What is the difference between signed and unsigned int

Question

What is the difference between signed and unsigned int?

Answer 1

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.)