有符号整数和无符号整数有什么区别

Question

What is the difference between signed and unsigned int?

Answer 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".

Answer 2

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.

Answer 3

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 .)

Answer 4

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.

Answer 5

In practice, there are two differences:

1. printing (eg with cout in C++ or printf in C): unsigned integer bit representation is interpreted as a nonnegative integer by print functions.
2. ordering : the ordering depends on signed or unsigned specifications.

this code can identify the integer using ordering criterion:

char a = 0;
a--;
if (0 < a)
    printf("unsigned");
else
    printf("signed");

char is considered signed in some compilers and unsigned in other compilers. The code above determines which one is considered in a compiler, using the ordering criterion. If a is unsigned, after a-- , it will be greater than 0 , but if it is signed it will be less than zero. But in both cases, the bit representation of a is the same. That is, in both cases a-- does the same change to the bit representation.