[英]How to pass an integer as an argument in Shell function and retrieve it
I have a small function like this - 我有一个这样的小功能-
#! /bin/sh
function hey() {
a=$1
echo $a
}
x=2
hey x
As per my understanding it should print 2
but it is printing x
. 根据我的理解,它应该打印
2
但它正在打印x
。 How to resolve it? 怎么解决呢?
Running you code in my bash (not sh) , it also prints x. 用我的bash(不是sh)运行代码,它也会打印x。 But if i change the last line to
但是如果我将最后一行更改为
hey $x
then it prints 2. 然后打印2。
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