Declaring an array of character pointers (arg passing)

Question

This is something that should be easy to answer, but is more difficult for me to find a particular right answer on Google or in K&R. I could totally be overlooking this, too, and if so please set me straight!

The pertinent code is below:

int main(){
    char tokens[100][100];
    char *str = "This is my string";
    tokenize(str, tokens);
    for(int i = 0; i < 100; i++){
        printf("%s is a token\n", tokens[i]);
    }
}
void tokenize(char *str, char tokens[][]){
    int i,j; //and other such declarations
    //do stuff with string and tokens, putting
    //chars into the token array like so:
    tokens[i][j] = <A CHAR>
}

So I realize that I can't have char tokens[][] in my tokenize function, but if I put in char **tokens instead, I get a compiler warning. Also, when I try to put a char into my char array with tokens[i][j] = <A CHAR>, I segfault.

Where am I going wrong? (And in how many ways... and how can I fix it?)

Thanks so much!

Answer 1

You would need to specify the size of the second dimension of the array:

#define SIZE 100
void tokenize(char *str, char tokens[][SIZE]);

This way, the compiler knows that when you say tokens[2][5] that it needs to do something like:

1. Find the address of tokens
2. Move 2 * SIZE bytes past the start
3. Move 5 more bytes past that address
4. ???
5. Profit!

As it stands, without the second dimension specified, if you said tokens[2][5] how would it know where to go?

Answer 2

You're close. Arrays and pointers aren't the same thing, even though it sometimes seems like they are. You can either make your two-dimensional array out of pointers:

 char **tokens = malloc(100 * sizeof(char *));
 for (i = 0; i < 100; i++)
     tokens[i] = malloc(100);

And then use:

void tokenize(char *str, char **tokens)

or you can specify the size of the array in your tokenize() function:

void tokenize(char *str, char tokens[][100])