Thread: Why is the output 0?!?!

Code:

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

int main(int argc, char *argv[])
{
    int i = 1;
    char buf[4];

    strcpy(buf, "AAAA");
    printf("%d\n", i);
    
    return 0;
}

Probably because you allocated 4 bytes for buff but wrote 5 bytes with the strcpy(). The 5th byte very likely over wrote part of the memory that i uses.

Indeed. Don't forget that strings in C have a null character assumed at the end.

Code:

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

int main(void) {
    int i = 1;
    char buf[4];

    strcpy(buf, "AAA\0");
    printf("%i\n", i);

    return 0;
}

Kleid-0 - the null character is assumed.

And infact your string literal still assigns 5 bytes.

YAArrrrr... I should have known. Thanks guys.

But when declaring variables, is the memory always assigned in contiguous blocks? Any links to more info on these kinds of details?

> Correct code should rely on neither of these.

Yeah, I am aware of this. I just wanted to know more about how exactly memory is allocated when declaring variables in C.

Ofcourse its not a good idea to use unallocated space.

Originally Posted by sean_mackrory

Kleid-0 - the null character is assumed.

dangit! Well at least the output was not 0!

Originally Posted by Dave_Sinkula

Does it?

Bah you're right it would stop at the first null character Shows you how often I use strcpy now that I'm using C++ almost full time

Yeah, I am aware of this. I just wanted to know more about how exactly memory is allocated when declaring variables in C.

This question cannot be directly answered since memory is not allocated by C, it is allocated by the OS. That said, I can explain how the memory most likely was allocated on your system to give you the results you got.

Code:

    int i = 1;
    char buf[4];

In your code, the above declarations are the ones that are requesting memory. You are requesting space for 1 integer, and 4 chars. On 99% of the computers in use today, integers are 4 bytes, and characters are 1 byte each. Keep in mind though that these memory sizes are not standard at all. An integer could be 8 bytes on some machines (new 64 bit processers for instance), and a char could be 2 bytes or some other values.

The memory allocated in your program is allocated on what's called the stack. First the 4 bytes for the integer are pushed onto the stack, then the 4 bytes for the char array. Here is a small diagram:

Code:

Top of stack on left, bottom of stack on right:
[ ][ ][ ][ ][ ][ ][ ][ ]
|  chars    | integer  |

Now first, i is assigned the value of 1. Now our stack looks like:

Code:

Top of stack on left, bottom of stack on right:
[ ][ ][ ][ ][1][0][0][0]
|  chars    | integer  |

Now you may ask yourself why is the number 1 stored as 0x1000 instead of 0x0001? The answer is that many processers store data in little endian format. This means that the least significant byte goes first.

Next you perform:

Code:

strcpy(buf, "AAAA");

Keep in mind strcpy automatically places a trailing null terminator at the end of the string. Now your stack looks like (assuming ASCII character set):

Code:

Top of stack on left, bottom of stack on right:
[41][41][41][41][0][0][0][0]
|    chars      | integer  |

Notice the trailing null terminator overwrote the least significant byte of your integer.

This explains a little bit about what goes on "under the hood". It doesn't actually have much to do with the C programming language, but I think it's good stuff to know