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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) 😉.
In C++ program I have some char buf[256]. The problem is here:
if (buf[pbyte] >= 0xFF)
buf[++pbyte] = 0x00;
This always returns false even when buf[pbyte] is equal to 255 AKA 0xFF as seen in immediate window and watch window. Thus the statement does not get executed. However, when I change this to below:
if (buf[pbyte] >= char(0xFF))
buf[++pbyte] = 0x00;
The program works; how come?
The literal 0xFF is treated as an int with the value 255.
When you compare a char to an int, the char is promoted to an int before the comparison.
On some platforms char is a signed value with a range like -128 to +127. On other platforms char is an unsigned value with a range like 0 to 255.
If your platform's char is signed, and its bit pattern is 0xFF, then it's probably -1. Since -1 is a valid int, the promotion stops there.
You end up comparing -1 to 255.
The solution is to eliminate the implicit conversions. You can write the comparison as:
if (buf[pbyte] == '\xFF') ...
Now both sides are chars, so they'll be promoted in the same manner and are directly comparable.
The problem is that char is signed on your system.
In the common 2s complement representation, a signed char with the "byte-value" 0xFF represents the integer -1, while 0xFF is an int with value 255. Thus, you are effectively comparing int(-1) >= int(255), which yields false. Keep in mind that they are compared as int because of arithmetic conversion rules, that is both operands are promoted ("cast implicitly") to int before comparing.
If you write char(0xFF) however, you do end up with the comparison -1 >= -1, which yields true as expected.
If you want to store numbers in the range [0,255], you should use unsigned char or std::uint8_t instead of char.
Due to the integer promotions in this condition
if (buf[pbyte] >= `0xFF`)
the two operands are converted to the type int (more precisely only the left operand is converted to an object of the type int because the right operand already has the type int). As it seems in your system the type char behaves as the type signed char then the value '\xFF' is a negative value equal to -1. Then this value is converted to an object of the type int you will get 0xFFFFFFFF (assuming that the type int occupies 4 bytes).
On the other hand the integer constant 0xFF is a positive value that has internal representation like 0x000000FF
Thus the condition in the if statement
if ( 0xFFFFFFFF >= `0x000000FF`)
yields false.
When you use the casting ( char )0xFF then the both operands have the same type and the same values.
Integer literals by default are converted to an int, assuming the value will fit into the int type, otherwise it is promoted to a long.
So in your code you are specifying 0xFF which is interpreted as an int type, i.e. 0x000000FF
Why does the following code print "?" ?
Also how can -1 be assigned to an unsigned char?
char test;
unsigned char testu; //isn't it supposed to hold values in range 0 - 255?
test = -1;
testu = -1;
cout<<"TEST CHAR = "<<test<<endl;
cout<<"TESTU CHAR = "<<testu<<endl;
unsigned simply affects how the internal representation of the number (chars are numbers, remember) is interpreted. So -1 is 1111 1111 in two's complement notation, which when put into an unsigned char changes the meaning (for the same bit representation) to 255.
The question mark is probably the result of your font/codepage not mapping the (extended) ASCII value 255 to a character it can display.
I don't think << discerns between an unsigned char and a signed char, since it interprets their values as ASCII codes, not plain numbers.
Also, it depends on your compiler whether chars are signed or unsigned by default; actually, the spec states there's three different char types (plain, signed, and unsigned).
When you assign a negative value to an unsigned variable, the result is that it wraps around. -1 becomes 255 in this case.
I don't know C or C++, but my intuition is telling me that it's wrapping -1 to 255 and printing ÿ, but since that's not in the first 128 characters it prints ? instead. Just a guess.
To test this, try assigning -191 and see if it prints A (or B if my math is off).
Signed/unsigned is defined by the use of the highest order bit of that number.
You can assign a negative integer to it. The sign bit will be interpreted in the signed case (when you perform arithmetics with it). When you treat it it like a character it will simply take the highest order bit as if it was an unsigned char and just produce an ASCII char beyond 127 (decimal):
unsigned char c = -2;
is equivalent to:
unsigned char c = 128;
WHEN the c is treated as a character.
-1 is an exception: it has all 8 bits set and is treated as 255 dec.
I was getting some odd behaviour out of a switch block today, specifically I was reading a byte from a file and comparing it against certain hex values (text file encoding issue, no big deal). The code looked something like:
char BOM[3] = {0};
b_error = ReadFile (iNCfile, BOM, 3, &lpNumberOfBytesRead, NULL);
switch ( BOM[0] ) {
case 0xef: {
// Byte Order Marker Potentially Indicates UTF-8
if ( ( BOM[1] == 0xBB ) && ( BOM[2] == 0xBF ) ) {
iNCfileEncoding = UTF8;
}
break;
}
}
Which didn't work, although the debug looked ok. I realized that the switch was promoting the values to integers, and once that clicked in place I was able to match using 0xffffffef in the case statement. Of course the correct solution was to make BOM[] unsigned and now everything promotes and compares as expected.
Can someone briefly explain what was going on in the char -> int promotion that produced 0xffffffef instead of 0x000000ef?
The sign of your (signed) char got extended to form a signed int. That is because of the way signed values are stored in binary.
Example
1 in binary char = 00000001
1 in binary int = 00000000 00000000 00000000 00000001
-1 in binary char = 11111111
-1 in binary int is NOT 00000000 00000000 00000000 11111111 but 11111111 11111111 11111111 11111111
if you convert back to decimal you should know up front whether you are dealing with signed or unsigned values because 11111111 might be -1 in signed and 255 in unsigned.
char must be signed on your platform, and what you are seeing is sign extension.
What hasn't been stated yet (as I type, anyway) is that it is unspecified whether or not char is singed. In your case - as was stated - char is signed, so any ASCII value above 127 is going to be interpreted as a negative.
"Can someone briefly explain what was
going on in the char -> int promotion
that produced 0xffffffef instead of
0x000000ef?"
Contrary to the four answers so far, it didn't.
Rather, you had a negative char value, which as a switch condition was promoted to the same negative int value as required by
C++98 §6.4.2/2
Integral promotions are performed.
Then with your 32-bit C++ compiler 0xffffffef was interpreted as an unsigned int literal, because it’s too large for a 32-bit int, by
C++98 2.13.1/2
If it is octal or hexadecimal and has no suffix, it has the first of these types in
which it can be represented: int, unsigned int, long int, unsigned long int.
Now, for the case label,
C++98 §6.4.2/2
The integral constant-expression (5.19) is implicitly converted to the promoted
type of the switch condition.
In your case, with signed destination type, the result of the conversion is formally implementation-defined, by
C++98 §4.7/3
If the destination type is signed, the value is unchanged if it can be represented
in the destination type (and bit-field width); otherwise, the value is
implementation-defined.
But in practice nearly all compilers use two's complement representation with no trapping, and so the implementation defined conversion is in your case that the bitpattern 0xffffffef is interpreted as two's complement specification of a negative value. You can calculate which value by 0xffffffef - 232, because we’re talking 32-bit representation here. Or, since this is just an 8-bit value that’s been sign extended to 32 bits, you can alternatively calculate it as 0xef - 28, where 0xef is the character code point.
Cheers & hth.,
Sign extension
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