Thread: 2 - D arrays & pointers

  1. #1
    Registered User
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    Nov 2009

    2 - D arrays & pointers

    Here is a code snippet :

    int a[2][2] = {
                        {7, 12},
                        {2, 7} 
    printf("\n%u %u %u", a + 1, *(a + 1), a[1]);
    All 3 would return the same value i.e., address of 1th 1 - D array.

    As far as I know, when the compiler encounters a reference to 2 - D array, in this case, for instance, 'a' it decays to int (*a)[2]. When I say a + 1, it moves to 1th 1 - D array, but when I say *(a + 1), instead of giving the output 2, it again gives me the address & not the value because all that is happening by saying *(a + 1) is, that it is being converted from pointer to array to just a pointer. Is that right ?

    The book I referred, seems to have a vague explanation, so I just need to confirm.

  2. #2
    C++ Witch laserlight's Avatar
    Join Date
    Oct 2003
    *(a + 1) is equivalent to a[1], so let's consider a[1]. Wait, there's not much to consider: it is just the element at index 1 for a, i.e., an array of 2 ints, which is convertible to a pointer to int.
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
    Look up a C++ Reference and learn How To Ask Questions The Smart Way

  3. #3
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