Dynamically allocated size

This is a discussion on Dynamically allocated size within the C++ Programming forums, part of the General Programming Boards category; Is there a way to determine the size of an object/variable that you have dynamically allocated memory to? For example: ...

  1. #1
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    Dynamically allocated size

    Is there a way to determine the size of an object/variable that you have dynamically allocated memory to? For example:

    Code:
    int foo() {
        int x;
        return sizeof(x);
    }
    returns 4, or the size of an int.

    Code:
    int foo() {
        char* data = new char[128];
        int size = ???;
        delete {} data;
        return size;
    }
    Is there a way to determine the size of the above dynamically allocated character array?

  2. #2
    C++ Witch laserlight's Avatar
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    To use new char[128] implies that you know that the size is 128.

    I would say that the solution is to keep track of the size of the dynamic array with another variable.

    Oh, and it is delete[] data, not delete{} data.
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  3. #3
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    Quote Originally Posted by laserlight
    I would say that the solution is to keep track of the size of the dynamic array with another variable.
    That's what I've always done. just wondering if there was some voodoo I wasn't aware of. Sorry bout the typo.

  4. #4
    C++ Witch laserlight's Avatar
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    That said, I would prefer std::string or std::vector<char> if I can.
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  5. #5
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    Code:
    size_t _msize( void *memblock );
    Don't quote me on that... ...seriously

  6. #6
    carry on JaWiB's Avatar
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    Quote Originally Posted by Brad0407 View Post
    Code:
    size_t _msize( void *memblock );
    I think this gives some good reasons not to use _msize.
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  7. #7
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    I'll let the function speak for itself:
    Code:
    #include <iostream>
    #include <malloc.h>
    
    int foo() {
        char* data = new char[128];
        int size = _msize(data);
        delete [] data;
        return size;
    }
    
    int main()
    {
    	std::cout << "foo() returns " << foo() << "\n";
    	return 0;
    }
    If you have a better idea, let me know.
    Last edited by Brad0407; 06-26-2007 at 11:13 AM.
    Don't quote me on that... ...seriously

  8. #8
    The larch
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    Try not to have plain pointers in the code. If, for some reason you don't think you can use a vector or any other standard container, put the pointer in a class that also has the size member.
    I might be wrong.

    Thank you, anon. You sure know how to recognize different types of trees from quite a long way away.
    Quoted more than 1000 times (I hope).

  9. #9
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    Quote Originally Posted by maverickbu View Post
    That's what I've always done. just wondering if there was some voodoo I wasn't aware of. Sorry bout the typo.
    By interrogating the size of the array and dividing it by the size of the type of data it holds you can use;

    Code:
    int size = sizeof(data)/sizeof(char);

  10. #10
    C++ Witch laserlight's Avatar
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    When you have a better idea, let me know.
    Code:
    #include <iostream>
    #include <vector>
    
    int foo() {
        std::vector<char> data(128);
        return data.size();
    }
    
    int main()
    {
    	std::cout << "foo() returns " << foo() << "\n";
    	return 0;
    }
    Or even:
    Code:
    #include <iostream>
    
    int foo() {
        return 128;
    }
    
    int main()
    {
    	std::cout << "foo() returns " << foo() << "\n";
    	return 0;
    }
    The point is that it is non-standard and non-portable, and consequently may not return the size as expected for this purpose.

    By interrogating the size of the array and dividing it by the size of the type of data it holds you can use;
    In this case it would just give the size of the pointer.
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  11. #11
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    Quote Originally Posted by indigo0086 View Post
    By interrogating the size of the array and dividing it by the size of the type of data it holds you can use;

    Code:
    int size = sizeof(data)/sizeof(char);
    size will always be 4 as long as pointers are stored in 32 bits of memory and characters in 8 bits.

    >> The point is that it is non-standard and non-portable, and consequently may not return
    >> the size as expected for this purpose.

    Amen
    Don't quote me on that... ...seriously

  12. #12
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    Ah, missed the vectors thing. Either way vectors are better.

  13. #13
    Cat without Hat CornedBee's Avatar
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    There are two more problems with _msize, aside from the obvious (it's compiler specific):
    1) new[] might be overloaded and not call malloc at all.
    2) new[] might allocate some additional bookkeeping memory that malloc (and thus _msize) is not aware of. The result would be that the result of _msize might not be evenly divisible by the item size, or the division may result in a higher item count than is actually available.
    All the buzzt!
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