A character constant has type int in C.
Now suppose my machine's local character set is Windows Latin-1 ( http://www.ascii-code.com/) which is a 256 character set so every char between single quotes, like 'x', is mapped to an int value between 0 and 255 right ?
Suppose plain char is signed on my machine and consider the following code:
char ch = 'â'
if(ch == 'â')
{
printf("ok");
}
Because of the integer promotion ch will be promoted into a negative
quantity of type int (cause it has a leading zero) and beingâ mapped to a positive
quantity ok will not be printed.
But I'm sure i'm missing something , can you help ?
Your C implementation has a notion of an execution character set. Moreover, if your program source code is read from a file (as it always is), the compiler has (or should have) a notion of a source character set. For example, in GCC you can tune those parameters on the command line. The combination of those two settings determines the integral value that is assigned to your literal â.
Actually, the initial assignment will not work as expected:
char ch = 'â';
There's an overflow here, and gcc will warn about it. Technically, this is undefined behavior, although for the very common single-byte char type, the behavior is predictable enough -- it's a simple integer overflow. Depending on your default character set, that's a multibyte character; I get decimal 50082 if I print it as an integer on my machine.
Furthermore, the comparison is invalid, again because char is too small to hold the value being compared, and again, a good compiler will warn about it.
ISO C defines wchar_t, a type wide enough to hold extended (i.e., non-ASCII) characters, along with wide character versions of many library functions. Code that must deal with non-ASCII text should use this wide character type as a matter of course.
In a case where char is signed:
When processing char ch = 'â', the compiler will convert â to 0xFFFFFFE2, and store 0xE2 in ch. There is no overflow, as the value is signed.
When processing if(ch == 'â'), the compiler will extend ch (0xE2) to integer (0xFFFFFFE2) and compare it to 'â' (0xFFFFFFE2 also), so the condition will be true.
Related
In my previous post I found a solution to do this using C++ strings, but I wonder if there would be a solution using char's in C as well.
My current solution uses str.compare() and size() of a character string as seen in my previous post.
Now, since I only use one (multibyte) character in the std::string, would it be possible to achieve the same using a char?
For example, if( str[i] == '¶' )? How do I achieve that using char's?
(edit: made a type on SO for comparison operator as pointed out in the comments)
How do I compare single multibyte character constants cross-platform in C?
You seem to mean an integer character constant expressed using a single multibyte character. The first thing to recognize, then, is that in C, integer character constants (examples: 'c', '¶') have type int, not char. The primary relevant section of C17 is paragraph 6.4.4.4/10:
An integer character constant has type int. The value of an integer character constant containing a single character that maps to a single-byte execution character is the numerical value of the representation of the mapped character interpreted as an integer. The value of an integer character constant containing more than one character (e.g.,’ab’ ), or containing a character or escape sequence that does not map to a single-byte execution character, is implementation-defined. If an integer character constant contains a single character or escape sequence, its value is the one that results when an object with type char whose value is that of the single character or escape sequence is converted to type int.
(Emphasis added.)
Note well that "implementation defined" implies limited portability from the get-go. Even if we rule out implementations defining perverse behavior, we still have alternatives such as
the implementation rejects integer character constants containing multibyte source characters; or
the implementation rejects integer character constants that do not map to a single-byte execution character; or
the implementation maps source multibyte characters via a bytewise identity mapping, regardless of the byte sequence's significance in the execution character set.
That is not an exhaustive list.
You can certainly compare integer character constants with each other, but if they map to multibyte execution characters then you cannot usefully compare them to individual chars.
Inasmuch as your intended application appears to be to locate individual mutlibyte characters in a C string, the most natural thing to do appears to be to implement a C analog of your C++ approach, using the standard strstr() function. Example:
char str[] = "Some string ¶ some text ¶ to see";
char char_to_compare[] = "¶";
int char_size = sizeof(char_to_compare) - 1; // don't count the string terminator
for (char *location = strstr(str, char_to_compare);
location;
location = strstr(location + char_size, char_to_compare)) {
puts("Found!");
}
That will do the right thing in many cases, but it still might be wrong for some characters in some execution character encodings, such as those encodings featuring multiple shift states.
If you want robust handling for characters outside the basic execution character set, then you would be well advised to take control of the in-memory encoding, and to perform appropriate convertions to, operations on, and conversions from that encoding. This is largely what ICU does, for example.
I believe you meant something like this:
char a = '¶';
char b = '¶';
if (a == b) /*do something*/;
The above may or may not work, if the value of '¶' is bigger than the char range, then it will overflow, causing a and b to store a different value than that of '¶'. Regardless of which value they hold, they may actually both have the same value.
Remember, the char type is simply a single-byte wide (8-bits) integer, so in order to work with multibyte characters and avoid overflow you just have to use a wider integer type (short, int, long...).
short a = '¶';
short b = '¶';
if (a == b) /*do something*/;
From personal experience, I've also noticed, that sometimes your environment may try to use a different character encoding than what you need. For example, trying to print the 'á' character will actually produce something else.
unsigned char x = 'á';
putchar(x); //actually prints character 'ß' in console.
putchar(160); //will print 'á'.
This happens because the console uses an Extended ASCII encoding, while my coding environment actually uses Unicode, parsing a value of 225 for 'á' instead of the value of 160 that I want.
What happens when we declare a character variable without using single quotes? For eg:
char ch=5;
char ch= a;
In C, char type is just an integer number, usually 8 bits wide, signed or unsigned depending on compiler.
"Characters" are just agreement on what these numbers mean, called "text encoding". 8 bit text encodings in common use today are based on ASCII. From there you can check that value 5 is non-printable control char (called ENQ, with historical meaning which has no practical relevance today except for curiosity), and printable character '5' has numeric value 53 in ASCII encoding.
Let's explain the two lines of code in your question:
char ch=5; defines variable ch of type char, and initializes it with value 5, which is a numeric literal. It is recognized as such literal, because it starts with a number.
char ch= a; defines variable ch of type char, and initializes it with value of variable a (and if there is no a defined yet, you get compiler error). a here is a symbol name (probably a variable name), recognized as such because it starts with letter character (underscore would also be ok).
I am recently reading The C Programming Language by Kernighan.
There is an example which defined a variable as int type but using getchar() to store in it.
int x;
x = getchar();
Why we can store a char data as a int variable?
The only thing that I can think about is ASCII and UNICODE.
Am I right?
The getchar function (and similar character input functions) returns an int because of EOF. There are cases when (char) EOF != EOF (like when char is an unsigned type).
Also, in many places where one use a char variable, it will silently be promoted to int anyway. Ant that includes constant character literals like 'A'.
getchar() attempts to read a byte from the standard input stream. The return value can be any possible value of the type unsigned char (from 0 to UCHAR_MAX), or the special value EOF which is specified to be negative.
On most current systems, UCHAR_MAX is 255 as bytes have 8 bits, and EOF is defined as -1, but the C Standard does not guarantee this: some systems have larger unsigned char types (9 bits, 16 bits...) and it is possible, although I have never seen it, that EOF be defined as another negative value.
Storing the return value of getchar() (or getc(fp)) to a char would prevent proper detection of end of file. Consider these cases (on common systems):
if char is an 8-bit signed type, a byte value of 255, which is the character ÿ in the ISO8859-1 character set, has the value -1 when converted to a char. Comparing this char to EOF will yield a false positive.
if char is unsigned, converting EOF to char will produce the value 255, which is different from EOF, preventing the detection of end of file.
These are the reasons for storing the return value of getchar() into an int variable. This value can later be converted to a char, once the test for end of file has failed.
Storing an int to a char has implementation defined behavior if the char type is signed and the value of the int is outside the range of the char type. This is a technical problem, which should have mandated the char type to be unsigned, but the C Standard allowed for many existing implementations where the char type was signed. It would take a vicious implementation to have unexpected behavior for this simple conversion.
The value of the char does indeed depend on the execution character set. Most current systems use ASCII or some extension of ASCII such as ISO8859-x, UTF-8, etc. But the C Standard supports other character sets such as EBCDIC, where the lowercase letters do not form a contiguous range.
getchar is an old C standard function and the philosophy back then was closer to how the language gets translated to assembly than type correctness and readability. Keep in mind that compilers were not optimizing code as much as they are today. In C, int is the default return type (i.e. if you don't have a declaration of a function in C, compilers will assume that it returns int), and returning a value is done using a register - therefore returning a char instead of an int actually generates additional implicit code to mask out the extra bytes of your value. Thus, many old C functions prefer to return int.
C requires int be at least as many bits as char. Therefore, int can store the same values as char (allowing for signed/unsigned differences). In most cases, int is a lot larger than char.
char is an integer type that is intended to store a character code from the implementation-defined character set, which is required to be compatible with C's abstract basic character set. (ASCII qualifies, so do the source-charset and execution-charset allowed by your compiler, including the one you are actually using.)
For the sizes and ranges of the integer types (char included), see your <limits.h>. Here is somebody else's limits.h.
C was designed as a very low-level language, so it is close to the hardware. Usually, after a bit of experience, you can predict how the compiler will allocate memory, and even pretty accurately what the machine code will look like.
Your intuition is right: it goes back to ASCII. ASCII is really a simple 1:1 mapping from letters (which make sense in human language) to integer values (that can be worked with by hardware); for every letter there is an unique integer. For example, the 'letter' CTRL-A is represented by the decimal number '1'. (For historical reasons, lots of control characters came first - so CTRL-G, which rand the bell on an old teletype terminal, is ASCII code 7. Upper-case 'A' and the 25 remaining UC letters start at 65, and so on. See http://www.asciitable.com/ for a full list.)
C lets you 'coerce' variables into other types. In other words, the compiler cares about (1) the size, in memory, of the var (see 'pointer arithmetic' in K&R), and (2) what operations you can do on it.
If memory serves me right, you can't do arithmetic on a char. But, if you call it an int, you can. So, to convert all LC letters to UC, you can do something like:
char letter;
....
if(letter-is-upper-case) {
letter = (int) letter - 32;
}
Some (or most) C compilers would complain if you did not reinterpret the var as an int before adding/subtracting.
but, in the end, the type 'char' is just another term for int, really, since ASCII assigns a unique integer for each letter.
This question already has answers here:
How to determine the result of assigning multi-character char constant to a char variable?
(5 answers)
Closed 9 years ago.
I have the following code in my program:
char ch='abcd';
printf("%c",ch);
The output is d.
I fail to understand why is a char variable allowed to take in 4 characters in its declaration without giving a compile time error.
Note: More than 4 characters is giving an error.
'abcd' is called a multicharacter constant, and will has an implementation-defined value, here your compiler gives you 'd'.
If you use gcc and compile your code with -Wmultichar or -Wall, gcc will warn you about this.
I fail to understand why is a char variable allowed to take in 4
characters in its declaration without giving a compile time error.
It's not packing 4 characters into one char. The multi-character const 'abcd' is of type int and then the compiler does constant conversion to convert it to char (which overflows in this case).
Assuming you know that you are using multi-char constant, and what it is.
I don't use VS these days, but my take on it is, that 4-char multi-char is packed into an int, then down-casted to a char. That is why it is allowed. Since the packing order of multi-char constant into an integer type is compiler-defined it can behave like you observe it.
Because multi-character constants are meant to be used to fill integer typed, you could try 8-byte long multi-char. I am not sure whether VS compiler supports it, but there is a good chance it is, because that would fit into a 64-bit long type.
It probably should give a warning about trying to fit a literal value too big for the type. It's kind of like unsigned char leet = 1337;. I am not sure, however, how does this work in VS (whether it fires a warning or an error).
4 characters are not being put into a char variable, but into an int character constant which is then assigned to a char.
3 parts of the C standard (C11dr §6.4.4.4) may help:
"An integer character constant is a sequence of one or more multibyte characters enclosed in single-quotes, as in 'x'."
"An integer character constant has type int."
"The value of an integer character constant containing more than one character (e.g., 'ab'), or containing a character or escape sequence that does not map to a single-byte execution character, is implementation-defined."
OP's code of char ch='abcd'; is the the assignment of an int to a char as 'abcd' is an int. Just like char ch='Z';, ch is assigned the int value of 'Z'. In this case, there is no surprise, as the value of 'Z' fits nicely in a char. In the 'abcd', case, the value does not fit in a char and so some information is lost. Various outcomes are possible. Typically on one endian platform, ch will have a value of 'a' and on another, the value of 'd'.
The 'abcd' is an int value, much like 12345 in int x = 12345;.
When the size(int) == 4, an int may be assigned a character constant such as 'abcd'.
When size(int) != 4, the limit changes. So with an 8-char int, int x = 'abcdefgh'; is possible. etc.
Given that an int is only guaranteed to have a minimum range -32767 to 32767, anything beyond 2 is non-portable.
The int endian-ness of even int = 'ab'; presents concerns.
Character constant like 'abcd' are typically used incorrectly and thus many compilers have a warning that is good to enable to flag this uncommon C construct.
A char in the C programming language is a fixed-size byte entity designed specifically to be large enough to store a character value from an encoding such as ASCII.
But to what extent are the integer values relating to ASCII encoding interchangeable with the char characters? Is there any way to refer to 'A' as 65 (decimal)?
getchar() returns an integer - presumably this relates directly to such values? Also, if I am not mistaken, it is possible in certain contexts to increment chars ... such that (roughly speaking) '?'+1 == '#'.
Or is such encoding not guaranteed to be ASCII? Does it depend entirely upon the particular environment? Is such manipulation of chars impractical or impossible in C?
Edit: Relevant: C comparison char and int
I am answering just the question about incrementing characters, since the other issues are addressed in other answers.
The C standard guarantees that '0' to '9' are consecutive, so you can increment a digit character (except '9') and get the next digit character, or do other arithmetic with them (C 1999 5.2.1 3).
The relationships between other characters are not guaranteed by the C standard, so you would need documentation from your specific C implementation (primarily the compiler) regarding this.
But to what extent are the integer values relating to ASCII encoding interchangeable with the char characters? Is there any way to refer to 'A' as 65 (decimal)?
In fact, you can't do anything else. char is just an integral type, and if you write
char ch = 'A';
then (assuming ASCII), ch will merely hold the integer value 65 - presenting it to the user is a different problem.
Or is such encoding not guaranteed to be ASCII?
No, it isn't. C doesn't rely on any specific character encoding.
Does it depend entirely upon the particular environment?
Yes, pretty much.
Is such manipulation of chars impractical or impossible in C?
No, you just have to be careful and know the standard quite well - then you'll be safe.
character literals like 'A' have type int .. they are completely interchangeable with their integer value. However, that integer value is not mandated by the C standard; it might be ASCII (and is for the vast majority of common implementations) but need not be; it is implementation defined. The mapping of integer values for characters does have one guarantee given by the Standard: the values of the decimal digits are continguous. (i.e., '1' - '0' == 1, ... '9' - '0' == 9).
Where the source code has 'A', the compiled object will just have the byte value instead. That's why it is allowed to do arithmetic with bytes (the type of 'A' is char, i.e. byte).
Of course, a character encoding (more accurately, a code page) must be applied to get that byte value, and that codepage would serve as the "native" encoding of the compiler for hard-coded strings and char values.
Loosely, you could think of char and string literals in C source as essentially being macros. On an ASCII system the "macro" 'A' would resolve to (char) 65, and on an EBCDIC system to (char) 193. Similarly, C strings compile down to zero-terminated arrays of chars (bytes). This logic affects the symbol table also, since the symbols are taken from the source in its native encoding.
So no, ASCII is not the only possibility for the encoding of literals in source code. But due to the restriction of single-quoted characters being chars, there is a guarantee that UTF-16 or other multi-byte encodings are excluded.