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Having following simple C++ code:
#include <stdio.h>
int main() {
char c1 = 130;
unsigned char c2 = 130;
printf("1: %+u\n", c1);
printf("2: %+u\n", c2);
printf("3: %+d\n", c1);
printf("4: %+d\n", c2);
...
return 0;
}
the output is like that:
1: 4294967170
2: 130
3: -126
4: +130
Can someone please explain me the line 1 and 3 results?
I'm using Linux gcc compiler with all default settings.
(This answer assumes that, on your machine, char ranges from -128 to 127, that unsigned char ranges from 0 to 255, and that unsigned int ranges from 0 to 4294967295, which happens to be the case.)
char c1 = 130;
Here, 130 is outside the range of numbers representable by char. The value of c1 is implementation-defined. In your case, the number happens to "wrap around," initializing c1 to static_cast<char>(-126).
In
printf("1: %+u\n", c1);
c1 is promoted to int, resulting in -126. Then, it is interpreted by the %u specifier as unsigned int. This is undefined behavior. This time the resulting number happens to be the unique number representable by unsigned int that is congruent to -126 modulo 4294967296, which is 4294967170.
In
printf("3: %+d\n", c1);
The int value -126 is interpreted by the %d specifier as int directly, and outputs -126 as expected (?).
In cases 1, 2 the format specifier doesn't match the type of the argument, so the behaviour of the program is undefined (on most systems). On most systems char and unsigned char are smaller than int, so they promote to int when passed as variadic arguments. int doesn't match the format specifier %u which requires unsigned int.
On exotic systems (which your target is not) where unsigned char is as large as int, it will be promoted to unsigned int instead, in which case 4 would have UB since it requires an int.
Explanation for 3 depends a lot on implementation specified details. The result depends on whether char is signed or not, and it depends on the representable range.
If 130 was a representable value of char, such as when it is an unsigned type, then 130 would be the correct output. That appears to not be the case, so we can assume that char is a signed type on the target system.
Initialising a signed integer with an unrepresentable value (such as char with 130 in this case) results in an implementation defined value.
On systems with 2's complement representation for signed numbers - which is ubiquitous representation these days - the implementation defined value is typically the representable value that is congruent with the unrepresentable value modulo the number of representable values. -126 is congruent with 130 modulo 256 and is a representable value of char.
A char is 8 bits. This means it can represent 2^8=256 unique values. A uchar represents 0 to 255, and a signed char represents -128 to 127 (could represent absolutely anything, but this is the typical platform implementation). Thus, assigning 130 to a char is out of range by 2, and the value overflows and wraps the value to -126 when it is interpreted as a signed char. The compiler sees 130 as an integer and makes an implicit conversion from int to char. On most platforms an int is 32-bit and the sign bit is the MSB, the value 130 easily fits into the first 8-bits, but then the compiler wants to chop of 24 bits to squeeze it into a char. When this happens, and you've told the compiler you want a signed char, the MSB of the first 8 bits actually represents -128. Uh oh! You have this in memory now 1000 0010, which when interpreted as a signed char is -128+2. My linter on my platform screams about this . .
I make that important point about interpretation because in memory, both values are identical. You can confirm this by casting the value in the printf statements, i.e., printf("3: %+d\n", (unsigned char)c1);, and you'll see 130 again.
The reason you see the large value in your first printf statement is that you are casting a signed char to an unsigned int, where the char has already overflowed. The machine interprets the char as -126 first, and then casts to unsigned int, which cannot represent that negative value, so you get the max value of the signed int and subtract 126.
2^32-126 = 4294967170 . . bingo
In printf statement 2, all the machine has to do is add 24 zeros to reach 32-bit, and then interpret the value as int. In statement one, you've told it that you have a signed value, so it first turns that to a 32-bit -126 value, and then interprets that -ve integer as an unsigned integer. Again, it flips how it interprets the most significant bit. There are 2 steps:
Signed char is promoted to signed int, because you want to work with ints. The char (is probably copied and) has 24 bits added. Because we're looking at a signed value, some machine instruction will happen to perform twos complement, so the memory here looks quite different.
The new signed int memory is interpreted as unsigned, so the machine looks at the MSB and interprets it as 2^32 instead of -2^31 as happened in the promotion.
An interesting bit of trivia, is you can suppress the clang-tidy linter warning if you do char c1 = 130u;, but you still get the same garbage based on the above logic (i.e. the implicit conversion throws away the first 24-bits, and the sign-bit was zero anyhow). I'm have submitted an LLVM clang-tidy missing functionality report based on exploring this question (issue 42137 if you really wanna follow it) 😉.
Up to 255, I can understand how the integers are stored in char and unsigned char ;
#include<stdio.h>
int main()
{
unsigned char a = 256;
printf("%d\n",a);
return(0);
}
In the code above I have an output of 0 for unsigned char as well as char.
For 256 I think this is the way the integer stored in the code (this is just a guess):
First 256 converted to binary representation which is 100000000 (totally 9 bits).
Then they remove the remove the leftmost bit (the bit which is set) because the char datatype only have 8 bits of memory.
So its storing in the memory as 00000000 , that's why its printing 0 as output.
Is the guess correct or any other explanation is there?
Your guess is correct. Conversion to an unsigned type uses modular arithmetic: if the value is out of range (either too large, or negative) then it is reduced modulo 2N, where N is the number of bits in the target type. So, if (as is often the case) char has 8 bits, the value is reduced modulo 256, so that 256 becomes zero.
Note that there is no such rule for conversion to a signed type - out-of-range values give implementation-defined results. Also note that char is not specified to have exactly 8 bits, and can be larger on less mainstream platforms.
On your platform (as well as on any other "normal" platform) unsigned char is 8 bit wide, so it can hold numbers from 0 to 255.
Trying to assign 256 (which is an int literal) to it results in an unsigned integer overflow, that is defined by the standard to result in "wraparound". The result of u = n where u is an unsigned integral type and n is an unsigned integer outside its range is u = n % (max_value_of_u +1).
This is just a convoluted way to say what you already said: the standard guarantees that in these cases the assignment is performed keeping only the bits that fit in the target variable. This norm is there since most platform already implement this at the assembly language level (unsigned integer overflow typically results in this behavior plus some kind of overflow flag set to 1).
Notice that all this do not hold for signed integers (as often plain char is): signed integer overflow is undefined behavior.
yes, that's correct. 8 bits can hold 0 to 255 unsigned, or -128 to 127 signed. Above that and you've hit an overflow situation and bits will be lost.
Does the compiler give you warning on the above code? You might be able to increase the warning level and see something. It won't warn you if you assign a variable that can't be determined statically (before execution), but in this case it's pretty clear you're assigning something too large for the size of the variable.
I'm writing a game server, and this might be an easy question, but I just want some clarification.
Why is it that a byte (char or unsigned char) can hold up to a value of 255 (0xFF, which I believe is 2 bytes)? When I use sizeof(unsigned char) the compiler tells me it is 1 byte.
Is it because (in ACSII) it is getting "converted" to a character?
Sorry for this poor explaination, I'm not really good at describing a question.
This touches on a bunch of subjects, including the historical meaning of a byte, the C definition of a char, and mathematics.
For starters, a byte has historically been a lot of things, but nowadays we nearly always mean an octet, which is 8 bits. As a play on words, there's also the nybble (or often nibble) which is half a byte (not called bite).
Mathematics tells us that with an ordered combination of 8 1-or-0 values, we get 28 = 256 combinations. Sometimes we use this unsigned, sometimes signed, but either way we want to have 0 in the range; so the unsigned range is 0..255. For the signed range, we have more options, of which two's complement is the most popular; in that case, we get one more negative value than positive, for a range of -128..+127.
C++ inherits char from C, where it is defined to have a sizeof of 1, to be the smallest addressable size (i.e. having distinct address values with &), and a minimal range of -128..127 or 0..255 depending on if it's signed or not. That boils down to requiring at least 8 bits, or one byte; exactly one byte if the machine supports it.
0xff is another way of writing 255. 0x is the C way of marking a hexadecimal constant, so each digit in it is 4 bits (for 16 possible digits), ergo the nibble. This translates to an unsigned octet with all bits set to 1.
If specific size matters to your code, there is a header stdint.h that defines types of minimal and exact sizes, for speed or size optimization.
Incidentally, ASCII is a 7-bit character set. Machines with 7-bit bytes are unusual nowadays, and wider character sets like ISO 8859-1 and UTF-8 are popular.
0xFF can be stored in 8 bits, which is one byte.
sizeof(char) is defined to always return 1, regardless of the actual size in bits of the underlying datatype (see 5.3.3.1 of the current standard). The sizes of all other dataypes are calculated relative to the size of a char.
When I use sizeof(unsigned char) the compiler tells me it is 1 byte.
The size of char [whether it is signed or unsigned ] is always 1 as mandated by the C++ Standard.
char size is always 1 but number of bits can differ, C define macro CHAR_BIT that have number of bits in char.
This mean maximum value that unsigned char can have is pow(2, CHAR_BIT) - 1.
More info there: What is CHAR_BIT?
Sizeof char or unsigned char is 1 Byte as per the standard.
Why different ranges if same size?
1 Byte = 8 bits or 2^8
2^8 = 256
Hence,
signed char range is from -128 to 127
unsigned char range is from 0 to 255
This is because in case of signed char one of the bits is used to store the sign, while since unsigned char cannot be -ve, that bit is utlized to increase the range.
255, 0xFF is one byte when represented as an unsigned char. You cannot represent 255 as a signed char.
1 byte is 8 bits so in case of
signed : (1 bit is used for sign so 2^7 = 128) it holds from -128 to 127
unsigned : (2^8 = 255) it holds from 0 to 255
What's the point of negative ASCII values?
int a = '«'; //a = -85 but as in ASCII table '<<' should be 174
There are no negative ASCII values. ASCII includes definitions for 128 characters. Their indexes are all positive (or zero!).
You're seeing this negative value because the character is from an Extended ASCII set and is too large to fit into the char literal. The value therefore overflows into the bit of your char (signed on your system, apparently) that defines negativeness.
The workaround is to write the value directly:
unsigned char a = 0xAE; // «
I've written it in hexadecimal notation for convention and because I think it looks prettier than 174. :)
This is an artefact of your compiler's char type being a signed integer type, and int being a wider signed integer type, and thus the character constant is considered a negative number and is sign-extended to the wider integer type.
There is not much sense in it, it just happens. The C standard allows for compiler implementations to choose whether they consider char to be signed or unsigned. Some compilers even have compile time switches to change the default. If you want to make sure about the signedness of the char type, explicitly write signed char or unsigned char, respectively.
Use an unsigned char to be extended to an int to avoid the negative int value, or open a whole new Pandora's box and enjoy wchar.
There is no such thing. ASCII is a table of characters, each character has an index, or a position, in the table. There are no "negative" indices.
Some compilers, though, consider char to be a signed integral data type, which is probably the reason for the confusion here.
If you print it as unsigned int, you will get the same bits interpreted as a unsigned (positive) value.
ASCII ranges 0..127, ANSI (also called 'extended ASCII') ranges 0..255.
ANSI range won't fit in a signed char (the default type for characters in most compilers).
Most compilers have an option like 'char' Type is Unsigned (GCC).
I had this artifact. When you use char as symbols you have no problem. But when you use it as integer (with isalpha(), etc.) and the ASCII code is greater then 127, then the 'char' interpret as 'signed char' and isalpha() return an exception. When I need use the 'char' as integer I cast the 'char' to unsigned:
isalpha((unsigned char)my_char);
#n0rd: koi8 codepage uses ascii from 128 to 255 and other national codepages: http://www.asciitable.com/
In a character representation, you have 8 bits (1 byte) allotted.
Out of this, the first bit is used to represent sign. In the case of unsigned character, it uses all 8 bits to represent a number allowing 0 to 255 where
128-255 are called extended ASCII.
Due to the representation in the memory as I have described, we have -1 having the same value as 255, char(-2)==char(254)
In C/C++, what an unsigned char is used for? How is it different from a regular char?
In C++, there are three distinct character types:
char
signed char
unsigned char
If you are using character types for text, use the unqualified char:
it is the type of character literals like 'a' or '0' (in C++ only, in C their type is int)
it is the type that makes up C strings like "abcde"
It also works out as a number value, but it is unspecified whether that value is treated as signed or unsigned. Beware character comparisons through inequalities - although if you limit yourself to ASCII (0-127) you're just about safe.
If you are using character types as numbers, use:
signed char, which gives you at least the -127 to 127 range. (-128 to 127 is common)
unsigned char, which gives you at least the 0 to 255 range.
"At least", because the C++ standard only gives the minimum range of values that each numeric type is required to cover. sizeof (char) is required to be 1 (i.e. one byte), but a byte could in theory be for example 32 bits. sizeof would still be report its size as 1 - meaning that you could have sizeof (char) == sizeof (long) == 1.
This is implementation dependent, as the C standard does NOT define the signed-ness of char. Depending on the platform, char may be signed or unsigned, so you need to explicitly ask for signed char or unsigned char if your implementation depends on it. Just use char if you intend to represent characters from strings, as this will match what your platform puts in the string.
The difference between signed char and unsigned char is as you'd expect. On most platforms, signed char will be an 8-bit two's complement number ranging from -128 to 127, and unsigned char will be an 8-bit unsigned integer (0 to 255). Note the standard does NOT require that char types have 8 bits, only that sizeof(char) return 1. You can get at the number of bits in a char with CHAR_BIT in limits.h. There are few if any platforms today where this will be something other than 8, though.
There is a nice summary of this issue here.
As others have mentioned since I posted this, you're better off using int8_t and uint8_t if you really want to represent small integers.
Because I feel it's really called for, I just want to state some rules of C and C++ (they are the same in this regard). First, all bits of unsigned char participate in determining the value if any unsigned char object. Second, unsigned char is explicitly stated unsigned.
Now, I had a discussion with someone about what happens when you convert the value -1 of type int to unsigned char. He refused the idea that the resulting unsigned char has all its bits set to 1, because he was worried about sign representation. But he didn't have to be. It's immediately following out of this rule that the conversion does what is intended:
If the new type is unsigned, the value is converted by repeatedly adding or subtracting one more than the maximum value that can be represented in the new type until the value is in the range of the new type. (6.3.1.3p2 in a C99 draft)
That's a mathematical description. C++ describes it in terms of modulo calculus, which yields to the same rule. Anyway, what is not guaranteed is that all bits in the integer -1 are one before the conversion. So, what do we have so we can claim that the resulting unsigned char has all its CHAR_BIT bits turned to 1?
All bits participate in determining its value - that is, no padding bits occur in the object.
Adding only one time UCHAR_MAX+1 to -1 will yield a value in range, namely UCHAR_MAX
That's enough, actually! So whenever you want to have an unsigned char having all its bits one, you do
unsigned char c = (unsigned char)-1;
It also follows that a conversion is not just truncating higher order bits. The fortunate event for two's complement is that it is just a truncation there, but the same isn't necessarily true for other sign representations.
As for example usages of unsigned char:
unsigned char is often used in computer graphics, which very often (though not always) assigns a single byte to each colour component. It is common to see an RGB (or RGBA) colour represented as 24 (or 32) bits, each an unsigned char. Since unsigned char values fall in the range [0,255], the values are typically interpreted as:
0 meaning a total lack of a given colour component.
255 meaning 100% of a given colour pigment.
So you would end up with RGB red as (255,0,0) -> (100% red, 0% green, 0% blue).
Why not use a signed char? Arithmetic and bit shifting becomes problematic. As explained already, a signed char's range is essentially shifted by -128. A very simple and naive (mostly unused) method for converting RGB to grayscale is to average all three colour components, but this runs into problems when the values of the colour components are negative. Red (255, 0, 0) averages to (85, 85, 85) when using unsigned char arithmetic. However, if the values were signed chars (127,-128,-128), we would end up with (-99, -99, -99), which would be (29, 29, 29) in our unsigned char space, which is incorrect.
signed char has range -128 to 127; unsigned char has range 0 to 255.
char will be equivalent to either signed char or unsigned char, depending on the compiler, but is a distinct type.
If you're using C-style strings, just use char. If you need to use chars for arithmetic (pretty rare), specify signed or unsigned explicitly for portability.
unsigned char takes only positive values....like 0 to 255
where as
signed char takes both positive and negative values....like -128 to +127
char and unsigned char aren't guaranteed to be 8-bit types on all platforms—they are guaranteed to be 8-bit or larger. Some platforms have 9-bit, 32-bit, or 64-bit bytes. However, the most common platforms today (Windows, Mac, Linux x86, etc.) have 8-bit bytes.
An unsigned char is an unsigned byte value (0 to 255). You may be thinking of char in terms of being a "character" but it is really a numerical value. The regular char is signed, so you have 128 values, and these values map to characters using ASCII encoding. But in either case, what you are storing in memory is a byte value.
In terms of direct values a regular char is used when the values are known to be between CHAR_MIN and CHAR_MAX while an unsigned char provides double the range on the positive end. For example, if CHAR_BIT is 8, the range of regular char is only guaranteed to be [0, 127] (because it can be signed or unsigned) while unsigned char will be [0, 255] and signed char will be [-127, 127].
In terms of what it's used for, the standards allow objects of POD (plain old data) to be directly converted to an array of unsigned char. This allows you to examine the representation and bit patterns of the object. The same guarantee of safe type punning doesn't exist for char or signed char.
unsigned char is the heart of all bit trickery. In almost all compilers for all platforms an unsigned char is simply a byte and an unsigned integer of (usually) 8 bits that can be treated as a small integer or a pack of bits.
In addition, as someone else has said, the standard doesn't define the sign of a char. So you have 3 distinct char types: char, signed char, unsigned char.
If you like using various types of specific length and signedness, you're probably better off with uint8_t, int8_t, uint16_t, etc simply because they do exactly what they say.
Some googling found this, where people had a discussion about this.
An unsigned char is basically a single byte. So, you would use this if you need one byte of data (for example, maybe you want to use it to set flags on and off to be passed to a function, as is often done in the Windows API).
An unsigned char uses the bit that is reserved for the sign of a regular char as another number. This changes the range to [0 - 255] as opposed to [-128 - 127].
Generally unsigned chars are used when you don't want a sign. This will make a difference when doing things like shifting bits (shift extends the sign) and other things when dealing with a char as a byte rather than using it as a number.
unsigned char takes only positive values: 0 to 255 while
signed char takes positive and negative values: -128 to +127.
quoted frome "the c programming laugage" book:
The qualifier signed or unsigned may be applied to char or any integer. unsigned numbers
are always positive or zero, and obey the laws of arithmetic modulo 2^n, where n is the number
of bits in the type. So, for instance, if chars are 8 bits, unsigned char variables have values
between 0 and 255, while signed chars have values between -128 and 127 (in a two' s
complement machine.) Whether plain chars are signed or unsigned is machine-dependent,
but printable characters are always positive.
signed char and unsigned char both represent 1byte, but they have different ranges.
Type | range
-------------------------------
signed char | -128 to +127
unsigned char | 0 to 255
In signed char if we consider char letter = 'A', 'A' is represent binary of 65 in ASCII/Unicode, If 65 can be stored, -65 also can be stored. There are no negative binary values in ASCII/Unicode there for no need to worry about negative values.
Example
#include <stdio.h>
int main()
{
signed char char1 = 255;
signed char char2 = -128;
unsigned char char3 = 255;
unsigned char char4 = -128;
printf("Signed char(255) : %d\n",char1);
printf("Unsigned char(255) : %d\n",char3);
printf("\nSigned char(-128) : %d\n",char2);
printf("Unsigned char(-128) : %d\n",char4);
return 0;
}
Output -:
Signed char(255) : -1
Unsigned char(255) : 255
Signed char(-128) : -128
Unsigned char(-128) : 128