问题
What is the difference between signed and unsigned int?
回答1:
As you are probably aware, int
s are stored internally in binary. Typically an int
contains 32 bits, but in some environments might contain 16 or 64 bits (or even a different number, usually but not necessarily a power of two).
But for this example, let's look at 4-bit integers. Tiny, but useful for illustration purposes.
Since there are four bits in such an integer, it can assume one of 16 values; 16 is two to the fourth power, or 2 times 2 times 2 times 2. What are those values? The answer depends on whether this integer is a signed int
or an unsigned int
. With an unsigned int
, the value is never negative; there is no sign associated with the value. Here are the 16 possible values of a four-bit unsigned int
:
bits value
0000 0
0001 1
0010 2
0011 3
0100 4
0101 5
0110 6
0111 7
1000 8
1001 9
1010 10
1011 11
1100 12
1101 13
1110 14
1111 15
... and Here are the 16 possible values of a four-bit signed int
:
bits value
0000 0
0001 1
0010 2
0011 3
0100 4
0101 5
0110 6
0111 7
1000 -8
1001 -7
1010 -6
1011 -5
1100 -4
1101 -3
1110 -2
1111 -1
As you can see, for signed int
s the most significant bit is 1
if and only if the number is negative. That is why, for signed int
s, this bit is known as the "sign bit".
回答2:
int
and unsigned int
are two distinct integer types. (int
can also be referred to as signed int
, or just signed
; unsigned int
can also be referred to as unsigned
.)
As the names imply, int
is a signed integer type, and unsigned int
is an unsigned integer type. That means that int
is able to represent negative values, and unsigned int
can represent only non-negative values.
The C language imposes some requirements on the ranges of these types. The range of int
must be at least -32767
.. +32767
, and the range of unsigned int
must be at least 0
.. 65535
. This implies that both types must be at least 16 bits. They're 32 bits on many systems, or even 64 bits on some. int
typically has an extra negative value due to the two's-complement representation used by most modern systems.
Perhaps the most important difference is the behavior of signed vs. unsigned arithmetic. For signed int
, overflow has undefined behavior. For unsigned int
, there is no overflow; any operation that yields a value outside the range of the type wraps around, so for example UINT_MAX + 1U == 0U
.
Any integer type, either signed or unsigned, models a subrange of the infinite set of mathematical integers. As long as you're working with values within the range of a type, everything works. When you approach the lower or upper bound of a type, you encounter a discontinuity, and you can get unexpected results. For signed integer types, the problems occur only for very large negative and positive values, exceeding INT_MIN
and INT_MAX
. For unsigned integer types, problems occur for very large positive values and at zero. This can be a source of bugs. For example, this is an infinite loop:
for (unsigned int i = 10; i >= 0; i --) [
printf("%u\n", i);
}
because i
is always greater than or equal to zero; that's the nature of unsigned types. (Inside the loop, when i
is zero, i--
sets its value to UINT_MAX
.)
回答3:
Sometimes we know in advance that the value stored in a given integer variable will always be positive-when it is being used to only count things, for example. In such a case we can declare the variable to be unsigned, as in, unsigned int num student;
. With such a declaration, the range of permissible integer values (for a 32-bit compiler) will shift from the range -2147483648 to +2147483647 to range 0 to 4294967295. Thus, declaring an integer as unsigned almost doubles the size of the largest possible value that it can otherwise hold.
回答4:
In laymen's terms an unsigned int is an integer that can not be negative and thus has a higher range of positive values that it can assume. A signed int is an integer that can be negative but has a lower positive range in exchange for more negative values it can assume.
回答5:
In practice, there are two differences:
- printing (eg with
cout
in C++ orprintf
in C): unsigned integer bit representation is interpreted as a nonnegative integer by print functions. - ordering: the ordering depends on signed or unsigned specification.
this code can identify the integer using ordering criterion:
char a = 0;
a--;
if (0 < a)
printf("unsigned");
else
printf("signed");
来源:https://stackoverflow.com/questions/5739888/what-is-the-difference-between-signed-and-unsigned-int