Hello, I'm writing a program that will be utilizing triple pointers but for some reason my test program isn't working as it should:
The result should be:Code:#include <stdio.h> #include <stdlib.h> int main() { char *tmp; char *ptr; char **ptr2; char ***ptr3; ptr = malloc(2); *ptr = 5; tmp = ptr + 1; *tmp = 6; *ptr2 = ptr; **ptr3 = ptr2; printf("%d\n", ***ptr3); printf("%d\n", ***(ptr3 + 1); return 0; }
5
6
Instead I get a segmentation fault.
how about referencing the pointer before pointing to an invalid location.Originally Posted by someprogr
Last edited by itCbitC; 01-06-2009 at 03:18 PM.
Well how about actually using an integer, since you are printing with %d.
Also, I don't think ptr3 actually contains an address you can dereference. But I do know that you can probably do this without triple indirection. What are you really doing?
That's because you have ptr3 pointing to the memory allocated as ptr - which holds the value 5 and 6 and two more bytes of "random" rubbish. So when the computer reads that memory it gets some address like 0x77180605 or 0x05061877 [which one you get depends on the order the processor reads memory to form a 32-bit value] (that's just a random guess, it can be any 65535 other values there). That doesn't look, to me, like an address that you would be able to read a pointer from, so it will fail there.
You may want to do this:
Not only will that compile without warning [something I can almost guarantee that your original code produces when you compile it], but it will also work correectly.Code:... ptr2 = &ptr; ptr3 = &ptr2; ...
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Mats
Compilers can produce warnings - make the compiler programmers happy: Use them!
Please don't PM me for help - and no, I don't do help over instant messengers.
well from my understanding *ptr2 has to point to a char * pointer type and so i pointed it at ptr because thats exactly what it is: a char * pointer.
Same for ptr2... **ptr2 points to a 32-bit address (in a 32-bit system) or four byte char *. Hence it should equal ptr2.
Pointer assignment assigns the value though, not the address of the actual pointer. For that you have to use the address-of (&) operator.
but what about plain old ptr2 and ptr3, where do you think they should point to?? see matsp's reply.
But a triple pointer is:
Each of those () will produce a 32-bit (or whatever size a pointer is) read before we get to the outermost memory access that reads a byte.Code:*(*(*ptr))
Oh, and your (ptr3 + 1), the +1 is in the wrong place - it should be *((**ptr3)+1).
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Mats
Compilers can produce warnings - make the compiler programmers happy: Use them!
Please don't PM me for help - and no, I don't do help over instant messengers.
A pointer is a variable that contains a memory address, not values. That's the first rule.
A pointer is a variable that "points" to a value. The type of that value is part of the pointer's type.
For example: int* p.
p is a pointer that points to int. The pointer's actual type is int*.
So if a pointer is a variable, and all variables store their contents in memory, then it must be possible to take the address of a pointer too, yes? Correct. If we want a pointer to pointer to a pointer, we get... a pointer to int*, or the actual type int**.
And so it continues.
Since a pointer is a variable, by default, it contains garbage (!!). So it must point to something valid first. This is your responsibility as a programmer.
