我不明白怎么execlp（）在Linux中运行 [英] I do not understand how execlp() works in Linux

问题描述

我花了最后2天试图理解 execlp（）系统调用，但还没有我在这里。让我直去的问题。

I have spent the last 2 days trying to understand the execlp() system call, but yet here I am. Let me get straight to the issue.

的手册页 execlp的声明系统调用为 INT execlp（为const char *文件，为const char *阿根廷，...）; 与描述：的常量字符精氨酸和随后的椭圆在execl的（），execlp（），和execle（）函数可以被认为是为arg0，ARG1，...，ARGN 的

The man page of execlp declares the system call as int execlp(const char *file, const char *arg, ...); with the description: The const char arg and subsequent ellipses in the execl(), execlp(), and execle() functions can be thought of as arg0, arg1, ..., argn.

然而，我看到了系统调用被称为像这样在我们的教材： execlp（/ bin / sh的，...，ls -l命令/斌/ ??.. ）; （以下简称......是我们要弄清楚的学生）。但是这个系统调用doesn't甚至像什么像手册页系统调用的声明。

Yet I see the system call being called like this in our text book: execlp("/bin/sh", ..., "ls -l /bin/??", ...); (the "..." are for us to figure out as students). However this system call doesn´t even resemble anything like the declaration on the man page of the system call.

我超困惑。任何帮助是AP preciated。

I am super confused. Any help is appreciated.

先谢谢了。

推荐答案

这个原型：

  int execlp(const char *file, const char *arg, ...);

说的execlp是一个可变参数的功能。这需要2 为const char * 。自变量的休息，如果有的话，是额外的参数交给程序，我们希望运行 - 也的char * - 所有这些都是C字符串（最后一个参数必须是一个空指针）

Says that execlp ìs a variable argument function. It takes 2 const char *. The rest of the arguments, if any, are the additional arguments to hand over to program we want to run - also char * - all these are C strings (and the last argument must be a NULL pointer)

因此​​，文件参数是要执行的可执行文件的路径名。 ARG 是我们要显示为的argv [0] 在可执行文件中的字符串。按照惯例，的argv [0] 仅仅是可执行文件的文件名，通常它被设置为相同的文件

So, the file argument is the path name of an executable file to be executed. arg is the string we want to appear as argv[0] in the executable. By convention, argv[0] is just the file name of the executable, normally it's set to the same as file.

的 ... 现在是额外的参数给可执行文件。

The ... are now the additional arguments to give to the executable.

假设你从一个命令/ shell运行这样的：

Say you run this from a commandline/shell:

$ ls

这会是 execlp（LS，LS，（字符*）NULL）;
或者，如果您运行

That'd be execlp("ls", "ls", (char *)NULL); Or if you run

$ ls -l /

这会是 execlp（LS，LS，-l，/，（字符*）NULL）;

所以到 execlp（/ bin / sh的，...，ls -l命令/斌/ ??，...）;

在这里，你要外壳，/ bin / sh的，而你给的shell命令来执行。这命令是ls -l命令/斌/ ??。您可以从一个命令/ shell手动运行：

Here you are going to the shell, /bin/sh , and you're giving the shell a command to execute. That command is "ls -l /bin/??". You can run that manually from a commandline/shell:

 $ ls -l /bin/??

现在，你怎么运行shell，并告诉它执行命令？你打开的文档/手册页的外壳和阅读。

Now, how do you run a shell and tell it to execute a command ? You open up the documentation/man page for your shell and read it.

要运行的是：

$ /bin/sh -c "ls -l /bin/??"

这成为

  execlp("/bin/sh","/bin/sh", "-c", "ls -l /bin/??", (char *)NULL);

附注：
在 /斌/ ?? 正在做模式匹配，这种匹配由壳完成，/它扩展到所有文件下的/ bin中有2个字符。如果你压根儿

Side note: The /bin/?? is doing pattern matching, this pattern matching is done by the shell, and it expands to all files under /bin/ with 2 characters. If you simply did

  execlp("ls","ls", "-l", "/bin/??", (char *)NULL);

也许什么都不会发生（除非有一个实际文件名为 /斌/ ?? ）因为没有外壳，除$ P $点和扩展/斌/？

Probably nothing would happen (unless there's a file actually named /bin/??) as there's no shell that interprets and expands /bin/??

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