I've spent some time poring over the standard references, but I've not been able to find an answer to the following:
is it technically guaranteed by the C/C++ standard that, given a signed integral type S and its unsigned counterpart U, the absolute value of each possible S is always less than or equal to the maximum value of U?
The closest I've gotten is from section 6.2.6.2 of the C99 standard (the wording of the C++ is more arcane to me, I assume they are equivalent on this):
For signed integer types, the bits of the object representation shall be divided into three
groups: value bits, padding bits, and the sign bit. (...) Each bit that is a value bit shall have the same value as the same bit in the object representation of the corresponding unsigned type (if there are M value bits in the signed type and Nin the unsigned type, then M≤N).
So, in hypothetical 4-bit signed/unsigned integer types, is anything preventing the unsigned type to have 1 padding bit and 3 value bits, and the signed type having 3 value bits and 1 sign bit? In such a case the range of unsigned would be [0,7] and for signed it would be [-8,7] (assuming two's complement).
In case anyone is curious, I'm relying at the moment on a technique for extracting the absolute value of a negative integer consisting of first a cast to the unsigned counterpart, and then the application of the unary minus operator (so that for instance -3 becomes 4 via cast and then 3 via unary minus). This would break on the example above for -8, which could not be represented in the unsigned type.
EDIT: thanks for the replies below Keith and Potatoswatter. Now, my last point of doubt is on the meaning of "subrange" in the wording of the standard. If it means a strictly "less-than" inclusion, then my example above and Keith's below are not standard-compliant. If the subrange is intended to be potentially the whole range of unsigned, then they are.
For C, the answer is no, there is no such guarantee.
I'll discuss types int and unsigned int; this applies equally to any corresponding pair of signed and unsigned types (other than char and unsigned char, neither of which can have padding bits).
The standard, in the section you quoted, implicitly guarantees that UINT_MAX >= INT_MAX, which means that every non-negative int value can be represented as an unsigned int.
But the following would be perfectly legal (I'll use ** to denote exponentiation):
CHAR_BIT == 8
sizeof (int) == 4
sizeof (unsigned int) == 4
INT_MIN = -2**31
INT_MAX = +2**31-1
UINT_MAX = +2**31-1
This implies that int has 1 sign bit (as it must) and 31 value bits, an ordinary 2's-complement representation, and unsigned int has 31 value bits and one padding bit. unsigned int representations with that padding bit set might either be trap representations, or extra representations of values with the padding bit unset.
This might be appropriate for a machine with support for 2's-complement signed arithmetic, but poor support for unsigned arithmetic.
Given these characteristics, -INT_MIN (the mathematical value) is outside the range of unsigned int.
On the other hand, I seriously doubt that there are any modern systems like this. Padding bits are permitted by the standard, but are very rare, and I don't expect them to become any more common.
You might consider adding something like this:
#if -INT_MIN > UINT_MAX
#error "Nope"
#endif
to your source, so it will compile only if you can do what you want. (You should think of a better error message than "Nope", of course.)
You got it. In C++11 the wording is more clear. §3.9.1/3:
The range of non-negative values of a signed integer type is a subrange of the corresponding unsigned integer type, and the value representation of each corresponding signed/unsigned type shall be the same.
But, what really is the significance of the connection between the two corresponding types? They are the same size, but that doesn't matter if you just have local variables.
In case anyone is curious, I'm relying at the moment on a technique for extracting the absolute value of a negative integer consisting of first a cast to the unsigned counterpart, and then the application of the unary minus operator (so that for instance -3 becomes 4 via cast and then 3 via unary minus). This would break on the example above for -8, which could not be represented in the unsigned type.
You need to deal with whatever numeric ranges the machine supports. Instead of casting to the unsigned counterpart, cast to whatever unsigned type is sufficient: one larger than the counterpart if necessary. If no large enough type exists, then the machine may be incapable of doing what you want.
Related
I know that in order to get the 4 least significant bytes of a number of type long I can cast it to int/unsigned int or use a bitwise AND (& 0xFFFFFFFF).
This code produces the following output:
#include <stdio.h>
int main()
{
long n = 0x8899AABBCCDDEEFF;
printf("0x%016lX\n", n);
printf("0x%016X\n", (int)n);
printf("0x%016X\n", (unsigned int)n);
printf("0x%016lX\n", n & 0xFFFFFFFF);
}
Output:
0x8899AABBCCDDEEFF
0x00000000CCDDEEFF
0x00000000CCDDEEFF
0x00000000CCDDEEFF
Does that mean that the two methods used are equivalent? If so, do they always produce the same output regardless of the platform/compiler?
Also, is there any catch or pitfall while casting to unsigned int rather than int for the purpose of this question?
Finally, why is the output the same if you change the number n to be an unsigned long instead?
The methods are definitely different.
According to integral conversion rules (cf, for example, this online c++11 standard), a conversion (e.g. through an explicit cast) from one integral type to another depends on whether the destination type is signed or unsigned. If the destination type is unsigned, one can rely on a "modulo 2n" truncation, whereas with signed destination types one could tap into implementation defined behaviour:
4.7 Integral conversions [conv.integral]
2 If the destination type is unsigned, the resulting value is the
least unsigned integer congruent to the source integer (modulo 2n
where n is the number of bits used to represent the unsigned type). [
Note: In a two's complement representation, this conversion is
conceptual and there is no change in the bit pattern (if there is no
truncation). — end note ]
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.
For your first question, as others have pointed out, the size of int and long is dependent on the platform, so the methods are not equivalent. In C data types, check that the types say "at least XX bits in size"
For the second question, it comes down to this: long and int are signed, meaning that one bit is reserved for sign (take a look also to two's complement). If you were the compiler, what can you do with negative values (especially the long ones)? As Stepahn Lechner mentioned, this is implementation defined (that is, is up to the compiler).
Finally, in the spirit of "your code must do what it says it does", the best thing to do if you need to do masks is to use masks (and, if you use masks, use unsigned types). Don't try to use cleaver answers. Believe me, they always bite you in the rear. I've dealt with a lot of legacy code to know that by heart.
What's the difference between casting a long to int versus using a bitwise AND in order to get the 4 least significant bytes?
Type. Casting makes the value an int. And'ing does not change the type.
Range. Depending on int,long range, a cast may not change the value at all.
IDB and UB. implementation defined behavior and undefined behavior are present with mixing signed-ness.
To "get" the 4 LSBytes, use & 0xFFFFFFFFu or cast to uint32_t.
OP's question is unnecessarily convoluted.
long n = 0x8899AABBCCDDEEFF; --> Converting a value outside the range of a signed integer type is implementation-defined.
Otherwise, the new type is signed and the value cannot be represented in it; either the
result is implementation-defined or an implementation-defined signal is raised.
C11 §6.3.1.3 3
printf("0x%016lX\n", n); --> Printing a long with a "%lX" outside the the common range of long/unsigned long is undefined behavior.
Let's go forward with unsigned long:
unsigned long n = 0x8899AABBCCDDEEFF; // no problem,
printf("0x%016lX\n", n); // no problem,
printf("0x%016X\n", (int)n); // problem, C11 6.3.1.3 3
printf("0x%016X\n", (unsigned int)n); // no problem,
printf("0x%016lX\n", n & 0xFFFFFFFF); // no problem,
The "no problem" are OK even is unsigned long is 32-bit or 64-bit. The output will differ, yet is OK.
Recall that int,long are not always 32,64 bit. (16,32), (32,32), (32,64) are common.
int is at least 16 bit.
long is at least that of int and at least 32 bit.
I have a use case where I need to convert signed values to unsigned, to make values sortable. I need this for char, short, int, long, and long long
By sortable, I mean that for signed type X, if (a < b) then converting to unsigned the converted(a) < converted(b). Note that in many cases, converting from a negative signed value directly to a unsigned value will make the value greater than 0 and breaks this constraint (two's complement implementations)
The simplest idea for a char is:
unsigned char convert(char x)
{
return (unsigned char)(x ^ 0x80); // flip sign to make it sortable
}
But this seems to be undefined behavior.
While it might be possible to convert to a larger type, add the types MIN value, and convert to the unsigned type, I'm not sure this is any more compliant, and won't work with long long
How can this be done without any undefined behavior for all types?
It seems safe to convert using memcpy, but it's not clear how to maintain sort order in a compliant way.
(Note that this is similar to: No compliant way to convert signed/unsigned of same size except I need the results to be maintain sort order)
You are doing it wrong, because flipping the sign-bit of a signed value isn't actually defined.
Let's use two-bit types:
00 01 10 11 Order for unsigned 0 1 2 3
10 11 00 01 Order for 2s complement -2 -1 0 1
11 (10 00) 01 Order for sign-magnitude -1 (-0 +0) 1
10 (11 00) 01 Order for 1s-complement -1 (-0 +0) 1
What you want to do is convert to unsigned (Which is always defined as value-preserving, with wrap-around), and then add a bias so the most-negative number becomes 0:
int x = whatever;
unsigned r = (unsigned)x - (unsigned)INT_MIN;
Take care: Signed overflow is not defined, so we avoided signed types.
Of course, it doesn't help if the unsigned type has fewer values than the signed one, which is allowed in general, though not for char.
And you need to take special care if you want to preserve a negative 0 as negative.
This is not possible if you want to remain fully portable.
The range of unsigned int is only specified to at least cover the non-negative values of int. The standard allows for implementations where UINT_MAX == INT_MAX. The same applies to all other non-fixed-width integer types.
Given that the range of unsigned int may be smaller than that of int, the pigeonhole principle applies: you have no way of redistributing all values of int to corresponding but different values of unsigned int unless unsigned int can store at least as many different values as int.
To quote N4140 (roughly C++14):
3.9.1 Fundamental types [basic.fundamental]
1 [...] For narrow character types, all bits of the object representation participate in the value representation. For unsigned narrow character types, all possible bit patterns of the value representation represent numbers. These requirements do not hold for other types. [...]
3 For each of the standard signed integer types, there exists a corresponding (but different) standard unsigned integer type: "unsigned char", "unsigned short int", "unsigned int", "unsigned long int", and "unsigned long long int", each of which occupies the same amount of storage and has the same alignment requirements (3.11) as the corresponding signed integer type47; that is, each signed integer type
has the same object representation as its corresponding unsigned integer type. [...] The range of non-negative values of a signed integer type is a
subrange of the corresponding unsigned integer type, and the value representation of each corresponding signed/unsigned type shall be the same. [...]
This guarantees that you don't have a problem for unsigned char. There is no possibility for unsigned char to have any kind of padding bits. It wouldn't make sense for unsigned char to have padding bits: given unsigned char c;, how would you access those padding bits? reinterpret_cast<unsigned char &>(c)? That obviously just gives you c. The only thing similar to padding bits that is possible for unsigned char is something that's completely transparent to the program, for instance when ECC memory is used.
For all the other non-fixed-width integer type, from short to long long, the standard meaning of "subrange" allows for an equal range.
I think I vaguely recall reading that there may have been ancient CPUs that did not provide any native unsigned operations. This would make it very tricky for implementations to properly implement unsigned division, unless they declared that the would-be-sign-bit of unsigned types would be treated as a padding bit. This way, they could simply use the CPU's signed division instruction for either signed or unsigned types.
To keep the ordering that you want, you must add the same amount to all values such that
a) their relative differences are unchanged and
b) all negative values are turned into nonnegative values.
Adding a consistent amount is the only way to do this. If all the values you are sorting are originally of the same signed type T, then the amount to add to ensure any negative value becomes nonnegative must be
"-numeric_limits::min()" or in other words you must subtract the minimum signed value, which is negative.
If you are bringing in different types into the same sort (e.g. sorting char values along with short, int, long, etc.), you might want to make the first step a conversion to the largest signed type you will handle. There is no loss of information to go from a smaller signed type to a larger signed type.
To avoid overflow problems, I would suggest doing the shift (i.e. subtracting the minimum) conditionally.
if (value < 0)
convert by first subtracting the minimum (making nonnegative) and then convert to the unsigned type (which is now completely safe)
else
first convert the already nonnegative value to the unsigned type (completely safe) and then add the same adjustment as a positive value i.e. add numeric_limits::max()+1
T for both is the original signed T. The expression "numeric_limits::max()+1" could be calculated and converted to the new destination type once and then used as a constant in type newT.
I would subtract numeric_limits<T>::min() from each value. This preserves the ordering property you want, and if the underlying representation is 2's complement (i.e., the only sane representation, and the one that is in practice what every non-museum-resident computer uses) will do what you expect, including for the boundary cases when the input value is equal to the most negative or most positive representable integer -- provided that the compiler uses a SUB instruction, and not an ADD instruction (since the positive value -numeric_limits<T>::min() is too large to represent).
Is this standard compliant? No idea. My guess is: Probably not. Feel free to edit if you know.
The formula x-(unsigned)INT_MIN will yield a suitable ranking on all machines where UINT_MAX > INT_MAX. For any pair of signed integers x and y, where x>=y,
(unsigned)x-(unsigned)y will equal the numerical value of x-y; so if y is
INT_MIN, then x>=y for all x, and the aformentioned formula will report the amount by which x is greater than INT_MIN, which is of course ranked the same as x.
The C and C++ standards both allow signed and unsigned variants of the same integer type to alias each other. For example, unsigned int* and int* may alias. But that's not the whole story because they clearly have a different range of representable values. I have the following assumptions:
If an unsigned int is read through an int*, the value must be within the range of int or an integer overflow occurs and the behaviour is undefined. Is this correct?
If an int is read through an unsigned int*, negative values wrap around as if they were casted to unsigned int. Is this correct?
If the value is within the range of both int and unsigned int, accessing it through a pointer of either type is fully defined and gives the same value. Is this correct?
Additionally, what about compatible but not equivalent integer types?
On systems where int and long have the same range, alignment, etc., can int* and long* alias? (I assume not.)
Can char16_t* and uint_least16_t* alias? I suspect this differs between C and C++. In C, char16_t is a typedef for uint_least16_t (correct?). In C++, char16_t is its own primitive type, which compatible with uint_least16_t. Unlike C, C++ seems to have no exception allowing compatible but distinct types to alias.
If an unsigned int is read through an int*, the value must be
within the range of int or an integer overflow occurs and the
behaviour is undefined. Is this correct?
Why would it be undefined? there is no integer overflow since no conversion or computation is done. We take an object representation of an unsigned int object and see it through an int. In what way the value of the unsigned int object transposes to the value of an int is completely implementation defined.
If an int is read through an unsigned int*, negative values wrap
around as if they were casted to unsigned int. Is this correct?
Depends on the representation. With two's complement and equivalent padding, yes. Not with signed magnitude though - a cast from int to unsigned is always defined through a congruence:
If the destination type is unsigned, the resulting value is the
least unsigned integer congruent to the source integer (modulo
2n where n is the number of bits used to represent the unsigned type). [ Note: In a two’s complement representation, this
conversion is conceptual and there is no change in the bit pattern (if
there is no truncation). — end note ]
And now consider
10000000 00000001 // -1 in signed magnitude for 16-bit int
This would certainly be 215+1 if interpreted as an unsigned. A cast would yield 216-1 though.
If the value is within the range of both int and unsigned int,
accessing it through a pointer of either type is fully defined and
gives the same value. Is this correct?
Again, with two's complement and equivalent padding, yes. With signed magnitude we might have -0.
On systems where int and long have the same range, alignment,
etc., can int* and long* alias? (I assume not.)
No. They are independent types.
Can char16_t* and uint_least16_t* alias?
Technically not, but that seems to be an unneccessary restriction of the standard.
Types char16_t and char32_t denote distinct types with the same
size, signedness, and alignment as uint_least16_t and
uint_least32_t, respectively, in <cstdint>, called the underlying
types.
So it should be practically possible without any risks (since there shouldn't be any padding).
If an int is read through an unsigned int*, negative values wrap around as if they were casted to unsigned int. Is this correct?
For a system using two's complement, type-punning and signed-to-unsigned conversion are equivalent, for example:
int n = ...;
unsigned u1 = (unsigned)n;
unsigned u2 = *(unsigned *)&n;
Here, both u1 and u2 have the same value. This is by far the most common setup (e.g. Gcc documents this behaviour for all its targets). However, the C standard also addresses machines using ones' complement or sign-magnitude to represent signed integers. In such an implementation (assuming no padding bits and no trap representations), the result of a conversion of an integer value and type-punning can yield different results. As an example, assume sign-magnitude and n being initialized to -1:
int n = -1; /* 10000000 00000001 assuming 16-bit integers*/
unsigned u1 = (unsigned)n; /* 11111111 11111111
effectively 2's complement, UINT_MAX */
unsigned u2 = *(unsigned *)&n; /* 10000000 00000001
only reinterpreted, the value is now INT_MAX + 2u */
Conversion to an unsigned type means adding/subtracting one more than the maximum value of that type until the value is in range. Dereferencing a converted pointer simply reinterprets the bit pattern. In other words, the conversion in the initialization of u1 is a no-op on 2's complement machines, but requires some calculations on other machines.
If an unsigned int is read through an int*, the value must be within the range of int or an integer overflow occurs and the behaviour is undefined. Is this correct?
Not exactly. The bit pattern must represent a valid value in the new type, it doesn't matter if the old value is representable. From C11 (n1570) [omitted footnotes]:
6.2.6.2 Integer types
For unsigned integer types other than unsigned char, the bits of the object representation shall be divided into two groups: value bits and padding bits (there need not be any of the latter). If there are N value bits, each bit shall represent a different power of 2 between 1 and 2N-1, so that objects of that type shall be capable of representing values from 0 to 2N-1 using a pure binary representation; this shall be known as the value representation. The values of any padding bits are unspecified.
For signed integer types, the bits of the object representation shall be divided into three groups: value bits, padding bits, and the sign bit. There need not be any padding bits; signed char shall not have any padding bits. There shall be exactly one sign bit. Each bit that is a value bit shall have the same value as the same bit in the object representation of the corresponding unsigned type (if there are M value bits in the signed type and N in the unsigned type, then M≤N). If the sign bit is zero, it shall not affect the resulting value. If the sign bit is one, the value shall be modified in one of the following ways:
the corresponding value with sign bit 0 is negated (sign and magnitude);
the sign bit has the value -2M (two's complement);
the sign bit has the value -2M-1 (ones' complement).
Which of these applies is implementation-defined, as is whether the value with sign bit 1 and all value bits zero (for the first two), or with sign bit and all value bits 1 (for ones' complement), is a trap representation or a normal value. In the case of sign and magnitude and ones' complement, if this representation is a normal value it is called a negative zero.
E.g., an unsigned int could have value bits, where the corresponding signed type (int) has a padding bit, something like unsigned u = ...; int n = *(int *)&u; may result in a trap representation on such a system (reading of which is undefined behaviour), but not the other way round.
If the value is within the range of both int and unsigned int, accessing it through a pointer of either type is fully defined and gives the same value. Is this correct?
I think, the standard would allow for one of the types to have a padding bit, which is always ignored (thus, two different bit patterns can represent the same value and that bit may be set on initialization) but be an always-trap-if-set bit for the other type. This leeway, however, is limited at least by ibid. p5:
The values of any padding bits are unspecified. A valid (non-trap) object representation of a signed integer type where the sign bit is zero is a valid object representation of the corresponding unsigned type, and shall represent the same value. For any integer type, the object representation where all the bits are zero shall be a representation of the value zero in that type.
On systems where int and long have the same range, alignment, etc., can int* and long* alias? (I assume not.)
Sure they can, if you don't use them ;) But no, the following is invalid on such platforms:
int n = 42;
long l = *(long *)&n; // UB
Can char16_t* and uint_least16_t* alias? I suspect this differs between C and C++. In C, char16_t is a typedef for uint_least16_t (correct?). In C++, char16_t is its own primitive type, which compatible with uint_least16_t. Unlike C, C++ seems to have no exception allowing compatible but distinct types to alias.
I'm not sure about C++, but at least for C, char16_t is a typedef, but not necessarily for uint_least16_t, it could very well be a typedef of some implementation-specific __char16_t, some type incompatible with uint_least16_t (or any other type).
It is not defined that happens since the c standard does not exactly define how singed integers should be stored. so you can not rely on the internal representation. Also there does no overflow occur. if you just typecast a pointer nothing other happens then another interpretation of the binary data in the following calculations.
Edit
Oh, i misread the phrase "but not equivalent integer types", but i keep the paragraph for your interest:
Your second question has much more trouble in it. Many machines can only read from correctly aligned addresses there the data has to lie on multiples of the types width. If you read a int32 from a non-by-4-divisable address (because you casted a 2-byte int pointer) your CPU may crash.
You should not rely on the sizes of types. If you chose another compiler or platform your long and int may not match anymore.
Conclusion:
Do not do this. You wrote highly platform dependent (compiler, target machine, architecture) code that hides its errors behind casts that suppress any warnings.
Concerning your questions regarding unsigned int* and int*: if the
value in the actual type doesn't fit in the type you're reading, the
behavior is undefined, simply because the standard neglects to define
any behavior in this case, and any time the standard fails to define
behavior, the behavior is undefined. In practice, you'll almost always
obtain a value (no signals or anything), but the value will vary
depending on the machine: a machine with signed magnitude or 1's
complement, for example, will result in different values (both ways)
from the usual 2's complement.
For the rest, int and long are different types, regardless of their
representations, and int* and long* cannot alias. Similarly, as you
say, in C++, char16_t is a distinct type in C++, but a typedef in
C (so the rules concerning aliasing are different).
I know that the following
unsigned short b=-5u;
evaluates to b being 65531 due to an underflow, but I don't understand if 5u is converted to a signed int before being transformed into -5 and then re-converted back to unsigned to be stored in b or -5u is equal to 0 - 5u (this should not be the case, -x is a unary operator)
5u is a literal unsigned integer, -5u is its negation.. Negation for unsigned integers is defined as subtraction from 2**n, which gets the same result as wrapping the result of subtraction from zero.
5u is a single token, an rvalue expression which has type unsigned int.
The unary - operator is applied to it according to the rules of unsigned
arithmetic (arithmetic modulo 2^n, where n is the number of bits in the
unsigned type). The results are converted to unsigned short; if they don't
fit (and they won't if sizeof(int) > sizeof(short)), the conversion will
be done using modulo arithmetic as well (modulo 2^n, where n is the number of
bits in the target type).
It's probably worth noting that if the original argument has type unsigned
short, the actual steps are different (although the results will always be
the same). Thus, if you'd have written:
unsigned short k = 5;
unsigned short b = -k;
the first operation would depend on the size of short. If shorts are smaller
than ints (often, but not always the case), the first step would be to promote
k to an int. (If the size of short and int are identical, then the first
step would be to promote k to unsigned int; from then on, everything
happens as above.) The unary - would be applied to this int, according to
the rules of signed integer arithmetic (thus, resulting in a value of -5). The
resulting -5 will be implicitly converted to unsigned short, using modulo
arithmetic, as above.
In general, these distinctions don't make a difference, but in cases where you
might have an integral value of INT_MIN, they could; on 2's complement
machines, -i, where i has type int and value INT_MIN is implementation
defined, and may result in strange values when later converted to unsigned.
ISO/IEC 14882-2003 section 4.7 says:
"If the destination type is unsigned, the resulting value is the least unsigned integer congruent to the source integer (modulo 2^n where n is the number of bits used to represent the unsigned type). [ Note: In a two’s complement representation, this conversion is conceptual and there is no change in the bit pattern (if there is no truncation). —end note ]
Technically not an underflow, just the representation of the signed value -5 when shown as an unsigned number. Note that signed and unsigned numbers have the same "bits" - they just display differently. If you were to print the value as a signed value [assuming it's extended using the sign bit to fill the remaining bits], it would show -5. [This assumes that it's a typical machine using 2s complement. The C standard doesn't require that signed and unsigned types are the same number of bits, nor that the computer uses 2s complement for representing signed numbers - obviously, if it's not using 2s complement, it won't match up to the value you've shown, so I made the assumption that yours IS a 2s complement machine - which is all common processors, such as x86, 68K, 6502, Z80, PDP-11, VAX, 29K, 8051, ARM, MIPS. But technically, it is not necessary for C to function correctly]
And when you use the unary operator -x, it has the same effect as 0-x [this applies for computers as well as math - it has the same result].
Consider the following code to set all bits of x
unsigned int x = -1;
Is this portable ? It seems to work on at least Visual Studio 2005-2010
The citation-heavy answer:
I know there are plenty of correct answers in here, but I'd like to add a few citations to the mix. I'll cite two standards: C99 n1256 draft (freely available) and C++ n1905 draft (also freely available). There's nothing special about these particular standards, they're just both freely available and whatever happened to be easiest to find at the moment.
The C++ version:
§5.3.2 ¶9: According to this paragraph, the value ~(type)0 is guaranteed to have all bits set, if (type) is an unsigned type.
The operand of ~ shall have integral or enumeration type; the result is the one’s complement of its operand. Integral promotions are performed. The type of the result is the type of the promoted operand.
§3.9.1 ¶4: This explains how overflow works with unsigned numbers.
Unsigned integers, declared unsigned, shall obey the laws of arithmetic modulo 2n where n is the number of bits in the value representation of that particular size of integer.
§3.9.1 ¶7, plus footnote 49: This explains that numbers must be binary. From this, we can infer that ~(type)0 must be the largest number representable in type (since it has all bits turned on, and each bit is additive).
The representations of integral types shall define values by use of a pure
binary numeration system49.
49) A positional representation for integers that uses the binary digits 0 and 1, in which the values represented by successive bits are additive, begin
with 1, and are multiplied by successive integral power of 2, except perhaps for the bit with the highest position. (Adapted from the American National
Dictionary for Information Processing Systems.)
Since arithmetic is done modulo 2n, it is guaranteed that (type)-1 is the largest value representable in that type. It is also guaranteed that ~(type)0 is the largest value representable in that type. They must therefore be equal.
The C99 version:
The C99 version spells it out in a much more compact, explicit way.
§6.5.3 ¶3:
The result of the ~ operator is the bitwise complement of its (promoted) operand (that is,
each bit in the result is set if and only if the corresponding bit in the converted operand is
not set). The integer promotions are performed on the operand, and the result has the
promoted type. If the promoted type is an unsigned type, the expression ~E is equivalent
to the maximum value representable in that type minus E.
As in C++, unsigned arithmetic is guaranteed to be modular (I think I've done enough digging through standards for now), so the C99 standard definitely guarantees that ~(type)0 == (type)-1, and we know from §6.5.3 ¶3 that ~(type)0 must have all bits set.
The summary:
Yes, it is portable. unsigned type x = -1; is guaranteed to have all bits set according to the standard.
Footnote: Yes, we are talking about value bits and not padding bits. I doubt that you need to set padding bits to one, however. You can see from a recent Stack Overflow question (link) that GCC was ported to the PDP-10 where the long long type has a single padding bit. On such a system, unsigned long long x = -1; may not set that padding bit to 1. However, you would only be able to discover this if you used pointer casts, which isn't usually portable anyway.
Apparently it is:
(4.7) If the destination type is unsigned, the resulting value is the least
unsigned integer congruent to the source integer (modulo 2n where n is
the number of bits used to represent the unsigned type). [Note: In a
two’s complement representation, this conversion is conceptual and
there is no change in the bit pattern (if there is no truncation).
It is guaranteed to be the largest amount possible for that type due to the properties of modulo.
C99 also allows it:
Otherwise, if the newtype is unsigned, the value is converted by repeatedly adding or subtracting one more than the maximum value that
can be represented in the newtype until the value is in the range of
the newtype. 49)
Which wold also be the largest amount possible.
Largest amount possible may not be all bits set. Use ~static_cast<unsigned int>(0) for that.
I was sloppy in reading the question, and made several comments that might be misleading because of that. I'll try to clear up the confusion in this answer.
The declaration
unsigned int x = -1;
is guaranteed to set x to UINT_MAX, the maximum value of type unsigned int. The expression -1 is of type int, and it's implicitly converted to unsigned int. The conversion (which is defined in terms of values, not representations) results in the maximum value of the target unsigned type.
(It happens that the semantics of the conversion are optimized for two's-complement systems; for other schemes, the conversion might involve something more than just copying the bits.)
But the question referred to setting all bits of x. So, is UINT_MAX represented as all-bits-one?
There are several possible representations for signed integers (two's-complement is most common, but ones'-complement and sign-and-magnitude are also possible). But we're dealing with an unsigned integer type, so the way that signed integers are represented is irrelevant.
Unsigned integers are required to be represented in a pure binary format. Assuming that all the bits of the representation contribute to the value of an unsigned int object, then yes, UINT_MAX must be represented as all-bits-one.
On the other hand, integer types are allowed to have padding bits, bits that don't contribute to the representation. For example, it's legal for unsigned int to be 32 bits, but for only 24 of those bits to be value bits, so UINT_MAX would be 2*24-1 rather than 2*32-1. So in the most general case, all you can say is that
unsigned int x = -1;
sets all the value bits of x to 1.
In practice, very very few systems have padding bits in integer types. So on the vast majority of systems, unsigned int has a size of N bits, and a maximum value of 2**N-1, and the above declaration will set all the bits of x to 1.
This:
unsigned int x = ~0U;
will also set x to UINT_MAX, since bitwise complement for unsigned types is defined in terms of subtraction.
Beware!
This is implementation-defined, as how a negative integer shall be represented, whether two's complement or what, is not defined by the C++ Standard. It is up to the compiler which makes the decision, and has to document it properly.
In short, it is not portable. It may not set all bits of x.