使用参数调用Unix外部命令 [英] Call Unix external commands with arguments

问题描述

我找到了一种不带参数(例如"ls"，"pwd")调用unix外部命令的方法.就像这样:

I've found a way to call unix external commands without arguments(ex. "ls", "pwd"). It goes like that:

//Child process
char cwd[1024];
getcwd(cwd, sizeof(cwd));
char *argv[] = {*args, NULL}//(ex.) {"ls", NULL}
char *env[] = {cwd, NULL};
//concat():method that connects 2 strings
char *command_source = concat("/bin/", *args);
execve(command_source, argv, env);
return 0;

我正在尝试转换此代码，以接受带有"ls -l"之类的参数的外部命令

I'm trying to convert this code in order to accept external commands with arguments like "ls -l"

推荐答案

假设您知道args中的参数个数，并且它是argcs:

Assuming that you know the number of arguments in args and that it's argcs:

...
char **argv = calloc(sizeof(char*), argcs+1); 
for (int i=0; i<argcs; i++) 
    argv[i]=args[i]; 
argv[argcs]=NULL; 
...

如果不是，则可以通过遍历数组搜索结尾的NULL来轻松确定argcs.

If not, you can easily determine argcs by iterating through the array searching for the ending NULL.

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