problems using memcpy

This is a discussion on problems using memcpy within the C Programming forums, part of the General Programming Boards category; hey, i have a problem trying to figure out what i'm doing wrong with memcpy, i get the following to ...

  1. #1
    Registered User
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    Jan 2008
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    3

    problems using memcpy

    hey, i have a problem trying to figure out what i'm doing wrong with memcpy,

    i get the following to errrors when i try to run this code.

    a4.c:222: error: incompatible type for argument 1 of `memcpy'
    a4.c:222: error: incompatible type for argument 2 of `memcpy'

    memArray is an array of memRec structures.


    Code:
    for(i=foundIndex; i<memLoc; i++) {
     memcpy(memArray[i],memArray[i+1],sizeof(memRec));
     }
    Could someone please tell me why my first and second arguments are incompatible types for memcpy and what they should be instead.

    Thanx a heap

  2. #2
    Kernel hacker
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    Farncombe, Surrey, England
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    Quote Originally Posted by newby01 View Post
    hey, i have a problem trying to figure out what i'm doing wrong with memcpy,

    i get the following to errrors when i try to run this code.

    a4.c:222: error: incompatible type for argument 1 of `memcpy'
    a4.c:222: error: incompatible type for argument 2 of `memcpy'

    memArray is an array of memRec structures.


    Code:
    for(i=foundIndex; i<memLoc; i++) {
     memcpy(memArray[i],memArray[i+1],sizeof(memRec));
     }
    Could someone please tell me why my first and second arguments are incompatible types for memcpy and what they should be instead.

    Thanx a heap
    You need the address of the memArray[i], not the memArray[i] itself.

    --
    Mats
    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

  3. #3
    and the hat of sweating
    Join Date
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    Toronto, ON
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    Assuming you have an array like this:
    Code:
    char str[100];
    str[1] is just a single char, not a char*

    (str + 1) is a char* because it takes the address of the first element of str and adds 1 to the pointer address, creating a new address.

    &str[1] is the address of the 2nd char in the array, so it's also a char*

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