sizeof on arrays

This is a discussion on sizeof on arrays within the C++ Programming forums, part of the General Programming Boards category; What should i do to make this code return the size of my dynamic allocated array. it just says that ...

  1. #1
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    Question sizeof on arrays

    What should i do to make this code return the size of my dynamic allocated array. it just says that its 4 bytes, but I want all of the array
    Code:
     
      floatarray = new float[7];
      for (k=0;k<7;k++) floatarray[k]=(k+1);
      cout << "size of floatarray" << sizeof(floatarray) << endl;

  2. #2
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    // cout << "size of floatarray" << strlen(floatarray) << endl;

    Kuphryn

  3. #3
    Me want cookie! Monster's Avatar
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    I don't think that's possible with dynamic arrays (only static).
    The sizeof(floatarray) returns the size of the pointer, not the array itself.

  4. #4
    and the hat of wrongness Salem's Avatar
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    As Monster says, you can only do this on an array (not an extern of an array, or an array generated by calling new)

    Do this
    Code:
    int floatarraysize = 7;
    floatarray = new float[floatarraysize];
    cout << "size of floatarray" << floatarraysize << endl;
    Your implementation may provide a NON-Standard way of doing this, but you'd have to read the manuals to find it (assuming it was even provided).

  5. #5
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    Interesting.

    I have never tried using strlen(pointertoArray).

    How about strlen(array[]) or strlen(array[0])?

    Kuphryn

  6. #6
    and the hat of wrongness Salem's Avatar
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    > I have never tried using strlen(pointertoArray).
    Since its only valid to try this on char arrays which have a \0 terminator, it isn't really applicable.

    In addition
    char arr[80] = "hello";

    sizeof(arr) is 80
    strlen(arr) is 5

    So it doesn't really help that much even when you have strings

  7. #7
    Banned borko_b's Avatar
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    if you use C memory allocation routines like:
    malloc calloc,realloc you could use
    special function called msize wich returns the size of the mem

    like:
    Code:
    #include <malloc.h>
        ....
       float array = (float*)malloc(10 * sizeof(float));
       ...
       /// do something 
       ...
       int memsz =0;
    #ifdef WIN32
       memsz = _msize(array); //Win32 specific name (ANSI name is msize)
    #else
       memsz = msize(array);
    #endif
    
      //free the mem
      free(array);
    you can try this with NEW and DELETE
    but i don't know the result (wether it will be succesfull or portable)

  8. #8
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    The size of an array is the size of the element multiplied with the number of places in the Array.

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    Thumbs up

    thx guys and gals. I had to put in a counter which i increased every time i increased the size of the dynamic array.

  10. #10
    Registered User Dual-Catfish's Avatar
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    Therefore, sizeof(array)/sizeof(array[0]) will give you the size of each individual element. Couldn't you derefrence the array if it's dynamic to find out how large it is?

  11. #11
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    thx guys and gals. I had to put in a counter which i increased every time i increased the size of the dynamic array.
    Once you do this, your dynamic array is incurring the same overhead as a Vector. Why not just use a Vector?
    Callou collei we'll code the way
    Of prime numbers and pings!

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