关于这个问题，我找到的最佳答案要感谢IBM引用的XDR标准:

Integer An XDR signed integer is a 32-bit piece of data that encodes an integer in the range [-2147483648,2147483647]. The integer is represented in two's complement notation. The most and least significant bytes are 0 and 3, respectively. The data description of integers is integer. Unsigned integer An XDR unsigned integer is a 32-bit piece of data that encodes a nonnegative integer in the range [0,4294967295]. It is represented by an unsigned binary number whose most and least significant bytes are 0 and 3, respectively. The data description of unsigned integers is unsigned.

参见维基百科上的XDR标准

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。

另一个区别是在不同大小的整数之间进行转换时。

例如，如果你从字节流中提取一个整数(简单来说就是16位)，使用无符号值，你可以这样做:

i = ((int) b[j]) << 8 | b[j+1]

(可能应该强制转换第二个字节，但我猜编译器会做正确的事情)

对于有符号的值，你必须担心符号扩展，并做:

i = (((int) b[i]) & 0xFF) << 8 | ((int) b[i+1]) & 0xFF

我将在x86上讨论硬件层面的差异。除非您正在编写编译器或使用汇编语言，否则这几乎无关紧要。但很高兴知道。

首先，x86原生支持这两个数字的有符号数的补表示。您可以使用其他表示，但这将需要更多的指令，通常是浪费处理器时间。

我所说的“原生支持”是什么意思?我的意思是，有一组指令用于无符号数，另一组用于有符号数。无符号数可以与有符号数位于相同的寄存器中，实际上，您可以混合有符号和无符号指令，而不用担心处理器。由编译器(或汇编程序员)来跟踪数字是否带符号，并使用适当的指令。

首先，2的补数具有加减法与无符号数相同的性质。这些数字是正还是负没有区别。(所以你只要继续做加法和减法就可以了，不用担心。)

当进行比较时，差异开始显现出来。X86有一种简单的区分方法:上面/下面表示无符号比较，而大于/小于表示有符号比较。(例如，JAE的意思是“高于或等于就跳”，没有符号。)

还有两组乘除指令用于处理有符号整数和无符号整数。

最后:如果你想检查溢出，你可以对有符号数和无符号数做不同的检查。

根据我们在课堂上学到的，有符号整数既可以表示正数也可以表示负数，而无符号整数只能表示非负数。

例如，看一个8位的数字:

无符号值为0到255

带符号的取值范围为-128 ~ 127

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。