[英]integer wrap-around not really understood
I have the following piece of code which is called approximately 100 times per second 我有以下代码,每秒大约调用100次
if (elapsed_centiseconds_mod_60000 < 59999) {
++elapsed_centiseconds_mod_60000;
} else {
elapsed_centiseconds_mod_60000 = 0;
}
if (elapsed_centiseconds_mod_60000 % 6000 == 0) {
++elapsed_minutes;
}
The types of the variables and constants are 变量和常量的类型是
volatile uint16_t elapsed_centiseconds_mod_60000;
volatile uint8_t start_minute_mod_10;
const int8_t calibration_second = 5;
Then exactly once per second the following is called 然后每秒恰好一次调用以下内容
int16_t get_phase_deviation(uint8_t current_second, uint8_t current_minute_mod_10) {
int32_t deviation=
(int32_t) elapsed_centiseconds_mod_60000 -
((int32_t) current_second + (int32_t)600 - (int32_t)calibration_second) * (int32_t) 100 -
((int32_t) current_minute_mod_10 + (int32_t) 60 - (int32_t)start_minute_mod_10)* (int32_t)6000;
deviation %= 60000;
return (int16_t)(deviation<=30000? deviation: deviation-60000);
}
The idea of the code is to detect relative frequency error of two oscillators. 代码的思想是检测两个振荡器的相对频率误差。 However the code does not work as expected.
但是代码不能按预期工作。 It will start to output a deviation of zero.
它将开始输出零偏差。 But once the elapsed centiseconds are one more than they should it will jump immediately to an output of 5537 which coincidences with 2^16 % 60000 + 1. I tried to cast the intermediate values to int32_t but still the issue sticks.
但是一旦经过的百分之一比它们应该的那么多,它将立即跳到5537的输出,这与2 ^ 16%60000 + 1重合。我试图将中间值转换为int32_t,但问题仍然存在。 Usually the issue appears after 100 seconds or 10 000 centiseconds.
通常问题出现在100秒或10 000厘秒之后。 I suspect some wrap around issue but I fail to see it.
我怀疑有一些问题,但我没有看到它。
Has anyone any idea which term is causing this and why? 有谁知道哪个术语导致这个以及为什么?
You are incorrectly casting the return value of get_phase_deviation()
as the range of deviation %= 60000
is -59999 to 59999
( not 0 to 59999
). 您错误地转换了
get_phase_deviation()
的返回值,因为deviation %= 60000
范围deviation %= 60000
是-59999 to 59999
( 不是 0 to 59999
-59999 to 59999
)。 You should change the line: 你应该改变这一行:
return (int16_t)(deviation<=30000? deviation: deviation-60000);
to correctly deal with deviation <= -30000 as is appropriate or change the return type to int32_t
. 正确处理偏差<= -30000或将返回类型更改为
int32_t
。
Looking at the specific cases to understand the numbers you were getting, assuming: 查看具体案例以了解您获得的数字,假设:
calibration_second = 0;
start_minute_mod_10 = 0;
Then for a perfect system get_phase_deviation()
is returning: 然后,对于一个完美的系统,
get_phase_deviation()
返回:
deviation = -420000; // As computed
deviation %= 60000; // == 0
return 0;
If you are 1 centisecond ahead this changes to: 如果您提前1厘米,则更改为:
deviation = -419999; // As computed
deviation %= 60000; // == -59999
return (int16_t) (-59999); // == 5537
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