help! (int*)+offset

This is a discussion on help! (int*)+offset within the C++ Programming forums, part of the General Programming Boards category; To start off, here's my code: Code: // setting stuff up std::cout.setf(std::ios::hex); std::cout.unsetf(std::ios::dec); int int_size = sizeof(int); int commands = ...

  1. #1
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    help! (int*)+offset

    To start off, here's my code:

    Code:
    // setting stuff up
    std::cout.setf(std::ios::hex);
    std::cout.unsetf(std::ios::dec);
    int int_size = sizeof(int);
    
    int commands = 5;
    int max_bytes = 3;
    
    int byte_array = (int)malloc(int_size*commands*max_bytes);
    int* pbyte_array = &byte_array;
    
    cout << (0x0012FF3C+(int_size*1)) << "\n\n";
    	
    for(int i=0x0;i<commands*max_bytes;i+=0x1)
    {
    	cout << pbyte_array << "\t+\t";
    	cout << i*int_size << "\t=\t";
    	cout << pbyte_array+(i*int_size) << "\n";
    }
    Okay so it compiles fine, but running gives me
    Code:
    12ff40
    
    0012FF3C	+	0	=	0012FF3C
    0012FF3C	+	4	=	0012FF4C
    0012FF3C	+	8	=	0012FF5C
    0012FF3C	+	c	=	0012FF6C
    .................
    0012FF3c	+	38	=	0013001C
    Basically it's adding 16 each time. If I change it to

    Code:
    cout << pbyte_array+1
    it adds 4 to the output (instead of 1 as it should). I notice if I change pbyte_array declaration to
    Code:
    int pbyte_array = (int)(&byte_array)
    then it works fine (although the hex output becomes lowercase o_O). I have checked the output of int_size and it's 4 as it should be. If I type the address directly into the loop it does the same thing as it does with the changed pbyte_array declaration.

    Basically I want a a pointer/address and the addition of an offset.
    I'm running Visual C++ 2008

    Thanks!

    Oh and also I couldn't get arrays to work properly oddly, which is why I created this code in the first place. It kept saying it couldn't convert (int*) to (int).
    Last edited by RobotGymnast; 01-06-2008 at 11:49 AM.

  2. #2
    and the Hat of Guessing tabstop's Avatar
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    sizeof(int) = 4. Since you have your array declared as type int, adding one to the pointer will take you to the next int, four bytes away. Since you add (i*int_size), you're adding 4, which means that you go four ints down the line, sixteen bytes away.

    PS: int* and int aren't the same thing. What were you trying to do.

  3. #3
    and the hat of wrongness Salem's Avatar
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    So use a type which has a size of 1 (say unsigned char), not 4 (as is the case with your ints).
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.
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  4. #4
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    Thanks, I got it working.. just changed it to
    Code:
    cout << pbyte_array+i << "\n";

    I know they aren't the same.. which is why it confused me, I was just going

    Code:
    int testarray[];
    testarray = new int[5];
    my compiler errors are

    Code:
    'testarray': unknown size.
    '=': cannot convert from 'int *' to 'int []'
    changing it to

    Code:
    int[] testarray
    didn't help either.. more errors

  5. #5
    and the Hat of Guessing tabstop's Avatar
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    Obvious way to do that:
    Code:
    int testarray[5];
    Way to do that for people who insist on new:
    Code:
    int *testarray;
    testarray = new int[5];
    You can't declare an array without giving it a size.

  6. #6
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    Okay thanks... I did get a warning with
    Code:
    int testarray[5];
    though. Thanks anyway though!

  7. #7
    C++まいる!Cをこわせ! Elysia's Avatar
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    You obviously don't know what pointers are and what the heap is or how to allocate memory.
    When allocating on the heap, we keep track of the memory by the use of pointers.
    Anyway, I suggest you stay away from pointer + n, because it's confusing and is obviously confusing you too. Use pointer[n] instead.

    Quote Originally Posted by RobotGymnast View Post
    Okay thanks... I did get a warning with
    Code:
    int testarray[5];
    though. Thanks anyway though!
    What warning?
    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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