无法访问argv [0]，如何获得程序名称? [英] Without access to argv[0], how do I get the program name?

问题描述

我知道程序名称是作为第一个参数传递的，下一个简单的示例会将其打印到标准输出中:

I know the program name is passed as the first argument, and next simple example will print it to the standard output :

#include <iostream>
int main ( int argc, char *argv[] )
{
  std::cout<<argv[0]<<std::endl;
}

有获取程序名称的函数吗?

Is there a function to get the program name?

编辑

我从外壳启动程序，上面的代码将始终打印程序名称(我使用的是fedora 9，但我确定它可以在其他发行版中使用).

I am starting the program from the shell, and the above code will always print the program name (I am using fedora 9, but I am sure it works in other distros).

我发现/proc/self/目录可能包含我要查找的内容，但是我找不到该目录中的确切内容.

I have found that /proc/self/ directory might contain what I am looking for, but I couldn't find what exactly in that directory.

推荐答案

不，没有这样的功能.Linux将程序名称存储在 __ progname 中，但这不是公共接口.如果您想将其用于警告/错误消息，请使用 err(3)函数.

No, there is no such function. Linux stores the program name in __progname, but that's not a public interface. In case you want to use this for warnings/error messages, use the err(3) functions.

如果要运行的程序的完整路径，请在/proc/self/exe 上调用 readlink :

If you want the full path of the running program, call readlink on /proc/self/exe:

char *program_path()
{
    char *path = malloc(PATH_MAX);
    if (path != NULL) {
        if (readlink("/proc/self/exe", path, PATH_MAX) == -1) {
            free(path);
            path = NULL;
        }
    }
    return path;
}

(我相信 __ progname 设置为 argv [0] 的基本名称.请确保查看glibc源代码.)

(I believe __progname is set to the basename of argv[0]. Check out the glibc sources to be sure.)

这篇关于无法访问argv [0]，如何获得程序名称?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！