Programming Forums > Application Development > C++

Klarre · Apr 22nd, 2007, 8:10 AM

I don't really understand the output of this piece of code. I expected the second row to print just 0x1 higher value than the first one. But it prints 0x4 higher. Why?

Thanks for your help!

/Klarre

(Toggle Plain Text)

#include <iostream>
#include <cstdlib>

int main()
{
	int* data = (int*)malloc(sizeof(int));

	std::cout << (&data + 0) << std::endl;
	std::cout << (&data + 1) << std::endl;

	free(data);

	return 0;
}

(Toggle Plain Text)

Ouput:
0x22ff74
0x22ff78

andro · Apr 22nd, 2007, 8:23 AM

Because an int is 4 bytes, not 1?

Why are you using malloc() instead of new?

Klarre · Apr 22nd, 2007, 8:36 AM

That seems pretty logic. So I guess it is impossible to start read on memory addresses that are not 4 bytes aligned then.

This is a simplified and modified version of a piece of code in a memory manager I am currently working on. I am using malloc since I have been overloading the new operator in the manager project and did not wanted this code to be memory managed by my memory manager. Therefor I chose to use malloc. In this case, of course, it has worked perfectly with new!

Thanks for your reply!

lectricpharaoh · Apr 22nd, 2007, 8:55 AM

Pointer math is scaled to the size of the item being pointed at. This is why pointers need to have a non-void type before you can do any pointer math.

It works like this because typically, the programmer will want to move in units of the pointed-at quantity. For example, if you're pointing at ints, you want to move to the next or previous int by incrementing or decrementing the pointer.

You could cast it to an int, do math, and cast it back (though this won't be portable). A safer option would be to cast to pointer to char, then do math, then cast back. However, some (many) architectures will perform much slower for non-aligned memory accesses, and a few will even generate some kind of fault, so be careful about these kinds of stunts.

Klarre · Apr 22nd, 2007, 9:25 AM

lectricpharaoh - If i interpret you answer correct, you mean that if I, in my example above, changes the int to a char instead, the output would be this:

(Toggle Plain Text)

Output:
0x22ff74
0x22ff76

This is not the case though. Even if I changes the data type to char, it will output the same values as for ints. Actually it increments with 4 even if I use a bigger data structure.

Have I misunderstand your reply, or is your statement incorrect?

lectricpharaoh · Apr 22nd, 2007, 9:43 AM

Quote:

Originally Posted by Klarre

lectricpharaoh - If i interpret you answer correct, you mean that if I, in my example above, changes the int to a char instead, the output would be this:

(Toggle Plain Text)

Output:
0x22ff74
0x22ff76

This is not the case though. Even if I changes the data type to char, it will output the same values as for ints. Actually it increments with 4 even if I use a bigger data structure.

Have I misunderstand your reply, or is your statement incorrect?

My bad. You need to output the pointers, not what they're pointing at- but if you're making them char pointers, you'll need to cast them to void pointers for the output (otherwise, the runtime library will interpret the char pointer as a string, and you'll get gibberish). In other words, you want to print the value of the pointer. This might be more what you're looking for, no?

(Toggle Plain Text)

#include <iostream>
#include <cstdlib>

int main()
{
  char *charData = (char*)malloc(sizeof(char));
  int *intData = (int*)malloc(sizeof(int));
  long long *longlongData = (long long*)malloc(sizeof(long long));

    std::cout << (void *)(charData + 0) << std::endl;
    std::cout << (void *)(charData + 1) << std::endl << std::endl;
    std::cout << intData + 0 << std::endl;
    std::cout << intData + 1 << std::endl << std::endl;
    std::cout << longlongData + 0 << std::endl;
    std::cout << longlongData + 1 << std::endl << std::endl;

    free(charData);
    free(intData);
    free(longlongData);

    return 0;
}

I get this output:

(Toggle Plain Text)

Klarre · Apr 22nd, 2007, 9:53 AM

Ah, that made sense. I shouldn't have use the & operator of course. Now it is clear. Thanks for your help!

/Klarre

Apr 22nd, 2007, 8:10 AM	#1
Klarre Game engine designer Join Date: May 2005 Location: Sweden Posts: 318 Rep Power: 4	Unexpected offset? I don't really understand the output of this piece of code. I expected the second row to print just 0x1 higher value than the first one. But it prints 0x4 higher. Why? Thanks for your help! /Klarre (Toggle Plain Text) #include <iostream> #include <cstdlib> int main() { int* data = (int)malloc(sizeof(int)); std::cout << (&data + 0) << std::endl; std::cout << (&data + 1) << std::endl; free(data); return 0; } #include <iostream> #include <cstdlib> int main() { int data = (int*)malloc(sizeof(int)); std::cout << (&data + 0) << std::endl; std::cout << (&data + 1) << std::endl; free(data); return 0; } (Toggle Plain Text) Ouput: 0x22ff74 0x22ff78 Ouput: 0x22ff74 0x22ff78 __________________ http://www.klarre.se

Apr 22nd, 2007, 8:36 AM	#3
Klarre Game engine designer Join Date: May 2005 Location: Sweden Posts: 318 Rep Power: 4	That seems pretty logic. So I guess it is impossible to start read on memory addresses that are not 4 bytes aligned then. This is a simplified and modified version of a piece of code in a memory manager I am currently working on. I am using malloc since I have been overloading the new operator in the manager project and did not wanted this code to be memory managed by my memory manager. Therefor I chose to use malloc. In this case, of course, it has worked perfectly with new! Thanks for your reply! __________________ http://www.klarre.se

Apr 22nd, 2007, 8:55 AM	#4
lectricpharaoh SEXY SHOELESS GOD OF WAR! Join Date: Jun 2005 Location: Wet west coast of Canada Posts: 1,292 Rep Power: 5	Pointer math is scaled to the size of the item being pointed at. This is why pointers need to have a non-void type before you can do any pointer math. It works like this because typically, the programmer will want to move in units of the pointed-at quantity. For example, if you're pointing at ints, you want to move to the next or previous int by incrementing or decrementing the pointer. You could cast it to an int, do math, and cast it back (though this won't be portable). A safer option would be to cast to pointer to char, then do math, then cast back. However, some (many) architectures will perform much slower for non-aligned memory accesses, and a few will even generate some kind of fault, so be careful about these kinds of stunts. __________________ Java? Rant? Me? Noooo....

Apr 22nd, 2007, 9:25 AM	#5
Klarre Game engine designer Join Date: May 2005 Location: Sweden Posts: 318 Rep Power: 4	lectricpharaoh - If i interpret you answer correct, you mean that if I, in my example above, changes the int to a char instead, the output would be this: (Toggle Plain Text) Output: 0x22ff74 0x22ff76 Output: 0x22ff74 0x22ff76 This is not the case though. Even if I changes the data type to char, it will output the same values as for ints. Actually it increments with 4 even if I use a bigger data structure. Have I misunderstand your reply, or is your statement incorrect? __________________ http://www.klarre.se

Apr 22nd, 2007, 9:53 AM	#7
Klarre Game engine designer Join Date: May 2005 Location: Sweden Posts: 318 Rep Power: 4	Ah, that made sense. I shouldn't have use the & operator of course. Now it is clear. Thanks for your help! /Klarre __________________ http://www.klarre.se

Apr 22nd, 2007, 8:23 AM	#2
andro Professional Programmer Join Date: Oct 2005 Location: California Posts: 321 Rep Power: 4	Because an int is 4 bytes, not 1? Why are you using malloc() instead of new?

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Programming Forums > Application Development > C++

Unexpected offset?