C：在无符号变量中执行有符号比较而不进行强制转换
我想要一个具有以下签名的函数：
bool signed_a_greater_than_signed_b(unsigned char a, unsigned char b);
当且仅当 a 中存储的位的 2 补码视图大于该位的 2 补码视图时，其输出应该为 1存储在b中。否则输出应为0。例如：
signed_a_greater_than_signed_b(0b10000000,any number) => 0
signed_a_greater_than_signed_b(0b01111111,any number other than 0b01111111) => 1
signed_a_greater_than_signed_b(0b00000000,0b00000001) => 0
signed_a_greater_than_signed_b(0b00000000,0b11111111) => 1
signed_a_greater_than_signed_b(0b00000000,0b00000000) => 0
该函数不进行任何隐式/显式转换（因为这些转换是 实现定义的，因此不可移植）
这样的实现之一是：
bool signed_a_greater_than_signed_b(unsigned char a, unsigned char b)
{
// if 'signed' a is positive then
// return 1 if a is greater than b or b is negative
// otherwise, if 'signed' a is negative then
// return 1 if a is greater than b and b is negative
if (a <= 0b01111111) return ((b < a) || (b > 0x01111111));
else return ((b < a) && (b > 0x01111111));
}
您能建议一种使用算术而不是条件来执行此计算的实现吗？如果必须
在比较中混合使用无符号/有符号变量，则可以使用一个条件，而 C 中的算术会导致灾难。该函数是如何规避该问题的示例。
我猜签名变量比较背后的程序集与我想要实现的函数类似（在不支持签名比较的架构上）
i want a function with the following signature:
bool signed_a_greater_than_signed_b(unsigned char a, unsigned char b);
its output should be 1 iff the 2's-complement view of the bits stored in a is greater than the 2's complement view of the bits stored in b. otherwise the output should be 0. for example:
signed_a_greater_than_signed_b(0b10000000,any number) => 0
signed_a_greater_than_signed_b(0b01111111,any number other than 0b01111111) => 1
signed_a_greater_than_signed_b(0b00000000,0b00000001) => 0
signed_a_greater_than_signed_b(0b00000000,0b11111111) => 1
signed_a_greater_than_signed_b(0b00000000,0b00000000) => 0
the function is not to have any implicit/explicit conversions (as these conversions are
implementation-defined, and thus not portable)
one such implementation is:
bool signed_a_greater_than_signed_b(unsigned char a, unsigned char b)
{
// if 'signed' a is positive then
// return 1 if a is greater than b or b is negative
// otherwise, if 'signed' a is negative then
// return 1 if a is greater than b and b is negative
if (a <= 0b01111111) return ((b < a) || (b > 0x01111111));
else return ((b < a) && (b > 0x01111111));
}
can you suggest an implementation that uses arithmetic rather than conditionals to perform this calculations? you may use one condition if you must
using a mix of un/signed variables in comparisons and arithmetic in C is a recipe for disaster. this function is an example of how to circumvent the problem.
i guess the assembly behind comparison of signed variables is similar to the function i want to implement (on architectures not supporting signed comparisons)
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评论（2）
假设 2 的补码：
其中
signbit显然是表示形式的 MSB。Assuming 2's complement:
where
signbitis obviously the MSB of the representation.您已经有了仅使用一种条件的解决方案。 ;)
由于您希望进行算术运算而不是条件，因此我认为目标是速度。使用查找表甚至比算术还要快。因为您使用的是 8 位字符，所以查找表并不意味着多余：您甚至不需要大小为 256x256 的表。表大小 256 完全足以存储
a的每个值的限制，指示值b可能必须导致 true（或 false）。每个函数调用只需要进行一次查表（a->limit）和一次比较（limit<> <代码>b）。You already have a solution using only one condition. ;)
As you would like to have arithmetic operations rather than conditionals, I assume that the goal is speed. And using a look-up table is even faster than arithmetic. Because you are using 8 bit chars, a look-up table means no overkill: You don't even need a table of size 256x256. A table size of 256 is perfectly adequate storing a limit for each value of
aindicating the value(s)bmay have to result in true (or false). Each function call only needs to perform one table look-up (a->limit) and one comparison (limit<>b).