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In C++, sizeof('a') == sizeof(char) == 1. This makes intuitive sense, since 'a' is a character literal, and sizeof(char) == 1 as defined by the standard.
In C however, sizeof('a') == sizeof(int). That is, it appears that C character literals are actually integers. Does anyone know why? I can find plenty of mentions of this C quirk but no explanation for why it exists.
discussion on same subject
"More specifically the integral promotions. In K&R C it was virtually (?)
impossible to use a character value without it being promoted to int first,
so making character constant int in the first place eliminated that step.
There were and still are multi character constants such as 'abcd' or however
many will fit in an int."
The original question is "why?"
The reason is that the definition of a literal character has evolved and changed, while trying to remain backwards compatible with existing code.
In the dark days of early C there were no types at all. By the time I first learnt to program in C, types had been introduced, but functions didn't have prototypes to tell the caller what the argument types were. Instead it was standardised that everything passed as a parameter would either be the size of an int (this included all pointers) or it would be a double.
This meant that when you were writing the function, all the parameters that weren't double were stored on the stack as ints, no matter how you declared them, and the compiler put code in the function to handle this for you.
This made things somewhat inconsistent, so when K&R wrote their famous book, they put in the rule that a character literal would always be promoted to an int in any expression, not just a function parameter.
When the ANSI committee first standardised C, they changed this rule so that a character literal would simply be an int, since this seemed a simpler way of achieving the same thing.
When C++ was being designed, all functions were required to have full prototypes (this is still not required in C, although it is universally accepted as good practice). Because of this, it was decided that a character literal could be stored in a char. The advantage of this in C++ is that a function with a char parameter and a function with an int parameter have different signatures. This advantage is not the case in C.
This is why they are different. Evolution...
I don't know the specific reasons why a character literal in C is of type int. But in C++, there is a good reason not to go that way. Consider this:
void print(int);
void print(char);
print('a');
You would expect that the call to print selects the second version taking a char. Having a character literal being an int would make that impossible. Note that in C++ literals having more than one character still have type int, although their value is implementation defined. So, 'ab' has type int, while 'a' has type char.
using gcc on my MacBook, I try:
#include <stdio.h>
#define test(A) do{printf(#A":\t%i\n",sizeof(A));}while(0)
int main(void){
test('a');
test("a");
test("");
test(char);
test(short);
test(int);
test(long);
test((char)0x0);
test((short)0x0);
test((int)0x0);
test((long)0x0);
return 0;
};
which when run gives:
'a': 4
"a": 2
"": 1
char: 1
short: 2
int: 4
long: 4
(char)0x0: 1
(short)0x0: 2
(int)0x0: 4
(long)0x0: 4
which suggests that a character is 8 bits, like you suspect, but a character literal is an int.
Back when C was being written, the PDP-11's MACRO-11 assembly language had:
MOV #'A, R0 // 8-bit character encoding for 'A' into 16 bit register
This kind of thing's quite common in assembly language - the low 8 bits will hold the character code, other bits cleared to 0. PDP-11 even had:
MOV #"AB, R0 // 16-bit character encoding for 'A' (low byte) and 'B'
This provided a convenient way to load two characters into the low and high bytes of the 16 bit register. You might then write those elsewhere, updating some textual data or screen memory.
So, the idea of characters being promoted to register size is quite normal and desirable. But, let's say you need to get 'A' into a register not as part of the hard-coded opcode, but from somewhere in main memory containing:
address: value
20: 'X'
21: 'A'
22: 'A'
23: 'X'
24: 0
25: 'A'
26: 'A'
27: 0
28: 'A'
If you want to read just an 'A' from this main memory into a register, which one would you read?
Some CPUs may only directly support reading a 16 bit value into a 16 bit register, which would mean a read at 20 or 22 would then require the bits from 'X' be cleared out, and depending on the endianness of the CPU one or other would need shifting into the low order byte.
Some CPUs may require a memory-aligned read, which means that the lowest address involved must be a multiple of the data size: you might be able to read from addresses 24 and 25, but not 27 and 28.
So, a compiler generating code to get an 'A' into the register may prefer to waste a little extra memory and encode the value as 0 'A' or 'A' 0 - depending on endianness, and also ensuring it is aligned properly (i.e. not at an odd memory address).
My guess is that C's simply carried this level of CPU-centric behaviour over, thinking of character constants occupying register sizes of memory, bearing out the common assessment of C as a "high level assembler".
(See 6.3.3 on page 6-25 of http://www.dmv.net/dec/pdf/macro.pdf)
I remember reading K&R and seeing a code snippet that would read a character at a time until it hit EOF. Since all characters are valid characters to be in a file/input stream, this means that EOF cannot be any char value. What the code did was to put the read character into an int, then test for EOF, then convert to a char if it wasn't.
I realize this doesn't exactly answer your question, but it would make some sense for the rest of the character literals to be sizeof(int) if the EOF literal was.
int r;
char buffer[1024], *p; // don't use in production - buffer overflow likely
p = buffer;
while ((r = getc(file)) != EOF)
{
*(p++) = (char) r;
}
I haven't seen a rationale for it (C char literals being int types), but here's something Stroustrup had to say about it (from Design and Evolution 11.2.1 - Fine-Grain Resolution):
In C, the type of a character literal such as 'a' is int.
Surprisingly, giving 'a' type char in C++ doesn't cause any compatibility problems.
Except for the pathological example sizeof('a'), every construct that can be expressed
in both C and C++ gives the same result.
So for the most part, it should cause no problems.
The historical reason for this is that C, and its predecessor B, were originally developed on various models of DEC PDP minicomputers with various word sizes, which supported 8-bit ASCII but could only perform arithmetic on registers. (Not the PDP-11, however; that came later.) Early versions of C defined int to be the native word size of the machine, and any value smaller than an int needed to be widened to int in order to be passed to or from a function, or used in a bitwise, logical or arithmetic expression, because that was how the underlying hardware worked.
That is also why the integer promotion rules still say that any data type smaller than an int is promoted to int. C implementations are also allowed to use one’s-complement math instead of two’s-complement for similar historical reasons. The reason that octal character escapes and octal constants are first-class citizens compared to hex is likewise that those early DEC minicomputers had word sizes divisible into three-byte chunks but not four-byte nibbles.
I don't know, but I'm going to guess it was easier to implement it that way and it didn't really matter. It wasn't until C++ when the type could determine which function would get called that it needed to be fixed.
This is only tangential to the language spec, but in hardware the CPU usually only has one register size -- 32 bits, let's say -- and so whenever it actually works on a char (by adding, subtracting, or comparing it) there is an implicit conversion to int when it is loaded into the register. The compiler takes care of properly masking and shifting the number after each operation so that if you add, say, 2 to (unsigned char) 254, it'll wrap around to 0 instead of 256, but inside the silicon it is really an int until you save it back to memory.
It's sort of an academic point because the language could have specified an 8-bit literal type anyway, but in this case the language spec happens to reflect more closely what the CPU is really doing.
(x86 wonks may note that there is eg a native addh op that adds the short-wide registers in one step, but inside the RISC core this translates to two steps: add the numbers, then extend sign, like an add/extsh pair on the PowerPC)
This is the correct behavior, called "integral promotion". It can happen in other cases too (mainly binary operators, if I remember correctly).
EDIT: Just to be sure, I checked my copy of Expert C Programming: Deep Secrets, and I confirmed that a char literal does not start with a type int. It is initially of type char but when it is used in an expression, it is promoted to an int. The following is quoted from the book:
Character literals have type int and
they get there by following the rules
for promotion from type char. This is
too briefly covered in K&R 1, on page
39 where it says:
Every char in an expression is
converted into an int....Notice that
all float's in an expression are
converted to double....Since a
function argument is an expression,
type conversions also take place when
arguments are passed to functions: in
particular, char and short become int,
float becomes double.
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.
I am trying to compare two chars with each other but treating them as integers. These are struct values in a linked list. I have printed out temp->next->variable and temp->variable and confirmed that the if statement should hold, ex: 3 > 2. But I'm thinking it might not work because they are char.
Will the fact that they are char values have an impact on the comparison?
if(temp->next->variable > temp->variable)
{
....
}
Chars are integers, of 8-bit length, so if you're always treating the chars as ints, this will have the exact behavior you desire. Be wary, as some compilers allow char to default to unsigned, some to signed, which may create issues seemingly at random. (If both numbers have the same highest order bit, it won't make a difference, otherwise the result will be opposite what you expect, if the signed-ness is also opposite what you expect.)
If you're treating them as characters, then this will give you their lexicographical comparison, which is based on the interal integer representation of that character. It may be good to check what your locale is -- whether your program will be using look up tables of ASCII or Unicode 8-bit, or anything else.
If you still get hidden issues, a common mistake is having many layered pointers, and even though the arrow [->] is used throughout, you may still need to apply a de-reference [*], or else you'll secretly be testing their relative locations in memory.
According to the C Standard (6.5.8 Relational operators)
3 If both of the operands have arithmetic type, the usual arithmetic
conversions are performed.
The usual arithmetic conversion includes the integer promotion that particularly means that objects of type char are converted to objects of type int
Take into account that type char can behave either as type unsigned char or signed char. So you can get different results of the comparison if sign bits of objects of type char are set.
In C++, sizeof('a') == sizeof(char) == 1. This makes intuitive sense, since 'a' is a character literal, and sizeof(char) == 1 as defined by the standard.
In C however, sizeof('a') == sizeof(int). That is, it appears that C character literals are actually integers. Does anyone know why? I can find plenty of mentions of this C quirk but no explanation for why it exists.
discussion on same subject
"More specifically the integral promotions. In K&R C it was virtually (?)
impossible to use a character value without it being promoted to int first,
so making character constant int in the first place eliminated that step.
There were and still are multi character constants such as 'abcd' or however
many will fit in an int."
The original question is "why?"
The reason is that the definition of a literal character has evolved and changed, while trying to remain backwards compatible with existing code.
In the dark days of early C there were no types at all. By the time I first learnt to program in C, types had been introduced, but functions didn't have prototypes to tell the caller what the argument types were. Instead it was standardised that everything passed as a parameter would either be the size of an int (this included all pointers) or it would be a double.
This meant that when you were writing the function, all the parameters that weren't double were stored on the stack as ints, no matter how you declared them, and the compiler put code in the function to handle this for you.
This made things somewhat inconsistent, so when K&R wrote their famous book, they put in the rule that a character literal would always be promoted to an int in any expression, not just a function parameter.
When the ANSI committee first standardised C, they changed this rule so that a character literal would simply be an int, since this seemed a simpler way of achieving the same thing.
When C++ was being designed, all functions were required to have full prototypes (this is still not required in C, although it is universally accepted as good practice). Because of this, it was decided that a character literal could be stored in a char. The advantage of this in C++ is that a function with a char parameter and a function with an int parameter have different signatures. This advantage is not the case in C.
This is why they are different. Evolution...
I don't know the specific reasons why a character literal in C is of type int. But in C++, there is a good reason not to go that way. Consider this:
void print(int);
void print(char);
print('a');
You would expect that the call to print selects the second version taking a char. Having a character literal being an int would make that impossible. Note that in C++ literals having more than one character still have type int, although their value is implementation defined. So, 'ab' has type int, while 'a' has type char.
using gcc on my MacBook, I try:
#include <stdio.h>
#define test(A) do{printf(#A":\t%i\n",sizeof(A));}while(0)
int main(void){
test('a');
test("a");
test("");
test(char);
test(short);
test(int);
test(long);
test((char)0x0);
test((short)0x0);
test((int)0x0);
test((long)0x0);
return 0;
};
which when run gives:
'a': 4
"a": 2
"": 1
char: 1
short: 2
int: 4
long: 4
(char)0x0: 1
(short)0x0: 2
(int)0x0: 4
(long)0x0: 4
which suggests that a character is 8 bits, like you suspect, but a character literal is an int.
Back when C was being written, the PDP-11's MACRO-11 assembly language had:
MOV #'A, R0 // 8-bit character encoding for 'A' into 16 bit register
This kind of thing's quite common in assembly language - the low 8 bits will hold the character code, other bits cleared to 0. PDP-11 even had:
MOV #"AB, R0 // 16-bit character encoding for 'A' (low byte) and 'B'
This provided a convenient way to load two characters into the low and high bytes of the 16 bit register. You might then write those elsewhere, updating some textual data or screen memory.
So, the idea of characters being promoted to register size is quite normal and desirable. But, let's say you need to get 'A' into a register not as part of the hard-coded opcode, but from somewhere in main memory containing:
address: value
20: 'X'
21: 'A'
22: 'A'
23: 'X'
24: 0
25: 'A'
26: 'A'
27: 0
28: 'A'
If you want to read just an 'A' from this main memory into a register, which one would you read?
Some CPUs may only directly support reading a 16 bit value into a 16 bit register, which would mean a read at 20 or 22 would then require the bits from 'X' be cleared out, and depending on the endianness of the CPU one or other would need shifting into the low order byte.
Some CPUs may require a memory-aligned read, which means that the lowest address involved must be a multiple of the data size: you might be able to read from addresses 24 and 25, but not 27 and 28.
So, a compiler generating code to get an 'A' into the register may prefer to waste a little extra memory and encode the value as 0 'A' or 'A' 0 - depending on endianness, and also ensuring it is aligned properly (i.e. not at an odd memory address).
My guess is that C's simply carried this level of CPU-centric behaviour over, thinking of character constants occupying register sizes of memory, bearing out the common assessment of C as a "high level assembler".
(See 6.3.3 on page 6-25 of http://www.dmv.net/dec/pdf/macro.pdf)
I remember reading K&R and seeing a code snippet that would read a character at a time until it hit EOF. Since all characters are valid characters to be in a file/input stream, this means that EOF cannot be any char value. What the code did was to put the read character into an int, then test for EOF, then convert to a char if it wasn't.
I realize this doesn't exactly answer your question, but it would make some sense for the rest of the character literals to be sizeof(int) if the EOF literal was.
int r;
char buffer[1024], *p; // don't use in production - buffer overflow likely
p = buffer;
while ((r = getc(file)) != EOF)
{
*(p++) = (char) r;
}
I haven't seen a rationale for it (C char literals being int types), but here's something Stroustrup had to say about it (from Design and Evolution 11.2.1 - Fine-Grain Resolution):
In C, the type of a character literal such as 'a' is int.
Surprisingly, giving 'a' type char in C++ doesn't cause any compatibility problems.
Except for the pathological example sizeof('a'), every construct that can be expressed
in both C and C++ gives the same result.
So for the most part, it should cause no problems.
The historical reason for this is that C, and its predecessor B, were originally developed on various models of DEC PDP minicomputers with various word sizes, which supported 8-bit ASCII but could only perform arithmetic on registers. (Not the PDP-11, however; that came later.) Early versions of C defined int to be the native word size of the machine, and any value smaller than an int needed to be widened to int in order to be passed to or from a function, or used in a bitwise, logical or arithmetic expression, because that was how the underlying hardware worked.
That is also why the integer promotion rules still say that any data type smaller than an int is promoted to int. C implementations are also allowed to use one’s-complement math instead of two’s-complement for similar historical reasons. The reason that octal character escapes and octal constants are first-class citizens compared to hex is likewise that those early DEC minicomputers had word sizes divisible into three-byte chunks but not four-byte nibbles.
I don't know, but I'm going to guess it was easier to implement it that way and it didn't really matter. It wasn't until C++ when the type could determine which function would get called that it needed to be fixed.
This is only tangential to the language spec, but in hardware the CPU usually only has one register size -- 32 bits, let's say -- and so whenever it actually works on a char (by adding, subtracting, or comparing it) there is an implicit conversion to int when it is loaded into the register. The compiler takes care of properly masking and shifting the number after each operation so that if you add, say, 2 to (unsigned char) 254, it'll wrap around to 0 instead of 256, but inside the silicon it is really an int until you save it back to memory.
It's sort of an academic point because the language could have specified an 8-bit literal type anyway, but in this case the language spec happens to reflect more closely what the CPU is really doing.
(x86 wonks may note that there is eg a native addh op that adds the short-wide registers in one step, but inside the RISC core this translates to two steps: add the numbers, then extend sign, like an add/extsh pair on the PowerPC)
This is the correct behavior, called "integral promotion". It can happen in other cases too (mainly binary operators, if I remember correctly).
EDIT: Just to be sure, I checked my copy of Expert C Programming: Deep Secrets, and I confirmed that a char literal does not start with a type int. It is initially of type char but when it is used in an expression, it is promoted to an int. The following is quoted from the book:
Character literals have type int and
they get there by following the rules
for promotion from type char. This is
too briefly covered in K&R 1, on page
39 where it says:
Every char in an expression is
converted into an int....Notice that
all float's in an expression are
converted to double....Since a
function argument is an expression,
type conversions also take place when
arguments are passed to functions: in
particular, char and short become int,
float becomes double.
In C++, sizeof('a') == sizeof(char) == 1. This makes intuitive sense, since 'a' is a character literal, and sizeof(char) == 1 as defined by the standard.
In C however, sizeof('a') == sizeof(int). That is, it appears that C character literals are actually integers. Does anyone know why? I can find plenty of mentions of this C quirk but no explanation for why it exists.
discussion on same subject
"More specifically the integral promotions. In K&R C it was virtually (?)
impossible to use a character value without it being promoted to int first,
so making character constant int in the first place eliminated that step.
There were and still are multi character constants such as 'abcd' or however
many will fit in an int."
The original question is "why?"
The reason is that the definition of a literal character has evolved and changed, while trying to remain backwards compatible with existing code.
In the dark days of early C there were no types at all. By the time I first learnt to program in C, types had been introduced, but functions didn't have prototypes to tell the caller what the argument types were. Instead it was standardised that everything passed as a parameter would either be the size of an int (this included all pointers) or it would be a double.
This meant that when you were writing the function, all the parameters that weren't double were stored on the stack as ints, no matter how you declared them, and the compiler put code in the function to handle this for you.
This made things somewhat inconsistent, so when K&R wrote their famous book, they put in the rule that a character literal would always be promoted to an int in any expression, not just a function parameter.
When the ANSI committee first standardised C, they changed this rule so that a character literal would simply be an int, since this seemed a simpler way of achieving the same thing.
When C++ was being designed, all functions were required to have full prototypes (this is still not required in C, although it is universally accepted as good practice). Because of this, it was decided that a character literal could be stored in a char. The advantage of this in C++ is that a function with a char parameter and a function with an int parameter have different signatures. This advantage is not the case in C.
This is why they are different. Evolution...
I don't know the specific reasons why a character literal in C is of type int. But in C++, there is a good reason not to go that way. Consider this:
void print(int);
void print(char);
print('a');
You would expect that the call to print selects the second version taking a char. Having a character literal being an int would make that impossible. Note that in C++ literals having more than one character still have type int, although their value is implementation defined. So, 'ab' has type int, while 'a' has type char.
using gcc on my MacBook, I try:
#include <stdio.h>
#define test(A) do{printf(#A":\t%i\n",sizeof(A));}while(0)
int main(void){
test('a');
test("a");
test("");
test(char);
test(short);
test(int);
test(long);
test((char)0x0);
test((short)0x0);
test((int)0x0);
test((long)0x0);
return 0;
};
which when run gives:
'a': 4
"a": 2
"": 1
char: 1
short: 2
int: 4
long: 4
(char)0x0: 1
(short)0x0: 2
(int)0x0: 4
(long)0x0: 4
which suggests that a character is 8 bits, like you suspect, but a character literal is an int.
Back when C was being written, the PDP-11's MACRO-11 assembly language had:
MOV #'A, R0 // 8-bit character encoding for 'A' into 16 bit register
This kind of thing's quite common in assembly language - the low 8 bits will hold the character code, other bits cleared to 0. PDP-11 even had:
MOV #"AB, R0 // 16-bit character encoding for 'A' (low byte) and 'B'
This provided a convenient way to load two characters into the low and high bytes of the 16 bit register. You might then write those elsewhere, updating some textual data or screen memory.
So, the idea of characters being promoted to register size is quite normal and desirable. But, let's say you need to get 'A' into a register not as part of the hard-coded opcode, but from somewhere in main memory containing:
address: value
20: 'X'
21: 'A'
22: 'A'
23: 'X'
24: 0
25: 'A'
26: 'A'
27: 0
28: 'A'
If you want to read just an 'A' from this main memory into a register, which one would you read?
Some CPUs may only directly support reading a 16 bit value into a 16 bit register, which would mean a read at 20 or 22 would then require the bits from 'X' be cleared out, and depending on the endianness of the CPU one or other would need shifting into the low order byte.
Some CPUs may require a memory-aligned read, which means that the lowest address involved must be a multiple of the data size: you might be able to read from addresses 24 and 25, but not 27 and 28.
So, a compiler generating code to get an 'A' into the register may prefer to waste a little extra memory and encode the value as 0 'A' or 'A' 0 - depending on endianness, and also ensuring it is aligned properly (i.e. not at an odd memory address).
My guess is that C's simply carried this level of CPU-centric behaviour over, thinking of character constants occupying register sizes of memory, bearing out the common assessment of C as a "high level assembler".
(See 6.3.3 on page 6-25 of http://www.dmv.net/dec/pdf/macro.pdf)
I remember reading K&R and seeing a code snippet that would read a character at a time until it hit EOF. Since all characters are valid characters to be in a file/input stream, this means that EOF cannot be any char value. What the code did was to put the read character into an int, then test for EOF, then convert to a char if it wasn't.
I realize this doesn't exactly answer your question, but it would make some sense for the rest of the character literals to be sizeof(int) if the EOF literal was.
int r;
char buffer[1024], *p; // don't use in production - buffer overflow likely
p = buffer;
while ((r = getc(file)) != EOF)
{
*(p++) = (char) r;
}
I haven't seen a rationale for it (C char literals being int types), but here's something Stroustrup had to say about it (from Design and Evolution 11.2.1 - Fine-Grain Resolution):
In C, the type of a character literal such as 'a' is int.
Surprisingly, giving 'a' type char in C++ doesn't cause any compatibility problems.
Except for the pathological example sizeof('a'), every construct that can be expressed
in both C and C++ gives the same result.
So for the most part, it should cause no problems.
The historical reason for this is that C, and its predecessor B, were originally developed on various models of DEC PDP minicomputers with various word sizes, which supported 8-bit ASCII but could only perform arithmetic on registers. (Not the PDP-11, however; that came later.) Early versions of C defined int to be the native word size of the machine, and any value smaller than an int needed to be widened to int in order to be passed to or from a function, or used in a bitwise, logical or arithmetic expression, because that was how the underlying hardware worked.
That is also why the integer promotion rules still say that any data type smaller than an int is promoted to int. C implementations are also allowed to use one’s-complement math instead of two’s-complement for similar historical reasons. The reason that octal character escapes and octal constants are first-class citizens compared to hex is likewise that those early DEC minicomputers had word sizes divisible into three-byte chunks but not four-byte nibbles.
I don't know, but I'm going to guess it was easier to implement it that way and it didn't really matter. It wasn't until C++ when the type could determine which function would get called that it needed to be fixed.
This is only tangential to the language spec, but in hardware the CPU usually only has one register size -- 32 bits, let's say -- and so whenever it actually works on a char (by adding, subtracting, or comparing it) there is an implicit conversion to int when it is loaded into the register. The compiler takes care of properly masking and shifting the number after each operation so that if you add, say, 2 to (unsigned char) 254, it'll wrap around to 0 instead of 256, but inside the silicon it is really an int until you save it back to memory.
It's sort of an academic point because the language could have specified an 8-bit literal type anyway, but in this case the language spec happens to reflect more closely what the CPU is really doing.
(x86 wonks may note that there is eg a native addh op that adds the short-wide registers in one step, but inside the RISC core this translates to two steps: add the numbers, then extend sign, like an add/extsh pair on the PowerPC)
This is the correct behavior, called "integral promotion". It can happen in other cases too (mainly binary operators, if I remember correctly).
EDIT: Just to be sure, I checked my copy of Expert C Programming: Deep Secrets, and I confirmed that a char literal does not start with a type int. It is initially of type char but when it is used in an expression, it is promoted to an int. The following is quoted from the book:
Character literals have type int and
they get there by following the rules
for promotion from type char. This is
too briefly covered in K&R 1, on page
39 where it says:
Every char in an expression is
converted into an int....Notice that
all float's in an expression are
converted to double....Since a
function argument is an expression,
type conversions also take place when
arguments are passed to functions: in
particular, char and short become int,
float becomes double.