Storing address of char pointer in a char pointer in C

Question

I am trying to achieve something in C and hope you guys can tell me where I went wrong.

So I have values that is an array of 6 const char pointers.

    int main() 
    {
      /* values is an array of 6 char pointers */
      const char *values[6] = 
       {
        "Two", 
        "Three", 
        "Four", 
        "Five", 
        "Six", 
        "Seven",
       };
       /* p is a pointer to a char pointer */     
       char *(*p) = &values[0];

    }

What I am doing here is to create p that is a pointer to a char pointer from my array and store the address of the first char pointer in my array into p.

Though the logic works, I keep getting a warning message saying:

warning: assignment from incompatible pointer type

What am I doing wrong?

i think its `values` not `value_names` – ntshetty Nov 27 '17 at 03:22 — ntshetty, Nov 27 '17 at 03:22
Fixed it, thanks for pointing that out. – user3702643 Nov 27 '17 at 03:22 — user3702643, Nov 27 '17 at 03:22

iBug · Accepted Answer · 2017-11-27T03:31:10.433

2

You can't cast a const pointer to non-const.

const char *(*p) = &values[0];
^~~~~

This is what you should have written.

edited Nov 27 '17 at 03:31

answered Nov 27 '17 at 03:20

iBug

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Yup that fixed it! Thanks for the help. Will accept the answer in 10 mins – user3702643 Nov 27 '17 at 03:23

Storing address of char pointer in a char pointer in C

1 Answers1