如何使用C程序中的选项来运行'ls' ?

I want to execute the command ls -a using execv() on a Linux machine as follows:

我想在Linux机器上执行命令ls -a使用execv():

char *const ptr={"/bin/sh","-c","ls","-a" ,NULL};
execv("/bin/sh",ptr);

However, this command does not list hidden files. What am I doing wrong?

但是，该命令不列出隐藏的文件。我做错了什么?

2 个解决方案

#1

I'm not sure why you're passing this via /bin/sh... but since you are, you need to pass all the arguments after -c as a single value because these are now to be interpreted by /bin/sh.

我不知道你为什么要通过/bin/ sh…但是由于您是这样的，所以您需要将-c之后的所有参数作为单个值传递，因为这些参数现在将由/bin/ sh来解释。

The example is to compare the shell syntax of

例子是比较shell语法

/bin/sh -c ls -a

来

/bin/sh -c 'ls -a'

The second works, but the first doesn't.

第二个行得通，但第一个行不通。

So your ptr should be defined as

所以ptr应该被定义为

char * const ptr[]={"/bin/sh","-c","ls -a" ,NULL};

#2

If you need to get the contents of a directory from a c program, then this is not the best way - you will effectively have to parse the output of ls, which is generally considered a bad idea.

如果您需要从c程序获取目录的内容，那么这不是最好的方法——您必须有效地解析ls的输出，这通常被认为是一个坏主意。

Instead you can use the libc functions opendir() and readdir() to achieve this.

相反，您可以使用libc函数opendir()和readdir()来实现这一点。

Here is a small example program that will iterate over (and list) all files in the current directory:

下面是一个小示例程序，它将遍历(并列出)当前目录中的所有文件:

#include <stdio.h>
#include <stdlib.h>
#include <errno.h>
#include <dirent.h>

int main (int argc, char **argv) {
    DIR *dirp;
    struct dirent *dp;

    dirp = opendir(".");
    if (!dirp) {
        perror("opendir()");
        exit(1);
    }

    while ((dp = readdir(dirp))) {
        puts(dp->d_name);
    }

    if (errno) {
        perror("readdir()");
        exit(1);
    }

    return 0;
}

Note the listing will not be sorted, unlike the default ls -a output.

注意，与默认的ls -a输出不同，清单不会被排序。

#1