Pointer to pointer question

Code:

---

#include <stdio.h>

int main(void)
{
        char *text[] = {"foo", "bar"};

        printf("%c\n", (*text)[0]);

        return 0;
}

---

This will display the character f. Now how would I display the first character of "the second" array of characters? I tried this but the program won't compile

Code:

---

#include <stdio.h>

int main(void)
{
        char *text[] = {"foo", "bar"};

        printf("%c\n", (*++text)[0]);

        return 0;
}

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Thanks.

printf( "%c\n", ( *(text+1) )[0] );

or

printf("%c\n", text[1][0]);

Like this?

Code:

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printf("%c\n", (*text)[4]);

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>> printf("%c\n", (*text)[4]);
That's a bad assumption to make. There's no guarantee that those string literals will be placed back-to-back in memory.

To expand on Antigloss's examples:

Code:

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printf("%c\n", **(text+1));
printf("%c\n", *text[1]);

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gg

The reason you can't do ++text is because text is an array. Likewise, you can't do this:

Code:

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{
  int a[5];

  a++;
}

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Quote:

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Originally Posted by Codeplug

>> printf("%c\n", (*text)[4]);
That's a bad assumption to make. There's no guarantee that those string literals will be placed back-to-back in memory.

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As a matter of fact, there is such a guarantee. This is initialised as an array, and the array elements (the strings, in this case) are guaranteed to be contiguous in memory.

However, a function that receives a raw pointer to pointer to char is not allowed to assume this. For example;

Code:

---

/*  #include appropriate standard headers */
void f(char **x)
{
      printf("%c\n", (*x) + 4);
      printf("%c\n", x[1][0]);
}

int main()
{
    char *text[] = {"foo", "bar"};
    char **text2;

    /*  If your compiler chokes on the following three lines, it is a C++ compiler.  A vanilla C compiler will not choke */

    text2 = malloc(2*sizeof(char *));
    text2[0] = malloc(4);
    text2[1] = malloc(4);
    strcpy(text2[0], "foo");
    strcpy(text2[1], "bar");
   
    f(text);
    f(text2);
    return 0;     
}

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The problem, for function f(), is that it cannot assume that the pointer to pointer it is given actually an array of strings, or that elements of that array are contiguous. The first printf() line will generate undefined behaviour when called with f(text2) in the above.

Quote:

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Originally Posted by grumpy

As a matter of fact, there is such a guarantee. This is initialised as an array, and the array elements (the strings, in this case) are guaranteed to be contiguous in memory.

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The array is of pointers.

Quote:

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As a matter of fact, there is such a guarantee. This is initialised as an array, and the array elements (the strings, in this case) are guaranteed to be contiguous in memory.

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Not the strings theselves, but their memory addresses. Which means that if the strings were at address 0x1234 and 0x5678 then those 2 values (the addresses) would be contiguous in memory for your array. But you can see that the strings themselves don't need to be right next to each other.