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Hi,
I would like to assign an address to an integer variable like in the following code but I am doing something wrong. Can someone tell me how this should be done correctly?
Thank you :)Code:int main()
{
char string[] = "string";
int x = &string;
return 0;
}
That won't work on 64-bit systems. The size of a pointer/address will be too large to fit into most ints in that case.
Either use the corresponding pointer type like it was intended, or use some special type to hold it (ie. Either size_t or probably more appropriately ptrdiff_t, which appears to have its own potential issues.)
Code:int main()
{
char string[] = "string";
int x = (int) &string;
return 0;
}
As noted, it will only work on 32-bit systems with a cast to int. A cast to a 64-bit type, unsigned long long or unsigned __int64 can work.
Remember that pointers are unsigned, as well.
It'll work on 16-bit systems too.
THIS (if we talk about casting pointer to int) will work on any system.
The reverse casting will NOT always bring the original pointer value however.
Thanks all :)
I don't understand why the casting has to take place. I would expect thatreturns an address which could then be assigned to the variable. So why doesn't it work this way?Code:&string
Because only pointers hold an address. Other variables do not hold addresses (they can, but that's another matter).
But it's fine to assign it to a proper pointer and use, as well:
And btw, no & before any sort of array or you'll get type**, which isn't what you want.Code:char string[] = "My string";
char* pStr = string;
printf("%x", pStr);
because "address" has a storage for it - called pointer
if you want some "general" storage - use void* pointer - it will be able to receive the value of any pointer without cast
int is a different type, that can even have different size
you do not expect to convert double to int without cast? this is the same story
I don't get it. So only pointers can hold addresses right? But this assigns the address of string to the x variable:
So why does it work this way?Code:int x = (int) string;
No, all types can hold addresses. But pointers is a special type that was designed to hold an address to something.
Your way works because an address is just a number anyway, so any integer variable can be able to hold it.
But the language defines that only pointers holds addresses, otherwise you need to explicitly tell the compiler you want a non-pointer to hold the address.
Happenstance.Quote:
Originally Posted by daYz
Code:#include <stdio.h>
#include <stdint.h>
#include <inttypes.h>
int main()
{
int i;
void *p = &i;
intptr_t ip = (intptr_t)p;
printf ("%" PRIxPTR "\n", ip);
return 0;
}
I need to go to bed. I'll continue with this tommorow. Thanks for your help people.
That is why I still don't understand why this would not work:Quote:
Originally Posted by Elysia
To me it looks like the address of string can be hold as a number by the integer variable.Code:int x = string;
Thanks :)
It just looks this way. Because you are using compiler where this two types have the same sizeQuote:
To me it looks like the address of string can be hold as a number
There are compilers where these two types will be of the different size
That's why standard complient compiler will always warn about such a cast
Addresses are treated as pointers in C, which is the whole reason it doesn't work without a cast.
Thanks guys :)
It is clear to me now.
Regards,
Ben
Yeah but what if the code had been this:Quote:
Originally Posted by daYz
A char is only a single byte. And this will compile fine, because it uses a cast. But a byte is obviously not enough to hold a pointer value. Unless you are working on an 8-bit system that has 256 bytes of memory :PCode:char x = (char)string;