/magnitude, then signed char is not suitable for int8_t. However, it could use an extended signed integer type that emulates twos complement to provide int8_t.
In 6.3.1.1 (1) (of the N1570 draft of the C11 standard), we can read
The rank of any standard integer type shall be greater than the rank of any extended integer type with the same width.
So the standard explicitly allows the presence of extended integer types of the same width as a standard integer type.
There is nothing in the standard prohibiting a
typedef implementation_defined_extended_8_bit-unsigned_integer_type uint8_t;
if that extended integer type matches the specifications for uint8_t (no padding bits, width of 8 bits), as far as I can see.
So yes, if the implementation provides such an extended integer type, uint8_t may be typedef'ed to that.
uint8_t may exist and be a distinct type from unsigned char.
One significant implication of this is in overload resolution; it is platform-dependent whether:
uint8_t by = 0;
std::cout << by;
uses
operator<<(ostream, char)
operator<<(ostream, unsigned char) or
operator<<(ostream, int)
int8_t and uint8_t differ only by REPRESENTATION and NOT the content(bits). int8_t uses lower 7 bits for data and the 8th bit is to represent "sign"(positive or negative). Hence the range of int8_t is from -128 to +127 (0 is considered a positive value).
uint8_t is also 8 bits wide, BUT the data contained in it is ALWAYS positive. Hence the range of uint8_t is from 0 to 255.
Considering this fact, char is 8 bits wide. unsigned char would also be 8 bits wide but without the "sign". Similarly short and unsigned short are both 16 bits wide.
IF however, "unsigned int" be 8 bits wide, then .. since C isn't too type-nazi, it IS allowed. And why would a compiler writer allow such a thing? READABILITY!
Is it possible to store more than a byte value to a char type?
Say for example char c; and I want to store 1000 in c. is it possible to do that?
Technically, no, you can't store more than a byte value to a char type. In C, a char and a byte are the same size, but not necessarily limited to 8 bits. Many standards bodies tend to use the term "octet" for an exactly-8-bit value.
If you look inside limits.h (from memory), you'll see the CHAR_BIT symbol (among others) telling you how many bits are actually used for a char and, if this is large enough then, yes, it can store the value 1000.
The range of values you can store in a C type depends on its size, and that is not specified by C, it depends on the architecture. A char type has a minimum of 8 bits. And typically (almost universally) that's also its maximum (you can check it in your limits.h).
Hence, in a char you will be able to store from -128 to 127, or from 0 to 255 (signed or unsigned).
The minimum size for a char in C is 8 bits, which is not wide enough to hold more than 256 values. It may be wider in a particular implementation such as a word-addressable architecture, but you shouldn't rely on that.
Include limits.h and check the value of CHAR_MAX.
Probably not. The C standard requires that a char can hold at least 8 bits, so you can't depend on being able to store a value longer than 8 bits in a char portably.
(* In most commonly-used systems today, chars are 8 bits).
Char's width is system-dependent. But assuming you're using something reasonably C99-compatible, you should have access to a header stdint.h, which defines types of the formats intN_t and uintN_t where N=8,16,32,64. These are guaranteed to be at least N bits wide. So if you want to be certain to have a type with a certain amount of bits (regardless of system), those are the guys you want.
Example:
#include <stdint.h>
uint32_t foo; /* Unsigned, 32 bits */
int16_t bar; /* Signed, 16 bits */
From The C Programming Language (Brian W. Kernighan), 2.7 TYPE CONVERSIONS, pg 43 :
"There is one subtle point about the
conversion of characters to integers.
... On some macines a char whose
leftmost bit is 1 will be converted to
a negative integer. On others, ... is
always positive. For portability,
specify signed or unsigned if
non-character data is to be stored in
char variables."
My questions are:
Why would anyone want to store
non-char data in char? (an example
where this is necessary will be real
nice)
Why does integer value of char
change when it is converted to int?
Can you elaborate more on this
portability issue?
In regards to 1)
People often use char arrays when they really want a byte buffer for a data stream. Its not great practice, but plenty of projects do it, and if you're careful, no real harm is done. There are probably other times as well.
In regards to 2)
Signed integers are often sign extended when they are moved from a smaller data type. Thus
11111111b (-1 in base 10) becomes 11111111 11111111 11111111 11111111 when expanded to 32 bits. However, if the char was intended to be unsigned +255, then the signed integer may end up being -1.
About portability 3)
Some machines regard chars as signed integers, while others interpret them as unsigned. It could also vary based on compiler implementation. Most of the time you don't have to worry about it. Kernighan is just trying to help you understand the details.
Edit
I know this is a dead issue, but you can use the following code to check if char's on your system are signed or unsigned:
#include <limits.h> //Include implementation specific constants (MAX_INT, et c.)
#if CHAR_MAX == SCHAR_MAX
// Plain "char" is signed
#else
// Plain "char" is unsigned
#endif
1) char is the size of a single byte in C, and is therefore used for storing any sort of data. For example, when loading an image into memory, the data is represented as an array of char. In modern code, typedefs such as uint8_t are used to indicate the purpose of a buffer more usefully than just char.
2 & 3) Whether or not char is signed or unsigned is platform dependent, so if a program depends on this behavior then it's best to specify one or the other explicitly.
The char type is defined to hold one byte, i.e. sizeof(char) is defined to be 1. This is useful for serializing data, for instance.
char is implementation-defined as either unsigned char or signed char. Now imagine that char means smallint. You are simply converting a small integer to a larger integer when you go from smallint to int. The problem is, you don't know whether that smallint is signed or unsigned.
I would say it's not really a portability issue as long as you follow The Bible (K&R).
unsigned char is often used to process binary data one byte at a time. A common example is UTF-8 strings, which are not strictly made up of "chars."
If a signed char is 8 bits and the top bit is set, that indicates that it's negative. When this is converted to a larger type, the sign is kept by extending the high bit to the high bit of the new type. This is called a "sign-extended" assignment.
1) Char is implemented as one byte across all systems so it is consistent.
2) The bit mentioned in you question is the one that is used in single byte integers for their singed-ness. When a int on a system is larger than one byte the signed flat is not affected when you convert char to int, other wise it is. ( there are also singed and unsigned chars)
3) Because of the consistence of the char implementation lots of libs use them like the Intel IPP (Intel Performance Primitives) libs and their cousins OpenCV.
Usually, in C, char to int conversion and vice versa is an issue because the stanard APIs for reading character input/writing character output use int's for the character arguments and return values. See getchar(), getc() and putchar() for example.
Also, since the size of a char is 1 byte, it is a convenient way to deal with arbitrary data as a byte stream.