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关于 time/rate 中取消返回 token 的问题（bug？）
source link: https://www.v2ex.com/t/877175
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官方包golang.org/x/time中 cancel 的部分，返回 token 的代码（ CancelAt ）
https://github.com/golang/time/blob/e5dcc9cfc0b9553953e355dde5bdf4ff9f82f742/rate/rate.go#L168
// calculate tokens to restore
// The duration between lim.lastEvent and r.timeToAct tells us how many tokens were reserved
// after r was obtained. These tokens should not be restored.
// 这行代码多减了一次最近申请的预支的 token ，用测试代码说明一下
restoreTokens := float64(r.tokens) - r.limit.tokensFromDuration(r.lim.lastEvent.Sub(r.timeToAct))
if restoreTokens <= 0 {
return
}
// advance time to now
now, _, tokens := r.lim.advance(now)
// calculate new number of tokens
tokens += restoreTokens
if burst := float64(r.lim.burst); tokens > burst {
tokens = burst
}
t0 := time.Now()
l := NewLimiter(1, 10)
l.ReserveN(t0, 5) //桶里还剩 -5 个 token
r := l.ReserveN(t0, 10) // 桶里还剩 -5 个 token
fmt.Printf("%+v\n", l)
// 最后结果还剩 8 个 token
l.ReserveN(t0, 2) // 桶里还有 -7 个
fmt.Printf("%+v\n", l)
//
r.CancelAt(t0) // -7 + (10 - 2) = 1
fmt.Printf("%+v\n", l)
l.ReserveN(t0, 8) // -7
fmt.Printf("%+v\n", l)
// 这样到了第七秒就只消费了 5+8+2=15 ，而取消之前应该是 5+10+2=17 ，取消之后 token 的总数变少了
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