Linux Shell 转移程序 - 这是什么？

Shell shift procedure - What is this?

提问by user1253622

In shell we have the command shift, but i saw on some example its giving shift 3

在 shell 中，我们有命令 shift，但我在一些例子中看到它给出了 shift 3

Why there is a number after shift ? and what its about ? what it does ?

为什么班次后有一个数字？它是关于什么的？它能做什么 ？

Example:

例子：

echo “arg1=   arg2= arg3=”
shift
echo “arg1=   arg2= arg3=”
shift   
echo “arg1=   arg2= arg3=”
shift  
echo “arg1=   arg2= arg3=”
shift

The output will be:

输出将是：

arg1= 1 arg2=2  arg3=3 
arg1= 2 arg2=3  arg3= 
arg1= 3 arg2=   arg3=
arg1=   arg2=   arg3=

But when i add that, it doesn't display it correctly.

但是当我添加它时，它没有正确显示。

回答by dogbane

Take a look at the manpage, which says:

看看手册页，它说：

shift [n]
    The  positional parameters from n+1 ... are renamed to  .... 
    If n is not given, it is assumed to be 1.

An Example script:

示例脚本：

#!/bin/bash
echo "Input: $@"
shift 3
echo "After shift: $@"

Run it:

运行：

$ myscript.sh one two three four five six

Input: one two three four five six
After shift: four five six

This shows that after shifting by 3, $1=four, $2=fiveand $3=six.

这表明在移位 3 之后，$1=four，$2=five和$3=six。

回答by Oliver Charlesworth

This would be answered simply by reading either the Bash manual, or typing man shift:

这可以通过阅读Bash 手册或键入man shift以下内容来回答：

      shift [n]

Shift the positional parameters to the left by n. The positional parameters from n+1 ... $# are renamed to $1 ... $#-n. Parameters represented by the numbers $# to $#-n+1 are unset. n must be a non-negative number less than or equal to $#. If n is zero or greater than $#, the positional parameters are not changed. If n is not supplied, it is assumed to be 1. The return status is zero unless n is greater than $# or less than zero, non-zero otherwise.

      shift [n]

将位置参数向左移动 n。来自 n+1 ... $# 的位置参数被重命名为 $1 ... $#-n。由数字 $# 到 $#-n+1 表示的参数未设置。n 必须是小于或等于 $# 的非负数。如果 n 为零或大于 $#，则不会更改位置参数。如果未提供 n，则假定为 1。除非 n 大于 $# 或小于零，否则返回状态为零，否则为非零。

回答by Zava

you use man bashto find the shiftbuiltin command:

你man bash用来查找shift内置命令：

shift [n]

The positional parameters from n+1 ... are renamed to $1 .... Parameters represented by the numbers $# down to $#-n+1 are unset. n must be a non-negative number less than or equal to $#. If n is 0, no parameters are changed. If n is not given, it is assumed to be 1. If n is greater than $#, the positional parameters are not changed. The return status is greater than zero if n is greater than $# or less than zero; otherwise 0.

移位 [n]

位置参数从 n+1 ... 重命名为 $1 .... 由数字 $# 到 $#-n+1 表示的参数未设置。n 必须是小于或等于 $# 的非负数。如果 n 为 0，则不更改任何参数。如果未给出 n，则假定为 1。如果 n 大于 $#，则不更改位置参数。如果 n 大于 $# 或小于零，则返回状态大于零；否则为 0。

回答by kalyan varma

Shift the positional parameters to the left by n. The positional parameters from n+1 ... $# are renamed to $1 ... $#-n. Parameters represented by the numbers $# to $#-n+1 are unset. n must be a non-negative number less than or equal to $#. If n is zero or greater than $#, the positional parameters are not changed. If n is not supplied, it is assumed to be 1. The return status is zero unless n is greater than $# or less than zero, non-zero otherwise.

将位置参数向左移动 n。来自 n+1 ... $# 的位置参数被重命名为 $1 ... $#-n。由数字 $# 到 $#-n+1 表示的参数未设置。n 必须是小于或等于 $# 的非负数。如果 n 为零或大于 $#，则不会更改位置参数。如果未提供 n，则假定为 1。除非 n 大于 $# 或小于零，否则返回状态为零，否则为非零。

1. List item

1. 项目清单

回答by Calculus

shifttreat command line arguments as a FIFO queue, it popleft element every time it's invoked.

shift将命令行参数视为 FIFO 队列，每次调用时都会弹出左元素。

array = [a, b, c]
shift equivalent to
array.popleft
[b, c]
, , can be interpreted as index of the array.

bash - The advantage of shift over reassign value straightforward - Stack Overflow

bash - 转移而不是直接重新分配值的优势 - VoidCC