size of array - why function gives size ONE only

This is a discussion on size of array - why function gives size ONE only within the C++ Programming forums, part of the General Programming Boards category; the user-defined function size_of_array(numA) is meant to give the size of numA which is 5 but it gives 1... I ...

  1. #1
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    size of array - why function gives size ONE only

    the user-defined function size_of_array(numA) is meant to give the size of numA which is 5 but it gives 1... I wonder where went wrong? can anyone help pls...
    insert
    Code:
    #include <iostream>
    using namespace std;
    int size_of_array(int numA[]);
    
    int main()
    {
    	int numA[]={0,1,2,3,4};
    	int i=0;
    	int total=0;
    
    	for(i=0; i<size_of_array(numA); i++)
    	{
    		total+=numA[i];
    	}
    	float average=total/i;
    	cout << "average is "<<average;
    	
    	return 0;
    }
    int size_of_array(int numA[])
    {
    		int num_of_elements=sizeof(numA)/sizeof(int);
    		return num_of_elements;
    }

  2. #2
    C++ Witch laserlight's Avatar
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    This has been answered elsewhere.
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  3. #3
    Registered User Dawnson's Avatar
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    Obviously, sizeof(numA) is 5, and sizeof(int) is 4(depending on the compiler). You have made 5/4 and that results in 1.

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    Quote Originally Posted by Dawnson View Post
    Obviously, sizeof(numA) is 5, and sizeof(int) is 4(depending on the compiler). You have made 5/4 and that results in 1.
    Wrong. sizeof(numA) is 4 here. Had numA been an array rather than a pointer, of 5 elements, sizeof(numA) would have been 20.
    Yes, depending on the compiler.

  5. #5
    DESTINY BEN10's Avatar
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    Quote Originally Posted by Dawnson View Post
    Obviously, sizeof(numA) is 5, and sizeof(int) is 4(depending on the compiler). You have made 5/4 and that results in 1.
    Absolutely wrong. If num[] would have contained 8 elements then according to you the answer should have been 2 which is incorrect. The correct reason is in the link given by laserlight. When you pass an array to a function it decomposes into pointer to the the first element and thus sizeof(num)==4 which is equal to sizeof(int).
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  6. #6
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    You could do this instead.
    Code:
    #include <iostream>
    using namespace std;
    
    template<typename T, int N>
    char (&array(T (&)[N]))[N];
    
    int main()
    {
    	int numA[]={0,1,2,3,4};
    	int i=0;
    	int total=0;
    
    	for(i=0; i<sizeof(array(numA)); i++)
    	{
    		total+=numA[i];
    	}
    	float average=total/i;
    	cout << "average is "<<average;
    	
    	return 0;
    }

  7. #7
    C++まいる!Cをこわせ!
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    Or how about using some typedefs? I am having trouble discerning the types that goes into that function. It's pure luck I can see it's a function, too.
    Code:
    template<typename T, int N>
    void myfunction(T (& name)[N])
    {
        //...
    }
    But I also believe something like this works, where N is the number of elements. You could do a sizeof on name and get the real size in this case.
    Or you could just use a vector and use the .size() method and multiple with the size of one element.
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    io.h certainly IS included in some modern compilers. It is no longer part of the standard for C, but it is nevertheless, included in the very latest Pelles C versions.
    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.

  8. #8
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    I think we need to wait til C++0x in order to really simplify that ugly function signature
    Code:
    template<typename T, int N> 
    typedef T (&array_t)[N];
    ** I assume that this will eventually be legal, although even Comeau with C++0x extensions doesn't recognise it right now, so perhaps I misunderstand what they actually mean by "adding support for template typedefs" :-)


    Yes I would prefer an STL container too; Personally I only find myself using these to initialise containers using const arrays as lookup tables.

    The reason I prefer the sizeof(char(&)[N]) approach is that the return type can be evaluated at compile time without ever actually calling the function, but I suppose a solution which actually calls a function would be OK aswell.
    Code:
    template<typename T, int N>
    int array_size( T (&dummy_arg)[N] )
    {
        return N;
    }
    Or, if a range is needed for an STL container or algorithm then maybe returning a pointer would be better
    Code:
    template<typename T, int N>
    T* end_of( T (&array)[N] )
    {
        return array+N;
    }

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