问题描述
我想要两个无界整数之间的差，每个整数都由一个uint32_t值表示，该值是取2 ^ 32为模的无界整数.如，例如，TCP序列号.请注意，模2 ^ 32 表示形式可以环绕0，而不像更多受限制的问题，即不允许环绕0 .
I want the difference between two unbounded integers, each represented by a uint32_t value which is the unbounded integer taken modulo 2^32. As in, for example, TCP sequence numbers. Note that the modulo 2^32 representation can wrap around 0, unlike more restricted questions that do not allow wrapping around 0.
假定基础无界整数之间的差在正常int的范围内.我想要这个带符号的差异值.换句话说，返回正常int范围内的值，该值等于两个uint32_t输入以模2 ^ 32为模的差.
Assume that the difference between the underlying unbounded integers are in the range of a normal int. I want this signed difference value. In other words, return a value within the normal int range that is equivalent to the difference of the two uint32_t inputs modulo 2^32.
例如，0 - 0xffffffff = 1，因为我们假定基础无界整数在int范围内.证明:如果A mod 2 ^ 32 = 0且B mod 2 ^ 32 = 0xffffffff，则(A = 0，B = -1)(mod 2 ^ 32)因此(AB = 1)(mod 2 ^ 32)和在int范围内，此模数类具有单个代表1.
For example, 0 - 0xffffffff = 1 because we assume that the underlying unbounded integers are in int range. Proof: if A mod 2^32 = 0 and B mod 2^32 = 0xffffffff, then (A=0, B=-1) (mod 2^32) and therefore (A-B=1) (mod 2^32) and in the int range this modulo class has the single representative 1.
我使用了以下代码:
static inline int sub_tcp_sn(uint32_t a, uint32_t b)
{
uint32_t delta = a - b;
// this would work on most systems
return delta;
// what is the language-safe way to do this?
}
这在大多数系统上都有效，因为它们对uint和int都使用modulo-2 ^ 32表示形式，并且普通的modulo-2 ^ 32减法是在此处生成的唯一合理的汇编代码.
This works on most systems because they use modulo-2^32 representations for both uint and int, and a normal modulo-2^32 subtraction is the only reasonable assembly code to generate here.
但是，我相信，如果delta>=0，则C标准仅定义上述代码的结果.例如，在这个问题上一个答案说:
However, I believe that the C standard only defines the result of the above code if delta>=0. For example on this question one answer says:
如何根据C标准将uint转换为int的2 ^ 32模数?
How should a modulo-2^32 conversion from uint to int be done according to the C standard?
注意:我希望答案代码不包含条件表达式，除非您可以证明它是必需的. (在代码说明中进行案例分析就可以了.)
Note: I would prefer the answer code not to involve conditional expressions, unless you can prove it's required. (case analysis in the explanation of the code is OK).
推荐答案
必须有一个标准函数可以执行此操作……但与此同时:
There must be a standard function that does this... but in the meantime:
#include <stdint.h> // uint32_t
#include <limits.h> // INT_MAX
#include <assert.h> // assert
static inline int sub_tcp_sn(uint32_t a, uint32_t b)
{
uint32_t delta = a - b;
return delta <= INT_MAX ? delta : -(int)~delta - 1;
}
请注意，如果结果无法表示，则为UB，但问题是确定的.
Note that it is UB in the case that the result is not representable, but the question said that was OK.
如果系统具有64位long long类型，则可以轻松地自定义和检查范围:
If the system has a 64-bit long long type, then the range can easily be customized and checked as well:
typedef long long sint64_t;
static inline sint64_t sub_tcp_sn_custom_range(uint32_t a, uint32_t b,
sint64_t out_min, sint64_t out_max)
{
assert(sizeof(sint64_t) == 8);
uint32_t delta = a - b;
sint64_t result = delta <= out_max ? delta : -(sint64_t)-delta;
assert(result >= out_min && result <= out_max);
return result;
}
例如sub_tcp_sn_custom_range(0x10000000, 0, -0xf0000000LL, 0x0fffffffLL) == -0xf00000000.
通过范围自定义，该解决方案在所有情况下都将范围损失降到最低，假设时间戳是线性的(例如，对0环绕没有特殊含义)，并且可以使用单个64位类型.
With the range customization, this solution minimizes range loss in all situations, assuming timestamps behave linearly (for example, no special meaning to wrapping around 0) and a singed 64-bit type is available.
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