C语言中如何检测包装计数器和大负值之间的差异 [英] How to detect the difference between a wrapping counter and large negative value in C language
问题描述
为我的无能道歉,因为这是我在这个论坛上的第一篇文章.我试图在以下代码的帮助下检测包装无符号 32 位计数器和大型负跳转之间的差异,但编译器给了我错误:
Apologies for my imbecility as this is my first post on this forum. I am trying to detect the difference between a wrapping unsigned 32-bit counter and a large negative Jump with the help of following code but the compiler give me error:
错误:由于数据类型范围有限[-Werror=type-limits]
这是我的代码片段:
#define MAX_BACKWARD_JUMP -4294959295 //UINT_MAX - 8000
#define MIN_BACKWARD_JUMP -3600
#define MAX_FORWARD_JUMP 4800000
signed int rtpDelta; //Signed 32-bit
unsigned int currRTPTs, prevRTPTs; //unsigned 32-bit
rtpDelta = currRTPTs - prevRTPTs;
if ((rtpDelta > MAX_BACKWARD_JUMP && rtpDelta < MIN_BACKWARD_JUMP)
|| (rtpDelta > MAX_FORWARD_JUMP))
{
printf("Delta in Timestamps too large
",rtpDelta);
}
这里的想法是在 RTP 时间戳中捕获无效的大增量.我们有从对等 RTP 客户端接收的当前时间戳和以前的时间戳.RTP 时间戳的无效值的边界限制为 -4294959295
The idea here is to catch the invalid large Deltas in RTP timestamps. We have a current TimeStamp and a previous Timestamp receiving from peer RTP client. The boundary limits for invalid values of RTP Timestamps are -4294959295 < rtpDelta < -3600 that is it should throw an Error if the Delta is less than -3600 and greater than -4294959295 because the values closer to UMAX_INT will be considered as roll-over. What am I doing wrong here?
推荐答案
考虑:
unsigned int LowerBound = -3600u, UpperBound = 4800000u;
unsigned int difference = currRTPTs - prevRTPTs;
观察到,由于包装,LowerBound
,-3600u
的值将是一个很大的正整数.现在,当数学差(无溢出计算)小于 -3600 一个合理的数量时,difference
的值将是一个大整数,它将小于 LowerBound代码>.此外,如果差异没有变得太大(在负方向上),那么
difference
将保持大于 UpperBound
.
Observe that, due to wrapping, the value of LowerBound
, -3600u
, will be a large positive integer. Now, when the mathematical difference (calculated without overflow) is less than -3600 by a reasonable amount, the value of difference
will be a large integer, and it will be less than LowerBound
. Additionally, if the difference does not become too great (in the negative direction), then difference
will remain greater than UpperBound
.
同样,如果差值大于 4,800,000 一个合理的数量,difference
的值将大于 UpperBound
.如果差异没有变得太大,那么它将保持小于 LowerBound
.
Similarly, if the difference is greater than 4,800,000 by a reasonable amount, the value of difference
will be greater than UpperBound
. If the difference does not become too much greater, then it will remain less than LowerBound
.
因此,在这两种情况下,当数学差异超出所需范围(但不是太多)时,difference
的值小于 LowerBound
且大于上界
:
Thus, in both cases, the value of difference
when the mathematical difference is outside the desired bounds (but not by too much) is less than LowerBound
and greater than UpperBound
:
if (difference < LowerBound && difference > UpperBound)
printf("Delta in timestamps is outside acceptable bounds.
");
请注意,当数学差异超过 -3600u
(即 4,294,967,296 - 3600)或小于 4,800,000 - 4,294,967,296 时,这将失败.因此,当差异在 [-4,290,167,296, 4,294,963,696] 时,测试有效.
Observe that this will fail when the mathematical difference exceeds -3600u
(which is 4,294,967,296 - 3600) or is less than 4,800,000 - 4,294,967,296. Thus, the test works when the difference is in [-4,290,167,296, 4,294,963,696].
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