arrays

This is a discussion on arrays within the C++ Programming forums, part of the General Programming Boards category; How do I turn a two dimensional array into a one dimensional array?...

  1. #1
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    arrays

    How do I turn a two dimensional array into a one dimensional array?

  2. #2
    Guest Sebastiani's Avatar
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    You don't have to. Hint: cast.
    Code:
    bool fun(bool value)
    {
        return std::pow(std::exp(1), std::complex<float>(0, 1) 
        * std::complex<float>(std::atan(1)*(1 << (value + 2))))
        .real() > 0;
    }

  3. #3
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    Here is what I have I dont know what to put in the for loops
    int val[height][width]
    int n,m;

    int main()
    {

    for( )
    for( )

    {val[]=(n+1)*(m+1);

    }
    return 0;
    {

  4. #4
    and the Hat of Guessing tabstop's Avatar
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    What are you trying to compute, here? I see no reason why you would expect, or for that matter want, val as a one-dimensional array.

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    Its just a problem for school I need to find out how to take a two dimensional array and turn it in to a one dimensional array

  6. #6
    Guest Sebastiani's Avatar
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    >> Here is what I have I dont know what to put in the for loops

    You put valid C statements in them and then check the output, for starters. But if you're too lazy to pick up a book and actually learn how to do that, well...
    Last edited by Sebastiani; 11-30-2008 at 07:15 PM. Reason: tone
    Code:
    bool fun(bool value)
    {
        return std::pow(std::exp(1), std::complex<float>(0, 1) 
        * std::complex<float>(std::atan(1)*(1 << (value + 2))))
        .real() > 0;
    }

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    Here is what the two dimensional array code looks like, I need to find out what goes in the for loops in the above code to turn it in to a one dimensional array.
    int val[height][width]
    int n,m;

    int main()
    {
    for(n=0; n<height;n++)
    for(m=0;m<width;m++)

    {val[n][m]=(n+1)*(m+1);
    }

    return 0;
    }
    Last edited by john5754; 11-30-2008 at 07:23 PM.

  8. #8
    Guest Sebastiani's Avatar
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    Ok, the array contents are *guaranteed* to be contiguous, so a simple cast is *really* all that is needed (ie: ** -> * ). Can you manage that?
    Code:
    bool fun(bool value)
    {
        return std::pow(std::exp(1), std::complex<float>(0, 1) 
        * std::complex<float>(std::atan(1)*(1 << (value + 2))))
        .real() > 0;
    }

  9. #9
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    what do you mean by cast

  10. #10
    Guest Sebastiani's Avatar
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    eg:

    Code:
    double d = 2.5;
    int i = int( d ); // c-style
    // - or -
    int j = static_cast< int >( d ); // c++-style
    It works with pointers, too, of course.
    Code:
    bool fun(bool value)
    {
        return std::pow(std::exp(1), std::complex<float>(0, 1) 
        * std::complex<float>(std::atan(1)*(1 << (value + 2))))
        .real() > 0;
    }

  11. #11
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    Quote Originally Posted by Sebastiani View Post
    Ok, the array contents are *guaranteed* to be contiguous, so a simple cast is *really* all that is needed (ie: ** -> * ). Can you manage that?
    Is that true everywhere? For example, dynamically allocated arrays, are they also guaranteed to be contiguous?

    My solution would have looked something like this

    Code:
    int old2darray[width][height];
    int new1darray[width * height];
    
    for (blah)
    {
       for(blah)
       {
          move data from oldarray to new array
       }
    }
    However, casting would be a much more efficient way to move the data. (Ok, you aren't really moving data)

  12. #12
    and the Hat of Guessing tabstop's Avatar
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    Quote Originally Posted by Cogman View Post
    Is that true everywhere? For example, dynamically allocated arrays, are they also guaranteed to be contiguous?
    Dynamic arrays are not guaranteed to be contiguous. (They almost always aren't, I would guess, if malloc/calloc writes extra bits to the allocated memory.)

  13. #13
    C++ Witch laserlight's Avatar
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    Quote Originally Posted by tabstop
    Dynamic arrays are not guaranteed to be contiguous.
    Just in case there is confusion, note that what tabstop means is that consecutively (or otherwise) allocated dynamic arrays are not guaranteed to be contiguous. However, like fixed size arrays, the contents of each dynamic array are contiguous.
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  14. #14
    C++まいる!Cをこわせ! Elysia's Avatar
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    Quote Originally Posted by tabstop View Post
    Dynamic arrays are not guaranteed to be contiguous. (They almost always aren't, I would guess, if malloc/calloc writes extra bits to the allocated memory.)
    That's new/delete for you.
    But anyone in their right mind wouldn't try to make dynamically allocated arrays using new and delete in C... if they can avoid it!
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    Quote Originally Posted by Salem View Post
    You mean it's included as a crutch to help ancient programmers limp along without them having to relearn too much.

    Outside of your DOS world, your header file is meaningless.

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    Here is what I got, It seems to work when I run it. See any problems?
    # define width 5
    # define height 3

    int jimmy[Height * Width]
    int n,m;
    int main()
    {
    for(n=0; n<height ; n++)
    for(m=0; m<width; m++)
    { jimmy[n]=(n+1)*(m+1);
    }
    return 0;
    {

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