1 /*
2 *
3 * Copyright 2020 gRPC authors.
4 *
5 * Licensed under the Apache License, Version 2.0 (the "License");
6 * you may not use this file except in compliance with the License.
7 * You may obtain a copy of the License at
8 *
9 * http://www.apache.org/licenses/LICENSE-2.0
10 *
11 * Unless required by applicable law or agreed to in writing, software
12 * distributed under the License is distributed on an "AS IS" BASIS,
13 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
14 * See the License for the specific language governing permissions and
15 * limitations under the License.
16 *
17 */
18 19 package grpcutil
20 21 import (
22 "strconv"
23 "time"
24 )
25 26 const maxTimeoutValue int64 = 100000000 - 1
27 28 // div does integer division and round-up the result. Note that this is
29 // equivalent to (d+r-1)/r but has less chance to overflow.
30 func div(d, r time.Duration) int64 {
31 if d%r > 0 {
32 return int64(d/r + 1)
33 }
34 return int64(d / r)
35 }
36 37 // EncodeDuration encodes the duration to the format grpc-timeout header
38 // accepts.
39 //
40 // https://github.com/grpc/grpc/blob/master/doc/PROTOCOL-HTTP2.md#requests
41 func EncodeDuration(t time.Duration) string {
42 // TODO: This is simplistic and not bandwidth efficient. Improve it.
43 if t <= 0 {
44 return "0n"
45 }
46 if d := div(t, time.Nanosecond); d <= maxTimeoutValue {
47 return strconv.FormatInt(d, 10) + "n"
48 }
49 if d := div(t, time.Microsecond); d <= maxTimeoutValue {
50 return strconv.FormatInt(d, 10) + "u"
51 }
52 if d := div(t, time.Millisecond); d <= maxTimeoutValue {
53 return strconv.FormatInt(d, 10) + "m"
54 }
55 if d := div(t, time.Second); d <= maxTimeoutValue {
56 return strconv.FormatInt(d, 10) + "S"
57 }
58 if d := div(t, time.Minute); d <= maxTimeoutValue {
59 return strconv.FormatInt(d, 10) + "M"
60 }
61 // Note that maxTimeoutValue * time.Hour > MaxInt64.
62 return strconv.FormatInt(div(t, time.Hour), 10) + "H"
63 }
64