Howto assign an address to an integer variable

This is a discussion on Howto assign an address to an integer variable within the C Programming forums, part of the General Programming Boards category; Hi, I would like to assign an address to an integer variable like in the following code but I am ...

  1. #1
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    Howto assign an address to an integer variable

    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?

    Code:
    int main()
    {
    
    char string[] = "string";
    
    int x = &string;
    
    
    return 0;
    
    }
    Thank you

  2. #2
    Deathray Engineer MacGyver's Avatar
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    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.)

  3. #3
    Aia
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    Code:
    int main()
    {
    
    char string[] = "string";
    
    int x = (int) &string;
    
    
    return 0;
    
    }
    When the eagles are silent, the parrots begin to jabber. ~Winston Churchill

  4. #4
    C++まいる!Cをこわせ!
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    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.
    Quote Originally Posted by Adak View Post
    io.h certainly IS included in some modern compilers. It is no longer part of the standard for C, but it is nevertheless, included in the very latest Pelles C versions.
    Quote Originally Posted by Salem View Post
    You mean it's included as a crutch to help ancient programmers limp along without them having to relearn too much.

    Outside of your DOS world, your header file is meaningless.

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    It'll work on 16-bit systems too.

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    CSharpener vart's Avatar
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    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.
    The first 90% of a project takes 90% of the time,
    the last 10% takes the other 90% of the time.

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    Thanks all

    I don't understand why the casting has to take place. I would expect that
    Code:
    &string
    returns an address which could then be assigned to the variable. So why doesn't it work this way?

  8. #8
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    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:
    Code:
    char string[] = "My string";
    char* pStr = string;
    printf("%x", pStr);
    And btw, no & before any sort of array or you'll get type**, which isn't what you want.
    Quote Originally Posted by Adak View Post
    io.h certainly IS included in some modern compilers. It is no longer part of the standard for C, but it is nevertheless, included in the very latest Pelles C versions.
    Quote Originally Posted by Salem View Post
    You mean it's included as a crutch to help ancient programmers limp along without them having to relearn too much.

    Outside of your DOS world, your header file is meaningless.

  9. #9
    CSharpener vart's Avatar
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    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
    The first 90% of a project takes 90% of the time,
    the last 10% takes the other 90% of the time.

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    I don't get it. So only pointers can hold addresses right? But this assigns the address of string to the x variable:

    Code:
    int x = (int) string;
    So why does it work this way?

  11. #11
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    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.
    Quote Originally Posted by Adak View Post
    io.h certainly IS included in some modern compilers. It is no longer part of the standard for C, but it is nevertheless, included in the very latest Pelles C versions.
    Quote Originally Posted by Salem View Post
    You mean it's included as a crutch to help ancient programmers limp along without them having to relearn too much.

    Outside of your DOS world, your header file is meaningless.

  12. #12
    Just Lurking Dave_Sinkula's Avatar
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    Quote Originally Posted by daYz View Post
    I don't get it. So only pointers can hold addresses right? But this assigns the address of string to the x variable:

    Code:
    int x = (int) string;
    So why does it work this way?
    Happenstance.
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    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;
    }

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    I need to go to bed. I'll continue with this tommorow. Thanks for your help people.

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    Quote Originally Posted by Elysia View Post
    Your way works because an address is just a number anyway, so any integer variable can be able to hold it.
    That is why I still don't understand why this would not work:

    Code:
    int x = string;
    To me it looks like the address of string can be hold as a number by the integer variable.

    Thanks

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