在bash中并行设置变量 [英] Set variables in parallel in bash

问题描述

这是一个示例程序:

#!/bin/bash

for x in {1..5}
do
  output[$x]=$(echo $x) &
done

wait

for x in {1..5}
do
  echo ${output[$x]}
done

我希望它能够运行并打印出分配给output数组的每个成员的值，但是它什么也不打印.正确删除&会分配变量.我必须使用不同的语法来并行实现吗?

I would expect this to run and print out the values assigned to each member of the output array, but it prints nothing. Removing the & correctly assigns the variables. Must I use different syntax to achieve this in parallel?

推荐答案

此

output[$x]=$(echo $x) &

将 whole 分配放在后台任务(子流程)中，这就是为什么您看不到结果的原因，因为它没有传播到父流程中.

puts the whole assignment in a background task (sub-process) and that's why you're not seeing the result, since it's not propogated to the parent process.

您可以使用

You can use wait to wait for subprocesses, but returning results (other than status codes) is going to be difficult. Perhaps you can write intermediate results to a file, and collect those results after all processes have finished ? (not nice, I appreciate)

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