Possible Duplicate:
signed to unsigned conversion in C - is it always safe?
Let's say I declare a variable of type unsigned int : unsigned int x = -1;
Now -1 in two's complement (assuming 32 bit machine) is 0xFFFFFFFF. Now when I assigned this value to x, did the value 0x7FFFFFFF get assigned to x?
If it were so, then printf ("%d",x); would have printed the decimal equivalent of 0x7FFFFFFF, right? But, clearly this isn't happening, as the value that gets printed is -1. What am I missing here?
Edit: I know that we can use the %u format specifier to print unsigned values. But that doesn't help answer the question above.
The "%d"
format is for (signed) int values. If you use it with an unsigned value, it could print something other than the actual value. Use "%u"
to see the actual value, or %x
to see it in hexadecimal.
In the declaration
unsigned int x = -1;
the expression -1
is of type int, and has the value -1. The initializer converts this value from int to unsigned int. The rules for signed-to-unsigned conversion say that the value is reduced modulo UINT_MAX + 1
, so -1
will convert to UINT_MAX
(which is probably 0xffffffff
or 4294967295
if unsigned int
is 32 bits).
You simply cannot assign a negative value to an object of an unsigned type. Any such value will be converted to the unsigned type before it's assigned, and the result will always be >= 0.