bash $* 在 shell 脚本中是什么意思
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/12413848/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
what does $* mean in a shell script
提问by Subin_Learner
What does $*exactly mean in a shell script?
$*在 shell 脚本中到底是什么意思？
For example consider the following code snippet
例如考虑以下代码片段
$JAVA_HOME/bin/java/com/test/Testclass $*
采纳答案by Ignacio Vazquez-Abrams
It means all the arguments passed to the script or function, split by word.
这意味着传递给脚本或函数的所有参数，按单词拆分。
It is usually wrong and should be replaced by "$@", which separates the arguments properly.
它通常是错误的，应该替换为"$@"，它可以正确分隔参数。
回答by ДМИТРИЙ МАЛИКОВ
It's easy to find answer by yourself: man bash→ /\$\*:
自己找答案很容易：man bash→ /\$\*：
Special Parameters
The shell treats several parameters specially. These parameters may only be referenced; assignment to them is not allowed.
- Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, it expands to a single word with the value of each parameter separated by the first character of the
IFSspecial variable. That is,"$*"is equivalent to"$1c$2c...", wherecis the first character of the value of theIFSvariable. IfIFSis unset, the parameters are separated by spaces. IfIFSis null, the parameters are joined without intervening separators.
特殊参数
shell 对几个参数进行了特殊处理。这些参数只能被引用；不允许分配给他们。
- 扩展到位置参数，从 1 开始。当扩展发生在双引号内时，它扩展为单个单词，每个参数的值由
IFS特殊变量的第一个字符分隔。也就是说，"$*"相当于"$1c$2c..."，其中c是IFS变量值的第一个字符。如果IFS未设置，则参数以空格分隔。如果IFS为 null，则连接参数而不插入分隔符。
回答by Jin Kim
$*expands to all parameters that were passed to that shell script.
$*扩展到传递给该 shell 脚本的所有参数。
$0= shell script's name
$0= shell 脚本的名称
$1= first argument
$1= 第一个参数
$2= second argument
...etc
$2= 第二个参数...等
$#= number of arguments passed to shellscript
$#= 传递给 shellscript 的参数数量
