我的bash脚本中有以下功能：
make() {
cd Python-3.2
make
}
当在此脚本中调用make时，将调用此函数，这将被递归。在函数内部进行调用实际上应该调用外部make实用程序。除了重命名我的make功能，实现这一点最干净的方法是什么？
您可以使用内置命令来抑制shell函数查找。
原文链接：https://www.f2er.com/bash/387695.htmlcommand: command [-pVv] command [arg ...]
Execute a simple command or display information about commands.
Runs COMMAND with ARGS suppressing shell function lookup,or display
information about the specified COMMANDs. Can be used to invoke commands
on disk when a function with the same name exists.
Options:
-p use a default value for PATH that is guaranteed to find all of
the standard utilities
-v print a description of COMMAND similar to the `type' builtin
-V print a more verbose description of each COMMAND
Exit Status:
Returns exit status of COMMAND,or failure if COMMAND is not found.
