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C:在无符号变量中执行有符号比较而不进行强制转换

[英]C: performing signed comparison in unsigned variables without casting

i want a function with the following signature: 我想要具有以下签名的功能:

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 . 如果a存储的位的2的补码视图大于b存储的位的2的补码视图,则其输出应为1 otherwise the output should be 0 . 否则输出应为0 for example: 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

the function is not to have any implicit/explicit conversions (as these conversions are implementation-defined, and thus not portable) 该函数不应进行任何隐式/显式转换 (因为这些转换是实现定义的,因此不可移植)

one such implementation is: 一种这样的实现是:

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));
}

can you suggest an implementation that uses arithmetic rather than conditionals to perform this calculations? 您可以建议使用算术运算而不是条件运算的实现吗? you may use one condition if you must 如果必须,可以使用一种条件

using a mix of un/signed variables in comparisons and arithmetic in C is a recipe for disaster. 在C语言中使用无符号/有符号变量的混合进行比较和算术是灾难的根源。 this function is an example of how to circumvent the problem. 此功能是如何解决此问题的示例。

i guess the assembly behind comparison of signed variables is similar to the function i want to implement (on architectures not supporting signed comparisons) 我猜想有符号变量比较后面的程序集类似于我要实现的功能(在不支持有符号比较的体系结构上)

Assuming 2's complement: 假设2的补码:

return (a^signbit) > (b^signbit);

where signbit is obviously the MSB of the representation. 其中signbit显然是表示形式的MSB。

you may use one condition if you must 如果必须,可以使用一种条件

You already have a solution using only one condition. 您已经有了仅使用一个条件的解决方案。 ;) ;)

As you would like to have arithmetic operations rather than conditionals, I assume that the goal is speed. 因为您希望使用算术运算而不是条件运算,所以我认为目标是速度。 And using a look-up table is even faster than arithmetic. 使用查找表甚至比算术更快。 Because you are using 8 bit chars, a look-up table means no overkill: You don't even need a table of size 256x256. 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。 A table size of 256 is perfectly adequate storing a limit for each value of a indicating the value(s) b may have to result in true (or false). 256的表的大小是完全足够存储用于的每个值的限制a指示值(S) b可能有导致真(或假)。 Each function call only needs to perform one table look-up ( a -> limit ) and one comparison ( limit <> b ). 每个函数调用仅需要执行一个查表( a - > limit )和一个比较( limit <> b )。

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