How can a int pointer stores more than one int

[kishore84]

Hi all,

This might be a small query but imp for me. I have put my ideas in comments in below prog.

My idea is if i am assigning memory for one int how can i store there 10 elements of int. Is this type of storing elements is safe.

int main(int argc, char* argv[])
{
 
    int*p = NULL;
    p = new int;  // here i guess i am assigning only 4 byte memory that is for one int
    int * q = NULL;
    abc(p);
 
    def(p);
 
    return 0;
}
 
void abc(int* p)
{
  for(int i = 0;i < 10;i++)
    {
 
        *(p+i)     = 10; // But here I can store 10 elements how other mem are allocated
    }
 
 
 
}
void def(int* p)
{
 
  for(int i = 1;i < 10;i++)
    {
 
      printf("values are : %d \r\n",*(p+i))    ; // also here i can print those value
    }
 
}
 
 

Please suggest.

Thanks

[Aftabpasha]

No, its not safe. You are allocating memory for only one integer, but you are storing 9 more values at next 9 integer locations from valid allocated address.
It may cause Segmentation Fault. You can't predict about it. Ideally they are invalid.
Kind Regards.
Aftab

[kishore84]

Thanks for ur reply. But i have seen often this typeof code wriiten by some proggramres(they are some exp people :)). and also same way we can delare a char* for only one char but can store more than one value . I myself seen they are working for some thousands byte. But don't why this is allowed at all.

I need some more words from u guys on this topic.....some elaboration will highly apreciated...


Thanks

[Aftabpasha]

It's certainly dangerous to work on unallocated memory. You can't predict when there will be Segmentation Fault. It may work sometimes because OS allocates memory in blocks for increasing efficiency. Though you want only 4 bytes, it may give you more (e.g. 512 bytes block). So your program still have some memory left as unused after using your 4 bytes. So, it will not show you error. But as I said its unpredictable and should be avoided. Handling memory safely is a good programming practice.
Kind Regards,
Aftab

[YarrOfDoom]

I think when you do this, you're writing at valid memory locations, but these locations aren't reserved for those ints. So it is very much possible you're overwriting other values there, or when you reserve space later, it gets put there, thus overwriting your ints.

[kishore84]

Ur answers are really helpfull. will all compilers allow this type of operations at all.
U people are right i think because this loop is not working when i incresed the i value. To be more specific for my machine it works upto 6600 (here int size is 4 bytes) some thing like below

for(int i = 0;i < 6600;i++)
    {
 
        *(p+i)     = 10; // it works upto 6600
        //q = p+i;
    }
 

I don't know what is this magical number 6600. After that it gives unhandled exception. so i guess it is aprox 25 kb. then what abt 512 byte (as earlier told by Aftabpasha).

please suggest.....



Thanks

[Banfa]

You should not try to explain the magical 6600 it is not really worth the effort. Once you wrote outside the area of allocated memory you invoked undefined behaviour. That means anything could happen, that it only appears to seg fault (raise an exception) after 6600 is fairly irrelevant, on a different platform or uder different circumstances it might seg fault at 2.

All compilers will allow this operation because if you examine each line individually there is nothing wrong with any given line of code, it is only when you combine them together and write outside the area of memory you allocated that you have a problem.

Try combining your code with additional memory allocations and writes (both before and after your allocation) and you will see more problems.

Basically never write to memory addresses that do not have memory (or hardware) explicitly assigned to them.

[kishore84]

Good insight in the subject. Thanks all for ur time.


thanks

[freelance programmer]

In embedded system , you may not get 10 bytes extra .

[Aftabpasha]

512 bytes was just an example. It mainly depends upon your platform, available memory, your requested memory etc. You can search for "Buddy Memory Allocation" or just "Dynamic Memory Allocation" in Wikipedia for more information.

Kind Regards,
Aftab

[weaknessforcats]

No one has yet mentioned decay of array. This is caused by C/C++ in passing arrays to functions. Only the address is passed so from inside the function, the number of elements is not known.
Read this insight: Arrays Revealed.