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POSwapping objects using pointers
 

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    <p>I'm trying to swap objects for a homework problem that uses void pointers to swap objects. The declaration of my function has to be:</p> <pre><code>void swap(void *a, void *b, size_t size); </code></pre> <p>I'm not looking for the exact code how to do it so I can figure it out by myself, but I'm not sure if I understand it correctly. I found that one problem is by doing:</p> <pre><code>void *temp; temp = a; a = b; b = temp; </code></pre> <p>only changes what the pointers point to. Is that correct? If it is correct, why doesn't swapping pointers actually change the contents between *a and *b. Because if your pointer points to something different, couldn't you dereference it and the objects would now be different? </p> <p>Similarly, just switching the values like:</p> <pre><code>void *temp; *temp = *a; *a = *b; *b = *temp; </code></pre> <p>Is not correct either, which I'm not sure why. Because again, it seems to me that the content is switched.</p> <p>Does swapping objects mean complete swapping of memory and value of what a pointer points to?</p> <p>So it seems like I have to use malloc to allocate enough space for my swap. If I allocate enough memory for one object, assuming they are the same size, I don't really see how it is different than the other two methods above. </p> <pre><code>void *temp = malloc(sizeof(pa)); // check for null pointer temp = a; // do something I'm not sure of since I don't quite get how allocating space is any // different than the two above methods??? </code></pre> <p>Thanks!</p>
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    1. COYou're close. Deferencing a `void *` isn't allowed though. Think about other ways you might move the information around, particularly think about the `size` parameter and what effect that has on your algorithm.
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      1. USCarl Norum
    2. COChanging the pointers is a nice try, but won't work because they were passed by value. Thus the caller will have the same value in the pointers before the call was made.
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      1. USThomas Matthews
 

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