Initializing array of pointers to char from a 2d char array

Question

I have a function

int executeCommand(char args[][TOKENSIZE]){
  char *argv[] = {args[0], args[1]}; // This works
  char *argv[] = {args}; // This doesn't
}

The args variable will not be of fixed length so I can't use the first line as it is. I don't want to use malloc or free due to some reasons. Is there any way of initializing this argv array from args without hard coding the number of entries.

Short answer: No. Long answer: Just pass the length of the args variable to the function so you know how many there are. — MateoConLechuga
– MateoConLechuga, Commented Sep 4, 2017 at 0:44
Why do you need to initialize it; it is a pointer. There's no need to copy the entire pointer array. — MateoConLechuga
– MateoConLechuga, Commented Sep 4, 2017 at 0:49

Lundin · Accepted Answer · 2018-04-24 09:42:42Z

2

The function needs to know the size somehow or it is mission impossible. One possible way is to use pointers to VLA:

int executeCommand(size_t n, char args[n][TOKENSIZE]){
  char* argv[n];
  for(size_t i=0; i<n; i++)
  {
    argv[i] = args[i];
  }
}

As for you why you can't simply use args instead of copying the array addresses, I have no idea.

A complete hard-copy of the data would look like

char argv[n][TOKENSIZE];
memcpy(argv, args, sizeof(char[n][TOKENSIZE]));

edited Apr 24, 2018 at 9:42

answered Apr 24, 2018 at 9:34

Lundin

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1 Comment

chux Over a year ago

If argv[] is used like the argv[] of main(), then argv[n] should be NULL, so char* argv[n+1]; ... argv[n] = NULL;. BTW: Could use memcpy(argv, args, sizeof argv); Yet OP's goal is a bit unclear. Nice answer to old question.

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