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  1. #1

    Thread Starter
    Fanatic Member
    Join Date
    Oct 2008

    Pointer initialization question

    If I use this code:
    int *x = 12345678
    Is this the same (when splitting it into 2 lines of code) as saying
    int *x //declare pointer variable
    *x = 12345678 //set the value at the memory address pointed to by x to be equal to 12345678
    Or is it the same as
    int *x //declare pointer variable
    x = 12345678 //make x point to memory address 12345678

  2. #2
    Frenzied Member
    Join Date
    Jun 2015

    Re: Pointer initialization question

    it would be the same as pointing x to memory location 12345678... if you put in the semicolons

    if we're talking C++, you should be getting a compiler error though.

    maybe it should be int *x = reinterpret_cast<int*>(12345678);
    Imagine what it would be like to set breakpoints in, or step through subclassing code;
    and then being able to hit stop/end/debug or continue, without crashing the IDE.

    VB6.tlb | Bulletproof Subclassing in the IDE (no thunks/assembly/DEP issues)

  3. #3
    Hyperactive Member 2kaud's Avatar
    Join Date
    May 2014

    Re: Pointer initialization question

    Note that statements in c/c++ are terminated with a ;

    int *x = 12345678;
    This means that x is of type pointer to integer (x is the address of a memory location at which location an integer is stored). The value of x is initialised to 12345678 (eg x points to the memory location 12345678 at which location an integer is stored). In this case, * means pointer to.

    int *x;
    *x = 12345678;
    This means that x is of type pointer to integer - and is uninitialised. At the memory location pointed to by the value of x, set to the value 12345678. As x is not initialised, x could take any value in the range of an integer. So some unknown location in memory is set to 12345678. In this case, the * in int *x means pointer to. The * in * x = means the contents of the memory pointed to (dereference).

    int *x; // declare pointer variable
    x = 12345678; // make x point to memory address 12345678

    Note that as pointed out in post #2 by dexwerx, you cannot directly assign a number to a pointer type. In this case 12345678 is of type int and x is of type pointer to int. Therefore a cast is required. In c++ this would be reinterpret_cast as in post #2.

    int *x = reinterpret_cast<int*>(12345678);
    auto x = reinterpret_cast<int*>(12345678);
    Or for c

    int *x = (int*)12345678;
    Note that if the value of x (the memory location) is not valid in the context of the program then a run-time exception will occur.

    Also note, that unless you are programming embedded-devices etc, you probably wouldn't initialise a memory pointer with a fixed value in this way. You would usually initialise with the address of an existing variable or with the value of a memory location obtained via a call to new, malloc() etc.
    Last edited by 2kaud; Feb 10th, 2018 at 05:37 AM.
    All advice is offered in good faith only. You are ultimately responsible for the effects of your programs and the integrity of the machines they run on. Anything I post, code snippets, advice, etc is licensed as Public Domain https://creativecommons.org/publicdomain/zero/1.0/

    C++17 Compiler: Microsoft VS2017 (15.7.1)

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