Is it safe to use -1 to set all bits to true?

Question

I've seen this pattern used a lot in C & C++.

unsigned int flags = -1;  // all bits are true

Is this a good portable way to accomplish this? Or is using 0xffffffff or ~0 better?

Answer 1

It is certainly safe, as -1 will always have all available bits set, but I like ~0 better. -1 just doesn't make much sense for an unsigned int. 0xFF... is not good because it depends on the width of the type.

Answer 2

I'm not sure using an unsigned int for flags is a good idea in the first place in C++. What about bitset and the like?

std::numeric_limit<unsigned int>::max() is better because 0xffffffff assumes that unsigned int is a 32-bit integer.

Answer 3

Practically: Yes

Theoretically: No.

-1 = 0xFFFFFFFF (or whatever size an int is on your platform) is only true with two's complement arithmetic. In practice, it will work, but there are legacy machines out there (IBM mainframes, etc.) where you've got an actual sign bit rather than a two's complement representation. Your proposed ~0 solution should work everywhere.

Answer 4

Frankly I think all fff's is more readable. As to the comment that its an antipattern, if you really care that all the bits are set/cleared, I would argue that you are probably in a situation where you care about the size of the variable anyway, which would call for something like boost::uint16_t, etc.

Answer 5

I recommend you to do it exactly as you have shown, since it's the most straight forward one. Initialize to -1 which will work always, independent of the actual sign representation, while ~ will sometimes have surprising behavior because you will have to have the right operand type. Only then you will get the most high value of an unsigned type.

For an example of a possibly surprise, consider this one:

unsigned long a = ~0u;

It won't necessarily store a pattern with all bits 1 into a. But it will first create a pattern with all bits 1 in an unsigned int, and then assign it to a. What happens when unsigned long has more bits is that not all of those are 1.

And consider this one, which will fail on a non-two's complement representation:

unsigned int a = ~0; // Should have done ~0u !

The reason for that is that ~0 has to invert all bits. Inverting that will yield -1 on a two's complement machine (which is the value we need!), but will not yield -1 on another representation. On a one's complement machine, it yields zero. Thus, on a one's complement machine, the above will initialize a to zero.

The thing you should understand is that it's all about values - not bits. The variable is initialized with a value. If in the initializer you modify the bits of the variable used for initialization, the value will be generated according to those bits. The value you need, to initialize a to the highest possible value, is -1 or UINT_MAX. The second will depend on the type of a - you will need to use ULONG_MAX for an unsigned long. However, the first will not depend on its type, and it's a nice way of getting the most highest value.

We are not talking about whether -1 has all bits one (it doesn't always have). And we're not talking about whether ~0 has all bits one (it has, of course).

But what we are talking about is what the result of the initialized flags variable is. And for it, only -1 will work with every type and machine.

Answer 6

I would not do the -1 thing. It's rather non-intuitieve (to me at least). Assigning signed data to an unsigned variable just seems to be a volation of the natural order of things.

In your situation, I always use 0xFFFF. (Use the right number of Fs for the variable size of course.)

[BTW, I very rarely see the -1 trick done in real-world code.]

Additionally, if you really care about the individual bits in a vairable, it would be good idea to start using the fixed-width uint8_t, uint16_t, uint32_t types.

Answer 7

As long as you have #include <stdint.h> as one of your includes, you should just use

unsigned int flags = UINT_MAX;

If you want a long's worth of bits, you could use

unsigned long flags = ULONG_MAX;

These values are guaranteed to have all the value bits of the result set to 1, regardless of how signed integers are implemented.

Answer 8

unsigned int flags = -1; is portable.

unsigned int flags = ~0; isn't portable because it relies on a two's-complement representation.

unsigned int flags = 0xffffffff; isn't portable because it assumes 32-bit ints.

If you want to set all bits in a way guaranteed by the C standard, use the first one.

Answer 9

A way which avoids the problems mentioned is to simply do:

unsigned int flags = 0;
flags = ~flags;

Portable and to the point.

Answer 10

On Intel's IA-32 processors it is OK to write 0xFFFFFFFF to a 64-bit register and get the expected results. This is because IA32e (the 64-bit extension to IA32) only supports 32-bit immediates. In 64-bit instructions 32-bit immediates are sign-extended to 64-bits.

The following is illegal:

mov rax, 0ffffffffffffffffh

The following puts 64 1s in RAX:

mov rax, 0ffffffffh

Just for completeness, the following puts 32 1s in the lower part of RAX (aka EAX):

mov eax, 0ffffffffh

And in fact I've had programs fail when I wanted to write 0xffffffff to a 64-bit variable and I got a 0xffffffffffffffff instead. In C this would be:

uint64_t x;
x = UINT64_C(0xffffffff)
printf("x is %"PRIx64"\n", x);

the result is:

x is 0xffffffffffffffff

I thought to post this as a comment to all the answers that said that 0xFFFFFFFF assumes 32 bits, but so many people answered it I figured I'd add it as a separate answer.

Answer 11

See litb's answer for a very clear explanation of the issues.

My disagreement is that, very strictly speaking, there are no guarantees for either case. I don't know of any architecture that does not represent an unsigned value of 'one less than two to the power of the number of bits' as all bits set, but here is what the Standard actually says (3.9.1/7 plus note 44):

The representations of integral types shall define values by use of a pure binary numeration system. [Note 44:]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.

That leaves the possibility for one of the bits to be anything at all.

Answer 12

Converting -1 into any unsigned type is guaranteed by the standard to result in all-ones. Use of ~0U is generally bad since 0 has type unsigned int and will not fill all the bits of a larger unsigned type, unless you explicitly write something like ~0ULL. On sane systems, ~0 should be identical to -1, but since the standard allows ones-complement and sign/magnitude representations, strictly speaking it's not portable.

Of course it's always okay to write out 0xffffffff if you know you need exactly 32 bits, but -1 has the advantage that it will work in any context even when you do not know the size of the type, such as macros that work on multiple types, or if the size of the type varies by implementation. If you do know the type, another safe way to get all-ones is the limit macros UINT_MAX, ULONG_MAX, ULLONG_MAX, etc.

Personally I always use -1. It always works and you don't have to think about it.

Answer 13

yes the representation shown is very much correct as if we do it the other way round u will require an operator to reverse all the bits but in this case the logic is quite straightforward if we consider the size of the integers in the machine

for instance in most machines an integer is 2 bytes = 16 bits maximum value it can hold is 2^16-1=65535 2^16=65536

0%65536=0 -1%65536=65535 which corressponds to 1111.............1 and all the bits are set to 1 (if we consider residue classes mod 65536) hence it is much straight forward.

I guess

no if u consider this notion it is perfectly dine for unsigned ints and it actually works out

just check the following program fragment

int main() {

unsigned int a=2;

cout<<(unsigned int)pow(double(a),double(sizeof(a)*8));

unsigned int b=-1;

cout<<"\n"<<b;

getchar();

return 0;

}

answer for b = 4294967295 whcih is -1%2^32 on 4 byte integers

hence it is perfectly valid for unsigned integers

in case of any discrepancies plzz report

Answer 14

I say:

int x;
memset(&x, 0xFF, sizeof(int));

This will always give you the desired result.