有符号整数和无符号整数之间的区别是:
Unsigned值可以是一个较大的正数,而不能是负数。 Unsigned版本使用前导位作为值的一部分,而signed版本使用最左边的位来识别数字是正还是负。 有符号整数可以同时包含正数和负数。
还有其他区别吗?
有符号整数和无符号整数之间的区别是:
Unsigned值可以是一个较大的正数,而不能是负数。 Unsigned版本使用前导位作为值的一部分,而signed版本使用最左边的位来识别数字是正还是负。 有符号整数可以同时包含正数和负数。
还有其他区别吗?
当前回答
Signed integers in C represent numbers. If a and b are variables of signed integer types, the standard will never require that a compiler make the expression a+=b store into a anything other than the arithmetic sum of their respective values. To be sure, if the arithmetic sum would not fit into a, the processor might not be able to put it there, but the standard would not require the compiler to truncate or wrap the value, or do anything else for that matter if values that exceed the limits for their types. Note that while the standard does not require it, C implementations are allowed to trap arithmetic overflows with signed values.
Unsigned integers in C behave as abstract algebraic rings of integers which are congruent modulo some power of two, except in scenarios involving conversions to, or operations with, larger types. Converting an integer of any size to a 32-bit unsigned type will yield the member corresponding to things which are congruent to that integer mod 4,294,967,296. The reason subtracting 3 from 2 yields 4,294,967,295 is that adding something congruent to 3 to something congruent to 4,294,967,295 will yield something congruent to 2.
Abstract algebraic rings types are often handy things to have; unfortunately, C uses signedness as the deciding factor for whether a type should behave as a ring. Worse, unsigned values are treated as numbers rather than ring members when converted to larger types, and unsigned values smaller than int get converted to numbers when any arithmetic is performed upon them. If v is a uint32_t which equals 4,294,967,294, then v*=v; should make v=4. Unfortunately, if int is 64 bits, then there's no telling what v*=v; could do.
鉴于标准的现状,我建议在需要与代数环相关的行为时使用无符号类型,在需要表示数字时使用有符号类型。不幸的是,C以这种方式进行了区分,但它们就是它们。
其他回答
是的,无符号整数可以存储大的值。 不,有不同的方式来表示正数和负数。 是的,有符号整数可以包含正数和负数。
除此之外,在C语言中,你不能溢出一个无符号整数;行为被定义为模算术。您可以溢出一个有符号整数,并且在理论上(尽管在当前主流系统上没有实践),溢出可能会触发一个错误(可能类似于除零错误)。
无符号整数比有符号整数更容易让您陷入特定的陷阱。陷阱来自于这样一个事实:虽然上面的1和3是正确的,但这两种类型的整数都可以被赋值到它可以“容纳”的范围之外,并且它将被无声地转换。
unsigned int ui = -1;
signed int si = -1;
if (ui < 0) {
printf("unsigned < 0\n");
}
if (si < 0) {
printf("signed < 0\n");
}
if (ui == si) {
printf("%d == %d\n", ui, si);
printf("%ud == %ud\n", ui, si);
}
运行此命令时,您将得到以下输出,尽管这两个值都赋值为-1,且声明方式不同。
signed < 0
-1 == -1
4294967295d == 4294967295d
在C语言中,有符号值和无符号值之间唯一保证的区别是有符号值可以为负、0或正,而无符号值只能为0或正。问题是C语言没有定义类型的格式(所以你不知道你的整数是2的补数)。严格来说,你提到的前两点是不正确的。
一般来说,这是正确的。在不知道你为什么要寻找差异的情况下,我想不出有符号和无符号之间的任何其他区别。