Thread: memory allocation

  1. #1
    Registered User
    Join Date
    Feb 2006

    Question memory allocation


    i am try to create a dynamic array of char

    this is what i have done

    char *temp;
    unsigned int tempSize = 0;
    tempSize = 8212;
    temp = (char *) malloc(tempSize);
    printf("size of temp %d\n", sizeof(temp));

    what gets printed is

    size of temp 4

    shouldnt the size of temp be 8212?

  2. #2
    Devil's Advocate SlyMaelstrom's Avatar
    Join Date
    May 2004
    Out of scope
    Nope. The pointer is still a pointer no matter how much memory you allocate to it. The only difference is, instead of pointing to nothing, it's pointing to 8212 bytes of memory. You can still access it like an array, the memory is there. Also you shouldn't, and in this case don't need to, cast malloc.
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  3. #3
    Registered User
    Join Date
    Feb 2006
    ok thanks =D

  4. #4
    Frequently Quite Prolix dwks's Avatar
    Join Date
    Apr 2005
    And you should also free() dynamically allocated memory.

    Also you shouldn't, and in this case don't need to, cast malloc.
    I can't think of too many cases that you would need to cast malloc(). Unless you've using Turbo C or something.

    sizeof(a pointer) will return 4 on 32 bit machines and 2 on 16 bit ones. It won't return the number of bytes allocated to that pointer, as you've found out. There is no way of knowing (that I'm aware of) how to figure out how much memory is allocated to a particular pointer. It's best just to keep the size for future reference.

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