写消耗的内存一定量的用户定义的时间bash shell脚本
问题说明
我想写消耗了用户定义的时间嵌入式设备上的内存量高一个bash shell脚本。我该怎么做它没有使用数组？
I am trying to write a bash shell script that consumes a high amount of RAM on an embedded device for a user defined time. How do I do it without using arrays ?
正确答案
即使传统猛砸阵列不支持，它仍然有可能创建使用内置在特定的壳评估命令类似数组变量。
Even if traditional Bash arrays are not supported, it may still be possible to create array-like variables using the eval command built into the particular shell.
以下示例脚本是基于一些脚本我在一个嵌入式Linux项目使用 BusyBox的什么时候。 BusyBox的使用 Almquist外壳 （也称为外壳，灰，SH），它不支持数组。
The following example script is based on some scripting I did when using BusyBox in an embedded Linux project. BusyBox uses the Almquist shell (also known as A Shell, ash, and sh), which does not support arrays.
#!/bin/ash
for index in 1 2 3 4 5; do
value=$(($index * 1024))
eval array$index=\"array[$index]: $value\"
done
for i in 1 3 5; do
eval echo \$array$i
done
<子>注意使用评估！
的输出：的
array[1]: 1024
array[3]: 3072
array[5]: 5120
根据特定的情况下，类似于以下就足够了脚本。
Depending on your particular scenario, a script similar to the following may suffice.
#!/bin/ash
echo "Provide sleep time in the form of NUMBER[SUFFIX]"
echo " SUFFIX may be 's' for seconds (default), 'm' for minutes,"
echo " 'h' for hours, or 'd' for days."
read -p "> " delay
echo "begin allocating memory..."
for index in $(seq 1000); do
value=$(seq -w -s '' $index $(($index 100000)))
eval array$index=$value
done
echo "...end allocating memory"
echo "sleeping for $delay"
sleep $delay
在我简短的测试，这个脚本消耗〜570M到575M〜物理内存 * 进行5分钟的指定时间段。
In my brief testing, this script consumed ~570M to ~575M physical memory* for the specified time period of 5 minutes.
*在不同的测试中使用监控顶部和memprof节目
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