Binary search of an array of pointers to structs using pointer arithmetic

This is a discussion on Binary search of an array of pointers to structs using pointer arithmetic within the C Programming forums, part of the General Programming Boards category; I have an array of structs and I want to B-Search these by a value inside of them. I am ...

  1. #1
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    Binary search of an array of pointers to structs using pointer arithmetic

    I have an array of structs and I want to B-Search these by a value inside of them. I am having trouble because I need to return the index (not address) after searching through the array of structs (typedef structs). This is what I have, and I am not sure my logic behind it is even right. Can somebody please help.

    Code:
    int findValue(StructType * const array[], int size, const String target) {
    	StructType * leftPtr = * array;
           	StructType * rightPtr = (* array) + size - 1; // error for the + sign
    	StructType * midPtr;
    	
    	while (strcmp((*leftPtr).value, (*rightPtr).value) <= 0) {	
    		midPtr = (leftPtr + rightPtr) / 2;
    		
    		while(strcmp((*midPtr).value, target) < 0) {
    			leftPtr = midPtr + 1;
    			midPtr = (leftPtr + rightPtr) / 2;
    		}
    		
    		if (strcmp((*midPtr).value, target) == 0)
    			return midPtr - array;
    	}
    	
    	return -1;
    	
    }
    Last edited by mgimbl; 07-02-2004 at 12:03 AM.

  2. #2
    & the hat of GPL slaying Thantos's Avatar
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    Given:
    Code:
    int array[10];
    int *p = array, index;
    /* some code that changes p */
    the easist way to get the index from the pointer is simply:
    Code:
    index = p - array;

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    well that solved one of my big questions .. and I now know what I have to return, thanks.

    But the other problem is I have StructType pointers and I need int pointers (I think) ..
    See my return statement .. if I did :
    Code:
    return midPtr - array;
    that would try to subtract two StructTypes instead of two ints ... see what I am saying?

  4. #4
    & the hat of GPL slaying Thantos's Avatar
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    you are right that the pointers themselves are of StructTypes but when you do the arithmatic on them the result is an integer.

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    here are my errors when trying to compile that code above:

    Code:
    [mgimbl@localhost project1]$ make
    gcc -c kennel.c
    kennel.c: In function `findDog':
    kennel.c:55: error: invalid operands to binary +
    kennel.c:59: error: invalid operands to binary +
    kennel.c:63: error: invalid operands to binary -
    make: *** [kennel.o] Error 1
    Last edited by mgimbl; 07-02-2004 at 12:03 AM.

  6. #6
    & the hat of GPL slaying Thantos's Avatar
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    Oh thats right you can not do addition between two pointers only subtraction.
    Its fairly simple to get around:
    Code:
    midPtr = ( (leftPtr-array) + (rightPtr-array)) / 2 + array;

  7. #7
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    sorry for being a major pain

    same errors are still present after your last suggestion ...
    I have linked my actual code.

    Code:
    int findDog(DogType * const kennel[], int numDogs, const String target) {
    	DogType * leftPtr = * kennel;
    	DogType * rightPtr = (* kennel) + (numDogs - 1);
    	DogType * midPtr;
    	
    	while (strcmp((*leftPtr).dogName, (*rightPtr).dogName) <= 0) {	
    		midPtr = (leftPtr + rightPtr) / 2;
    		
    		while(strcmp((*midPtr).dogName, target) < 0) {
    			leftPtr = midPtr + 1;
    			midPtr = (leftPtr + rightPtr) / 2 + kennel;
    		}
    		
    		if (strcmp((*midPtr).dogName, target) == 0)
    			return midPtr - kennel; 
    	}
    	
    	return -1;
    	
    }
    Last edited by mgimbl; 07-02-2004 at 12:53 AM.

  8. #8
    & the hat of GPL slaying Thantos's Avatar
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    Like I said in the last post you can't add two pointers together. So
    Code:
    midPtr = (leftPtr + rightPtr) / 2;
    should be
    Code:
    midPtr = ( (leftPtr-kennel) + (rightPtr-kennel)) / 2 + kennel;

  9. #9
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    oh I am sorry ... I didnt update that piece of code ... these are my error now:

    kennel.c: In function `findDog':
    kennel.c:55: error: invalid operands to binary -
    kennel.c:55: error: invalid operands to binary -
    kennel.c:59: error: invalid operands to binary -
    kennel.c:59: error: invalid operands to binary -
    kennel.c:63: error: invalid operands to binary -
    make: *** [kennel.o] Error 1

    Still talking about me subtracting them ...

  10. #10
    Yes, my avatar is stolen anonytmouse's Avatar
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    There is one golden rule with pointer arithmetic - "Don't use pointer arithmetic!".
    Code:
    int findValue(StructType * const array[], size_t size, const String target)
    {
    	size_t mid_index   = 0;
    	size_t left_index  = 0;
    	size_t right_index = size - 1;
    
    	while (strcmp(array[left_index]->value, array[right_index]->value) <= 0)
    	{
    		mid_index = (left_index + right_index) / 2;
    
    		while(strcmp(array[mid_index]->value, target) < 0)
    		{
    			left_index = mid_index + 1;
    			mid_index  = (left_index + right_index) / 2;
    		}
    
    		if (strcmp(array[mid_index]->value, target) == 0)
    			return (int) mid_index;
    	}
    	
    	return -1;
    }
    Things to think about:
    - I think the binary search algo also moves the right_index when appropriate.
    - strcmp is expensive, can you remove any calls to strcmp?
    - What happens if a negligent coder passes unsorted data to your function?
    Last edited by anonytmouse; 07-02-2004 at 05:36 AM.

  11. #11
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    Quote Originally Posted by C+++C_forever
    > midPtr = ( (leftPtr-kennel) + (rightPtr-kennel)) / 2 + kennel;

    if i am not wrong, its because kennel is a pointer to a pinter. yoo should dereference the kernels..

    Anyway, i actually post to ask my question on binary search. The complexity of it is logN ( = log base 10 N ) ok? But shouldn't it be log base 2 N ?

    Yes, the complexity of a binary search is O(log2 (N)).
    As an example, if you have 127 elements you want to search
    through, you can guarantee a decision one way or another in
    7 or less steps because log2(127) + 1 = 7.

  12. #12
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    log2(N) = log10(N) / log10(2) = k*log10(N) where k is a constant.

    so O(log2(N)) = O(log10(N)) = O(ln(N)) = ......

    since constants are omitted in O() notation.

    {the equals sign may not be the best representation to
    use. I'm trying to say that they are equivalent I guess}
    Last edited by DavT; 07-02-2004 at 08:28 AM.
    DavT
    -----------------------------------------------

  13. #13
    Code Goddess Prelude's Avatar
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    >- I think the binary search algo also moves the right_index when appropriate
    The traditional algorithm for binary search modifies both the right_index and the left_index:
    Code:
    low := 0
    high := n - 1
    while low <= high do
      mid := (low + high) / 2
      if x < array[mid] then
        high := mid - 1
      else if x > array[mid] then
        low := mid + 1
      else
        return mid
      endif
    loop
    
    return -1
    The same changes are made for a pointer based rather than index based implementation.

    >- What happens if a negligent coder passes unsorted data to your function?
    Checking to see if the data is already sorted can be expensive. This is a difficult tradeoff that most people usually make undefined and expect the client code to be smart enough to use it properly.
    My best code is written with the delete key.

  14. #14
    & the hat of GPL slaying Thantos's Avatar
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    There is one golden rule with pointer arithmetic - "Don't use pointer arithmetic!".
    Right and there is never a time to use goto or continue or 3+ dimensional arrays.
    if i am not wrong, its because kennel is a pointer to a pinter. yoo should dereference the kernels..
    Good catch, its amazing what I can miss when I am haven't had enough sleep
    Last edited by Thantos; 07-02-2004 at 09:43 AM.

  15. #15
    Code Goddess Prelude's Avatar
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    >Right and there is never a time to use goto or continue or 3+ dimensional arrays.
    You forgot the smiley after that statement.
    My best code is written with the delete key.

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