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AugOmin

关于 time/rate 中取消返回 token 的问题(bug?)

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  •   AugOmin · Sep 2, 2022 · 1985 views
    This topic created in 1370 days ago, the information mentioned may be changed or developed.

    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
    	}
    
    

    测试代码

    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)
    
    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
    
    1 replies    2024-08-30 20:02:38 +08:00
    oluoluo
        1
    oluoluo  
       Aug 30, 2024
    确实好像是 bug
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