C语言 C: execvp() 和命令行参数
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C: execvp() and command line arguments
提问by user3295674
So I'm writing a program where the arguments are as follows:
所以我正在编写一个程序，其中的参数如下：
program start emacs file.c
or even
甚至
program wait
In essence, the first argument (argv[0]) is the program name, followed by user inputs.
本质上，第一个参数 (argv[0]) 是程序名称，然后是用户输入。
Inside my code, I invoke execvp. Thing is, I'm not entirely sure I'm invoking the right arguments.
在我的代码中，我调用了 execvp。问题是，我并不完全确定我是否在引用正确的论点。
if (pid == 0) {
execvp(argv[1], argv); //Line of interest
exit(1);
}
are argv[1] and argv the correct arguments to pass for the functionality described above? I looked at the man page and they make sense but might not be correct for this case. Thank you!
argv[1] 和 argv 是为上述功能传递的正确参数吗？我查看了手册页，它们很有意义，但对于这种情况可能不正确。谢谢！
回答by Evans
In your main, argv will be like this in the first example:
在您的主文件中，第一个示例中的 argv 将如下所示：
argv[0] = "program";
argv[1] = "start";
argv[2] = "emacs";
argv[3] = "file.c";
argv[4] = NULL;
In execv you want to execute the program "start" with args "emacs file.c", right?. Then the first parameter should be argv[1] - "start" and the second one an array with this strings: {"start", "emacs", "file.c", NULL}. If you use argv, you include the "program" string in argv[0].
在 execv 中，您想使用 args “emacs file.c” 执行程序“start”，对吗？那么第一个参数应该是 argv[1] - "start"，第二个参数是一个带有这个字符串的数组：{"start", "emacs", "file.c", NULL}。如果使用 argv，则在 argv[0] 中包含“程序”字符串。
You can create a new array and copy these parameters or use the address of argv[1] like this:
您可以创建一个新数组并复制这些参数或使用 argv[1] 的地址，如下所示：
execvp(argv[1], &argv[1]); //Line of interest
回答by Guilherme Salazar
My understanding is that you want to take a specific action based on the second command-line argument (argv[1]). If the second argument is 'start', your program should start the executable named argv[2]with the arguments provided thereafter (right?). In this case, you should provide execvpwith the executable name (argv[2]) [1] and a list of arguments, which by convention starts with the name of the executable (argv[2]).
我的理解是您希望根据第二个命令行参数 ( argv[1])执行特定操作。如果第二个参数是“start”，你的程序应该启动以后面argv[2]提供的参数命名的可执行文件（对吗？）。在这种情况下，您应该提供execvp可执行文件名称 ( argv[2]) [1] 和参数列表，按照惯例，参数列表以可执行文件的名称 ( ) 开头argv[2]。
execvp(argv[2], &argv[2])would implement what we have described in the last paragraph (assuming this is what you intended to do).
execvp(argv[2], &argv[2])将实现我们在上一段中描述的内容（假设这是您打算做的）。
[1] execvpexpects 2 arguments as you know. The first is a filename; if the specified filename does not contain a slash character (/), execvpwill do a lookup in the PATH environment variable (which contains a list of directories where executable files reside) to find the executable's fully-qualified name. The second argument is a list of command-line arguments that will be available to the program when it starts.
[1]execvp如您所知，需要 2 个参数。第一个是文件名；如果指定的文件名不包含斜杠字符 (/)，execvp将在 PATH 环境变量（其中包含可执行文件所在的目录列表）中查找可执行文件的完全限定名称。第二个参数是程序启动时可用的命令行参数列表。
回答by millinon
The only thing that might be an issue is that argv[0]in argvpassed to execvpwon't match argv[1](the first argument). Otherwise, it looks okay.
这可能是一个问题的唯一的事情就是argv[0]在argv传递给execvp不匹配argv[1]（第一个参数）。否则，它看起来没问题。
Imagine calling program cat file.txt. In your program, argv will be {"program", "cat", "file.txt", NULL}. Then, in cat, even though the binary called will be cat, argv will still be {"program", "cat", "file.txt", NULL}.
想象一下调用program cat file.txt. 在您的程序中， argv 将为{"program", "cat", "file.txt", NULL}. 然后，在 cat 中，即使调用的二进制文件是 cat，argv 仍然是{"program", "cat", "file.txt", NULL}.
Since cattries to open and read each argument as a file, the first file it'll try to open is cat(argv[1]), which isn't the desired behavior.
由于cat尝试将每个参数作为文件打开和读取，它将尝试打开的第一个文件是cat( argv[1])，这不是所需的行为。
The simple solution is to use execvp(argv[1], argv+1)- this essentially shifts the argument array to the left by one element.
简单的解决方案是使用execvp(argv[1], argv+1)- 这实质上是将参数数组向左移动一个元素。
