Normal iterator pointer behavior

This is a discussion on Normal iterator pointer behavior within the C++ Programming forums, part of the General Programming Boards category; I attempted to write a simple function that would apply an input function to all elements of an 'array'. The ...

  1. #1
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    Normal iterator pointer behavior

    I attempted to write a simple function that would apply an input function to all elements of an 'array'. The array is part of a complex class built to handle matrices from a third party. I had thought the normal behavior of the iterator is that derefence operator *iter can be used to both access and set the values of the underlying object. However this code leaves the original array (rm) unchanged.
    Code:
    /* apply a function to all elements of the array */
    void rapp2self(cvm::rmatrix &rm, double (*dfunc)(double))
    {
    	cvm::rmatrix::iterator q ;
    	for (q = rm.begin() ; q != rm.end() ; ++q) {
    		*q = (*dfunc) (*q) ;
    	}
    
    }
    The cvm library is an external matrix lib. that uses the std libraries to generate it's internal arrays (or so I thought). I suppose the *q itself could be overloaded here causing surprising results. I'm not even sure I have the source for the cvm library to check this. For a 'correctly' implemented iterator can I assume that I can assign to the object q points to via *q?

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    C++ Witch laserlight's Avatar
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    For a 'correctly' implemented iterator can I assume that I can assign to the object q points to via *q?
    Yes, since iterators follow pointer semantics.

    Incidentally, have you considered using std::for_each or some other appropriate generic algorithm?
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    Cat without Hat CornedBee's Avatar
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    If the code leaves the original array unchanged, then it seems that the iterators are not conforming. But you'd have to show the iterator code.
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    Quote Originally Posted by laserlight View Post
    Yes, since iterators follow pointer semantics.

    Incidentally, have you considered using std::for_each or some other appropriate generic algorithm?
    I considered it, but my C++ reference made it look like more trouble than it was worth. I'll take a look at it again when I get a chance. Does it result in fewer characters type into my source code?

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    C++ Witch laserlight's Avatar
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    Does it result in fewer characters type into my source code?
    In this case, I think it is likely, but that should not really be an important factor to begin with. If you are going to write a function to apply a function to the elements of a container, you might as well use a generic function that applies a function (or function object) to a range.
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    Cat without Hat CornedBee's Avatar
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    Incidently, your function is effectively for_each, except for the begin() and end() calls. So, it's a non-generic range version of for_each.
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    Quote Originally Posted by laserlight View Post
    In this case, I think it is likely, but that should not really be an important factor to begin with. If you are going to write a function to apply a function to the elements of a container, you might as well use a generic function that applies a function (or function object) to a range.
    I agree with this wholeheartedly how does this work in C++?
    Is C++ so equipped with appropriate libraries?

    For example can I do something like this for some specially defined vectorized log function, or an overloaded log function?
    Code:
    std::vector<double> myvec ;
    ... code to initialize my vec ...
    
    std::vector<double>  logvec = log(myvec) ;
    That would be sweet. Come to think of it, it wouldn't be so hard to overload these functions now would it?

    By the way for some reason this code works just fine:
    Code:
    /* apply a function to all elements of the array  and
       return a copy of a new array */
    cvm::rmatrix &rapply(cvm::rmatrix &rm, double (*dfunc)(double))
    {
    	cvm::rmatrix::iterator q, i ;
    	cvm::rmatrix *x = new (GC) cvm::rmatrix(rm.msize(), rm.nsize()) ;
    	i = x->begin() ;
    	for (q = rm.begin() ; q != rm.end() ; ++q) {
    		*i = (*dfunc) (*q) ;
    		++i ;
    	}
    	return(*x) ;
    
    }
    So if I assign data via the pointer (i) to an allocated variable and return it. No problem. I could of course write vectorized versions of various functions via overloading, using this code as a template I suppose.

    However in the original case (rapp2self()) of passing an rmatrix by reference and trying to update it's values internally fails for some reason. I'll have to assume I'm doing something wrong, but the call I'm using to the rapp2self() function looks like this:

    Code:
    rapp2self(theta_nm_aoa, prin_value) ;
    where theta_nm_aoa is of type rmatrix. Not much to it really. Walking through the code with a debugger suggests internally that rapp2self is doing everything it's supposed to. It's only in the scope of the callee, where theta_nm_aoa remains unchanged. It's got to be something I'm doing that's stupid, but I haven't found it yet.

  8. #8
    Captain Crash brewbuck's Avatar
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    It's not always true that an iterator is assignable. For instance, std::map<K, V>::iterator is not assignable, even though it is non-const. However, i->second is assignable. See footnote [1] in http://www.sgi.com/tech/stl/Map.html

    However in this case, it seems like it makes sense to allow the assignment, so this was probably just an oversight on the part of the class designer.

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    C++ Witch laserlight's Avatar
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    I agree with this wholeheartedly how does this work in C++?
    Is C++ so equipped with appropriate libraries?

    For example can I do something like this for some specially defined vectorized log function, or an overloaded log function?
    Yes, with std::transform. Come to think of it, I think std::transform is more appropriate than std::for_each here.
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    Cat without Hat CornedBee's Avatar
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    Well, vectorized? Depends. GCC 4.3 has an experimental parallel implementation of some algorithms, using OpenMP. But for vectorization you'll have to rely on the compiler's loop unrolling. (No reason that wouldn't work, though, if the involved elements compile down well enough.)
    All the buzzt!
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