initializing multi-dimensional arrays

This is a discussion on initializing multi-dimensional arrays within the C++ Programming forums, part of the General Programming Boards category; Hi, I am confused about how multi-dimensional arrays are initialized. For instance, if I want to do something to the ...

  1. #1
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    initializing multi-dimensional arrays

    Hi,
    I am confused about how multi-dimensional arrays are initialized. For instance, if I want to do something to the effect of:
    Code:
    int a[2][3];
    for (int i = 0; i < 2; ++i) {
            for (int r = 0; r < 3; ++r) {
                   a[i][r] = i+r;
            }
    }
    I would imagine that I can initialize a[2][3] like this:
    Code:
    int a[2][3] = {
            { 0, 1 },
            { 1, 2 },
            { 2, 3 }
    };
    since a[2][3] can be seen as "an array of 3 elements, each element is an array of 2 elements"?

    that is apparently incorrect as my compiler (gcc 4.1) gives me:
    Code:
    a.cpp:6: error: too many initializers for ‘int [2][3]’
    and writing the declaration as
    Code:
     int a[3][2];
    fixes it.

    Can you please point out what am I doing wrong?
    Thank you very much

  2. #2
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    So it would be
    Code:
    int a[2][3] = {
            { 0, 1, 2 },  // 3 in each row, to match the minor dimension
            { 3, 4, 5 },  // 2 rows, to match the major dimension
    };
    You had a 3rd row in the "too many" case.
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    Thank you for your help.

    I guess there's something conceptually wrong with my understanding of multi-dimensional arrays then...

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    >> since a[2][3] can be seen as "an array of 3 elements, each element is an array of 2 elements"?

    a[2][3] should be seen as an array of 2 elements, each of which is an array of 3 elements. Just make that simple switch in your head and you should be good.

  5. #5
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    a[2][3] should be seen as an array of 2 elements, each of which is an array of 3 elements. Just make that simple switch in your head and you should be good.
    Thank you for your reply. I think I get it now.

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