Passing an array of strings as parameter to a function in C

Question

I want a simple function that receives a string and returns an array of strings after some parsing. So, this is my function signature:

int parse(const char *foo, char **sep_foo, int *sep_foo_qty) {
    int i;
    char *token;
    ...
    strcpy(sep_foo[i], token); /* sf here */
    ...
}

Then I call it like this:

char sep_foo[MAX_QTY][MAX_STRING_LENGTH];
char foo[MAX_STRING_LENGTH];
int sep_foo_qty, error;

...

error = parse(foo, sep_foo, &sep_foo_qyt);

...

This way I get a warning during compilation:

warning: passing argument 2 of 'parse' from incompatible pointer type

And then a segmentation fault during execution in the line marked with /* sf here */

What is wrong in my C code?

Thanks in advance

Answer 1

The warning is exactly right. Your function wants an array of pointers. You're giving it an array of arrays.

Expected:

 sep_foo:
 +------+       +-----+
 |char**|--> 0: |char*|-->"string1"
 +------+       +-----+
             1: |char*|-->"string2"
                +-----+
*sep_foo_qty-1: |...  |
                +-----+

What you provided:

           sep_foo:
           +--------------------------------+
        0: | char[MAX_STRING_LENGTH]        |
           +--------------------------------+
        1: | char[MAX_STRING_LENGTH]        |
           +--------------------------------+
MAX_QTY-1: | ...                            |
           +--------------------------------+

An array with elements of type X can "decay" into a pointer-to-X, or X*. But the value of X isn't allowed to change in that conversion. Only one decay operation is allowed. You'd need it to happen twice. In your case, X is array-of-MAX_STRING_LENGTH-chars. The function wants X to be pointer-to-char. Since those aren't the same, the compiler warns you. I'm a bit surprised it was just a warning since nothing good can come from what the compiler allowed to happen.

In your function, you could write this code:

char* y = NULL;
*sep_foo = y;

That's legal code since sep_foo is a char**, so *sep_foo is a char*, and so is y; you can assign them. But with what you tried to do, *sep_foo wouldn't really be a char*; it would be pointing to an array of char. Your code, in effect, would be attempting to do this:

char destination[MAX_STRING_LENGTH];
char* y = NULL;
destination = y;

You can't assign a pointer into an array, and so the compiler warns that the call is no good.

There are two ways to solve this:

• Change the way you declare and allocate sep_foo on the calling side so it matches what the function expects to receive:

  char** sep_foo = calloc(MAX_QTY, sizeof(char*));
  for (int i = 0; i < MAX_QTY; ++i)
    sep_foo[i] = malloc(MAX_STRING_LENGTH);
  

  or, equivalently

  char* sep_foo[MAX_QTY];
  for (int i = 0; i < MAX_QTY; ++i)
    sep_foo[i] = malloc(MAX_STRING_LENGTH);
  

• Change the prototype of the function to accept what you're really giving it:

  int parse(const char *foo, char sep_foo[MAX_QTY][MAX_STRING_LENGTH], int *sep_foo_qty);