问题
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.
回答1:
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.
回答2:
Use %u instead of %d in order to print unsigned values. Then you should see 0xFFFFFFFF.
回答3:
What is happening is that you convert the value first to unsigned int, assigning 0xffffffff to x. Then using printf("%d\n") you will convert the value back to signed int still keeping the value of 0xffffffff. Thus printing -1.
来源：https://stackoverflow.com/questions/7152759/what-happens-when-i-assign-a-negative-value-to-an-unsigned-int