encode_duration.go (1846B)
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 }