如何将char * a []转换为char **？

Question

I'm trying to return a char** while working with an array of char*

Is there any way I could do this effectively?

  char *pos;
  if ((pos=strchr(line, '\n')) != 0) 
     *pos = '\0';

  //parses command in input (max of (arg_num - 2) flags)
  int arg_num = count_tokens( line, delim);

  char *args[arg_num]; // = malloc( sizeof(char) * strlen(line)); //[     delim_count + 2 ];  //plus the last seat for NULL

  //puts cmd && flags in args
  int i = 0;
  for(; i < arg_num; i++) 
    //strcpy(args[i], strsep( &line, " "));
    args[i] = strsep( &line, " ");

  char** a = args;
  return a;

EDIT: If I'm using char**, how can I get it to work like an array, in terms of setting and accessing the strings (with strsep). I'm kind of confused on the whole concept of a char**

Answer 1

args is a local array on the stack. When the function returns it gets destroyed, so that you cannot return it from the function.

To survive the return from the function you need to allocate it on the heap. Instead of char *args[arg_num]; do char** args = malloc(sizeof *args * arg_num); .

Answer 2

args is a local variable and will be destroyed when you return , so you clearly cannot return a pointer to it. You have to think of something else.

You could allocate args dynamically:

char **args = malloc(sizeof(*args) * arg_num);

Then args is of the right type to begin with.