有没有办法在bash中实现/使用lambda函数?我在想像这样的东西:
$ someCommand | xargs -L1 (lambda function)
发布于 2013-12-19 16:46:15
如果你想要真正的函数,而不仅仅是管道或while循环(例如,如果你想传递它们,就像它们是数据一样),我就不会做lambdas,而是定义一个重复出现的虚拟名称的虚拟函数,以便立即使用,然后丢弃。如下所示:
# An example map function, to use in the example below.
map() { local f="$1"; shift; for i in "$@"; do "$f" "$i"; done; }
# Lambda function [λ], passed to the map function.
λ(){ echo "Lambda sees $1"; }; map λ *
就像在正确的函数式语言中一样,不需要传递参数,因为您可以将它们包装在闭包中:
# Let’s say you have a function with three parameters
# that you want to use as a lambda:
# (As in: Partial function application.)
trio(){ echo "$1 Lambda sees $3 $2"; }
# And there are two values that you want to use to parametrize a
# function that shall be your lambda.
pre="<<<"
post=">>>"
# Then you’d just wrap them in a closure, and be done with it:
λ(){ trio "$pre" "$post" "$@"; }; map λ *
我认为它甚至比这里提出的所有其他解决方案都要短。
发布于 2009-03-01 16:01:12
那这个呢?
somecommand | xargs -d"\n" -I{} echo "the argument is: {}"
(假设每个参数都是一行,否则更改分隔符)
发布于 2020-10-12 06:04:41
#!/bin/bash
function customFunction() {
eval $1
}
command='echo Hello World; echo Welcome;'
customFunction "$command"
总帐
https://stackoverflow.com/questions/448126
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