问题
As we know that the range of char is -128 to 127.
The 2's compliment of -128 and the binary of 128 is same, which is
10000000
So why the range of char is -128 to 127 but not -127 to 128 .
Where as in case of int , 128 and -128 both are different.
回答1:
In twos-complement notation, whenever the high-order bit is 1, the number is negative. So the biggest positive number is
01111111 = 127
and the smallest negative number is
10000000 = -128
The same thing happens for int, but its range is much larger. The biggest positive int is
01111111 11111111 11111111 11111111 = 2147483647
and the smallest negative int is
10000000 00000000 00000000 00000000 = -2147483648
As you can see, in both cases, the smallest negative is 1 more than the biggest positive.
You might be wondering, If there are the same number of bits other than the sign bit, why are there more negative numbers than positives? That's because you're not counting 0. It has 0 in the high-order bit, so it's considered positive in this notation. When you include that, there are 128 negative chars, and 128 non-negative chars, and they balance equally.
回答2:
A character has 8 bits. So the bit pattern for -128 and 128 are the same when you have 8 bits to deal with. (ignoring the sign bit)
Now the int. It has more bits so the MSB is not -128 but a larger -ve number depending on the platform.
It does exhibit the same problem at a "higher" number
回答3:
Look 128 is
10000000
But by default char is signed compiler treats first bit as sign bit and treats the no. as negative and calculates its 2s complement ( because MSB is 1)
2s complement is
01111111
+ 1
10000000
So - 128 is stored To store +128 compiler have to consider 9 bits that is to interpret
010000000
Here MSB is 0 for positive no. But size is of 9 bits this shows why char only accomodate upto +127 and if we want to store +128 then -128 will be stored as explained above
来源：https://stackoverflow.com/questions/21951828/confusion-regarding-range-of-char