What does this double pointer do in c programming?

Question

    #include <stdio.h>

#define LENGTH 3

char *words[LENGTH];

int main( int argc, char *argv[] )
{
     char *pc;    // a pointer to a character
     char **ppc;  // a pointer to a pointer character

     printf( "\n------------Multiple indirection example\n" );


 // initialize our array string

   words[0] = "zero";
   words[1] = "one";
   words[2] = "two";

// print each element
   for ( int i = 0; i < LENGTH; ++i )
   {
      printf( "%s \n", words[i] );
   }


   ppc = words; // initialize the pointer to pointer to a character


   for( int i = 0; i < LENGTH; ++i)   // loop over each string
   {
       ppc = words + i;
       pc = *ppc;
           while (*pc != 0) {                   // process each character in a string
            printf("%c ", *pc);             // print out each character of the string individually
            pc += 1;
            }
        printf("\n");

   }




return 0;
}

The part that says " char **ppc " declares a char double pointer. Then it says ppc = word . What is the difference between a regular pointer variable being assigned to words and this double pointer variable? I also don't get why a regular pointer variable would be assigned to have the value of a double pointer variable. Can someone explain this?

Nitpick: a double pointer usually means a pointer to a double type variable. The term you want here is a pointer to pointer. — Sourav Ghosh
– Sourav Ghosh, Commented May 12, 2023 at 14:07
There's absolutely no reason to use a global variable here. In fact, it complicates what could be char *words[] = { "zero", "one", "two", NULL } where LENGTH is no longer necessary either. — tadman
– tadman, Commented May 12, 2023 at 14:08

Some programmer dude · Accepted Answer · 2023-05-12 14:12:21Z

There are two things you need to learn:

All arrays naturally decay to pointers to their first element. So for example as an expression word is really the same as &word[0]. Since word is an array of char * elements, a pointer to such an element must then be char **.
For any pointer (or array) p and index i, the expression p[i] is exactly the same as *(p + i). From this follows that p[0] is the same as *(p + 0) which is the same as *p.

If we now put it all together we start with

ppc = words;

which according to point 1 above must the same as

ppc = &words[0];

It makes ppc point to the first element of the array words.

Then we have

ppc = words + i;

which will then be equal to

ppc = &words[i];

I.e. it makes ppc point to the i:th element in the array words.

Finally we have

pc = *ppc;

which is the same as

pc = ppc[0];

But since ppc points to the i:th element of the array words, the last assignment is really the same as

pc = words[i];

The variable ppc isn't really needed in this program, other than as an example.

To illustrate it, lets "draw" it out...

We have the words array:

+---+     +--------+
| 0 | --> | "zero" |
+---+     +--------+
| 1 | --> | "one"  |
+---+     +--------+
| 2 | --> | "two"  |
+---+     +--------+

In the first iteration of the loop, when i == 0, then we have

ppc = &words[0];

which leads to

+-----+     +---+     +--------+
| ppc | --> | 0 | --> | "zero" |
+-----+     +---+     +--------+
            | 1 | --> | "one"  |
            +---+     +--------+
            | 2 | --> | "two"  |
            +---+     +--------+

Then we have the second iteration, when i == 1 which then give

ppc = &words[1];

which looks like

            +---+     +--------+
            | 0 | --> | "zero" |
+-----+     +---+     +--------+
| ppc | --> | 1 | --> | "one"  |
+-----+     +---+     +--------+
            | 2 | --> | "two"  |
            +---+     +--------+

And exactly the same for the third and last iteration of the loop.

Relenor · Accepted Answer · 2023-05-12 14:08:13Z

0

if you wanted an array of char* (array of words) you will need a char** pointer

so for example: char** wordsArr = {"hello","word"} is an array of char*, and you can treat each cell as a string.

it is different from char* words = "hello world" because if we wanted to print just the first word, printf("%s",words) will print "hello world" where printf("%s",wordsArr[0]) will print "hello"

answered May 12, 2023 at 14:08

Relenor

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