Program output when using pointers is odd

I get an odd output when I run a program I wrote. This program is just a basic pointer excersize that I thought up.

Code:

---

#include <stdio.h>
#include <string.h>

int main(int argc, char** argv)
{
        char** buffarray;
        char b[20] = "hello world";
        buffarray[0] = b;
        printf("%s\n", buffarray[0]);
        return 0;
}

---

this is the output

Quote:

---

thomas@skringla:~/projects$ ./test
hello world
6554:
6554: runtime linker statistics:
6554: final number of relocations: 84
6554: final number of relocations from cache: 3
thomas@skringla:~/projects$

---

when compliled there isn't even a warning.
my compliler version in case that helps

Code:

---

thomas@skringla:~/projects$ gcc -v
Using built-in specs.
Target: x86_64-linux-gnu
Configured with: ../src/configure -v --enable-languages=c,c++,fortran,objc,obj-c++,treelang --prefix=/usr --enable-shared --with-system-zlib --libexecdir=/usr/lib --without-included-gettext --enable-threads=posix --enable-nls --with-gxx-include-dir=/usr/include/c++/4.2 --program-suffix=-4.2 --enable-clocale=gnu --enable-libstdcxx-debug --enable-objc-gc --enable-mpfr --enable-checking=release --build=x86_64-linux-gnu --host=x86_64-linux-gnu --target=x86_64-linux-gnu
Thread model: posix
gcc version 4.2.4 (Ubuntu 4.2.4-1ubuntu3)

---

Well to start with you have not allocated any memory at all for buffarray.

Following on to Mr. 27's comment, you could assign buffarray (if it was a single char*) to b just fine -- that wouldn't require anything extra. But buffarray[0] tries to follow the value of buffarray, which is currently pointing to The Land of Uninitialized Pointers, and then trying to assign b to whatever address is found there.

Isn't the land of uninitialized pointer in new jersey?

Quote:

---

Originally Posted by skringla

Isn't the land of uninitialized pointer in new jersey?

---

Yes.

Quote:

---

Originally Posted by MK27

Well to start with you have not allocated any memory at all for buffarray.

---

Then what am I still doing wrong?

Code:

---

#include <stdio.h>
#include <string.h>

int main(int argc, char** argv)
{
        char** buffarray;
        char* bb;
        char b[20] = "hello world";
        bb = b;
        buffarray[0] = bb;
        printf("%s\n", buffarray[0]);
        return 0;
}

---

It would seem that new jersey has called again. some where in the area of uninititialized pointers.

Quote:

---

lookup 0x0000000000400000 0x00000000000002b8 -> 0x00007ffacee51000 0x00000000000650b0 /1 puts
hello world
9813:
9813: runtime linker statistics:
9813: final number of relocations: 84
9813: final number of relocations from cache: 3
thomas@skringla:~/projects$

---

Quote:

---

Originally Posted by skringla

Then what am I still doing wrong?

Code:

---

#include <stdio.h>
#include <string.h>

int main(int argc, char** argv)
{
        char** buffarray;
        char* bb;
        char b[20] = "hello world";
        bb = b;
        buffarray[0] = bb;
        printf("%s\n", buffarray[0]);
        return 0;
}

---

It would seem that new jersey has called again. some where in the area of uninititialized pointers.

---

buffarray[0] still doesn't exist. buffarray points to New Jersey; trying to dereference it (with the [0] notation) will then call forth the demons.

This makes the char pointer bb point to the first char in the array b:

Code:

---

bb = b;

---

Well and good. Now, this makes the char pointer buffarray[0] point to what bb points to:

Code:

---

buffarray[0] = bb;

---

Oh wait, buffarray is a pointer to a char pointer (that presumably is the first char pointer in an array of char pointers)... but it does not point to anything (yet). As such, buffarray[0] accesses a pointer that does not exist, and that is a mistake.

doesn't that defeat the purpose of a pointer to a pointer, the whole idea is that buffarray is a array of pointers pointing to whole strings and not just individual characters. Like when you use argv[1] to access the first argument of the program you are running. It doesn't give back just the "-" but the whole argument "-l" (say program "ls -l")

Quote:

---

Originally Posted by skringla

doesn't that defeat the purpose of a pointer to a pointer, the whole idea is that buffarray is a array of pointers pointing to whole strings and not just individual characters. Like when you use argv[1] to access the first argument of the program you are running. It doesn't give back just the "-" but the whole argument "-l" (say program "ls -l")

---

No, each level of pointer needs to point to a valid memory location.

In this case, there is no memory for "bufferarray".

If you do something like this:

Code:

---

char *backbuffer[10];
char **bufferarray = backbuffer;

---

then you can use bufferarray[0..9].

--
Mats

Quote:

---

Originally Posted by skringla

doesn't that defeat the purpose of a pointer to a pointer, the whole idea is that buffarray is a array of pointers pointing to whole strings and not just individual characters. Like when you use argv[1] to access the first argument of the program you are running. It doesn't give back just the "-" but the whole argument "-l" (say program "ls -l")

---

argv though is properly allocated. This would be valid:

Code:

---

#include<stdlib.h>
#include <stdio.h>
#include <string.h>

int main(int argc, char** argv)
{
        char** buffarray = malloc(sizeof(char*));;
        char b[20] = "hello world";
        buffarray[0] = b;
        printf("&#37;s\n", buffarray[0]);
        return 0;
}

---

Now, you have 1 spaces of memory for buffarray. You can thus use buffarray[0]
Again, you wouldn't be able to use buffarray[0][0] prior to buffaray = b, because for that you would need

Code:

---

buffarray[0] = malloc(sizeof(char));

---

Allocating memory is different than assigning a value to a pointer. Another way to go would be something like

Code:

---

char* buffarray[20];

---

Now you allocate an array of 20 char*

This is how it looks in the memory when you do char** buffarray

Address Value Name
0x12312 (undefined) buffarray

If you do buffarray = malloc(sizeof(char*));

Address Value Name
0x12312 0x12358 buffarray
0x12358 undef (no name, just value returned from malloc)

Now, buffarray[0] means *(buffarray + 0), that is *(0x12312 + 0) thus *(0x12312). Which is 0x12358 (the value of the address 0x12312, known as buffarray).
That address exists, it is allocated, so you can use it. When you do buffarray[0] = b then you will have:

Address Value Name
0x12312 0x12358 buffarray
0x12358 value of b (no name, just value returned from malloc)


P.S. Haha. In New Jersey. We all know that's a lie parents tell to their children. Something like Santa Claus

You forgot to free.
But I like to do it the easy way:

Code:

---

char array[] = "My 1337 z0mg text!";
char* p;
char** p2p = &p;
p2p[0] = array;

---

Tada!

Code:

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const char[] = "The Land of Initialized Pointers";

---

a small detail, but const illustrates more the string literal

If you understand the difference between your initial code and Elysia's code then you understand why you got errors :)

You only really need const if you aren't going to modify the array or if it's a string literal, but since you are really initializing a char array, you don't really need const there.

I think I have it now. I am going to try more complex pointer stuff now that I understand them better. Thank you to every one who responded and especially Elysia because she made me understand it.