Thread: Quick double pointer question

When I try to assign strings to a double pointer, it only prints out the last string when I try to print everything out, and then it segfaults.

Code:

#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <assert.h>
main(){  
  char **test = calloc(3,sizeof(char));
  *test = "Test";
  (*test)++;
  *test = "This";
  (*test)++;
  *test = "Code";
  int a;
  for(a=0;a<3;a++){
      printf("%s\n",test[a]);
  }
  for(a=0;a<3;a++){
      free(test[a]);
  }
  free(test);
return 0;
}

Am I assigning something the wrong way?
Also, I am trying to avoid using array notation in order to practice, at least for the assigning of the strings.

This is wrong:

Code:

char **test = calloc(3,sizeof(char));

You probably wanted to allocate space for 3 pointers to char, so it should have been:

Code:

char **test = calloc(3, sizeof(char*));

but you can avoid such mistakes to begin with by writing:

Code:

char **test = calloc(3, sizeof(*test));

This is wrong:

Code:

  for(a=0;a<3;a++){
      free(test[a]);
  }

You did not use malloc/calloc to allocate, so you should not use free.

Oh, and this is not good practice:

Code:

*test = "Test";

"Test" will be converted to a const char*, hence you are assigning a const char* to a char*. This risks modifying the string literal through the pointer to non-const char, resulting in undefined behaviour.

Also, where are your header inclusions?

Oh, and another thing: why are you incrementing the char* instead of the char** when you want to traverse to the next char*?

Besides this, main should be explicitly declared as returning an int.

You probably missed my last edit:

Originally Posted by laserlight

Oh, and another thing: why are you incrementing the char* instead of the char** when you want to traverse to the next char*?

Besides this, main should be explicitly declared as returning an int.

The second one shouldn't be an actual problem here, though it is still wrong in modern C.

Originally Posted by Otto45

I added the headers to my code above, I had them in mine but I forgot to put them above. Also, I tried your suggestions and I still have the same issue

You're also not actually traversing the three pointers in your array of pointers, you're incrementing the value of the first of the three pointers, and never touching the second or third pointer.

What is your current code?

Well, now that you have changed test, writing test[a] and free(test) is wrong. You should create another char** to save the original value of the pointer.

Also, instead of an unsigned char**, you probably want a const char**

Originally Posted by Otto45

Code:

#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <assert.h>
int main(){
  unsigned char **test = calloc(3,sizeof(char*));
  *test = "Test";
  test++;
  *test = "This";
  test++;
  *test = "Code";
  int a;
  for(a=0;a<3;a++){
      printf("%s\n",test[a]);
  }
  free(test);
return 0;
}

Now that you increment test, you're correctly traversing your array... but you've lost your pointer to the original array. That is, when the for loop begins, test is pointing to the third element of the array, rather than the first.

There are better ways to do what you want:

Code:

  unsigned char **test = calloc(3,sizeof(char*));
  *test = "Test";
  *(test+1) = "This";
  *(test+2) = "Code";

Or the more common exact equivalent:

Code:

  unsigned char **test = calloc(3,sizeof(char*));
  test[0] = "Test";
  test[1] = "This";
  test[2] = "Code";

Originally Posted by Otto45

Ah, I get it. I thought that once I started using the array notation in the for loop, it would automatically go to the correct position for test[0], etc.

Not if test no longer holds that value.

All a pointer really is (under the hood) is a memory address, which is just a number of a certain size appropriate for your operating system & compiler. When you do test++, you change the value of that number, just like you would change the value of any number that you incremented.

It can't find the "correct" location of test[0] unless test has the same address it initially had, which is the address to the first of your three character pointers.