bash 变量的值翻倍 - 为什么？

Question

我有一个简单的 shell 脚本设置为每 X 秒捕获一次图像。出于某种原因，X 的值似乎在每次循环中都加倍。

#!/bin/bash

# basic setup for time-lapse

SECONDS=1

while true
do
    DATE=$(date +"%Y-%m-%d_%H%M%S")
    filename=${DATE}_img.jpg
#   fswebcam -r 1280x720  --timestamp "%a %Y-%b-%d %H:%M (%Z)"  /home/pi/JPGS/$filename
    date
    echo "pausing for ${SECONDS} seconds"
    sleep $SECONDS
    date
    echo "====="
done

这是我得到的输出。 SECONDS 的值未在循环内操作，所以我对这里发生的事情感到困惑。另外，原来的间隔是 30 秒，为了测试我把它改成了 1 秒，日期调用也是为了测试/调试。

Sun Mar  3 17:51:57 CST 2019
pausing for 1 seconds
Sun Mar  3 17:51:58 CST 2019
=====
Sun Mar  3 17:51:58 CST 2019
pausing for 2 seconds
Sun Mar  3 17:52:00 CST 2019
=====
Sun Mar  3 17:52:00 CST 2019
pausing for 4 seconds
Sun Mar  3 17:52:04 CST 2019
=====
Sun Mar  3 17:52:04 CST 2019
pausing for 8 seconds
Sun Mar  3 17:52:12 CST 2019
=====
Sun Mar  3 17:52:12 CST 2019
pausing for 16 seconds
Sun Mar  3 17:52:28 CST 2019
=====
Sun Mar  3 17:52:28 CST 2019
pausing for 32 seconds
Sun Mar  3 17:53:00 CST 2019
=====
Sun Mar  3 17:53:00 CST 2019
pausing for 64 seconds
Sun Mar  3 17:54:04 CST 2019
=====
Sun Mar  3 17:54:04 CST 2019
pausing for 128 seconds

我在这里错过了什么？

这是在树莓派下

Answer 1

为 $SECONDS 选择一个不同的名称。

$SECONDS 是一个内置的 shell 变量。它扩展为自 shell 启动以来的秒数。

来自Bash manual :

'SECONDS'

This variable expands to the number of seconds since the shell was started. Assignment to this variable resets the count to the value assigned, and the expanded value becomes the value assigned plus the number of seconds since the assignment.