Sorting a list with class for complex types [modified]

[FarPointer]

Change it in the for loop also . Regards, FarPointer Blog:FARPOINTER

[FarPointer]

how did u bring those mydata in-side the angular brackets . and that horizontal slash in prev post. Regards, FarPointer Blog:FARPOINTER

[Harold_Wishes]

I decided to pick this apart and comment out the for loop because it is still not obvious to me where the problem is. The for loop seems to declare a pointer and initialize it to point to the beginning of the list and simply prints each item on screen until the pointer reaches the end. That aside I notice there appears to be some conversion type error in the line above with mylist.sort(MyDataSortPredicate); See error message below. :sigh: c:\Documents and Settings\WoodallH\Desktop\C++\Project 6 Nesty tag\Sort.cpp(33) : error C2664: 'void __thiscall std::list >::sort(struct std::greater)' : cannot convert parameter 1 from ' bool (const class MyData &,const class MyData &)' to 'struct std::greater' No constructor could take the source type, or constructor overload resolution was ambiguous Error executing cl.exe. Sort.exe - 1 error(s), 0 warning(s)

mylist.sort(MyDataSortPredicate);

// Dump the list to check the result
/* for (list::const_iterator citer = mylist.begin();
citer != mylist.end(); ++citer)
{
cout << (*citer).m_iData << endl;
}*/

[David Crow]

FarPointer wrote:

and that horizontal slash in prev post.

Horizontal rule.

---

"The largest fire starts but with the smallest spark." - David Crow

"Judge not by the eye but by the heart." - Native American Proverb

[Zac Howland]

Harold_Wishes wrote:

// Dump the list to check the result for (list::const_iterator citer = mylist.begin(); citer != mylist.end(); ++citer) { cout << (*citer).m_iData << endl; }

Don't write your own loop for this.

// declared somewhere
void printData(const MyData& data)
{
	cout << data.m_iData << endl;
}

// use instead of loop
for_each(myList.begin(), myList.end(), printData);

If you decide to become a software engineer, you are signing up to have a 1/2" piece of silicon tell you exactly how stupid you really are for 8 hours a day, 5 days a week Zac

[Harold_Wishes]

This worked. But I still have commented out the code that is suppose to do the sort---> mylist.sort(MyDataSortPredicate);

C:\Documents and Settings\WoodallH\Desktop\C++\Project 6 Nesty tag\Sort.cpp(38) : error C2664: 'void __thiscall std::list<class MyData,class std::allocator<class MyData> >::sort(struct std::greater<class MyData>)' : cannot convert parameter 1 from '
bool (const class MyData &,const class MyData &)' to 'struct std::greater<class MyData>'
No constructor could take the source type, or constructor overload resolution was ambiguous
Error executing cl.exe.

-- modified at 15:03 Monday 10th July, 2006

[David Crow]

See here for the STL fix.

---

"The largest fire starts but with the smallest spark." - David Crow

"Judge not by the eye but by the heart." - Native American Proverb

[Zac Howland]

You can either use the general sort algorithm (std::sort) or write a specialized less<MyData>() functor that looks something like:

struct std::less
{
	bool operator()(const MyData& lhs, const MyData& rhs)
	{
		return lhs.m_iData < rhs.m_iData;
	}
};

And then call mylist.sort(std::less<MyData>). If you decide to become a software engineer, you are signing up to have a 1/2" piece of silicon tell you exactly how stupid you really are for 8 hours a day, 5 days a week Zac

[earl]

Harold, You'll do much better if you get (1) a copy of Stroustrup and (2) a copy of Josuttis. Read them in that order. Because what you want to do is trivial and will be explained by the first book; putting in the time for the second will be worth it. earl

[Harold_Wishes]

Are these C++ authors of C++ books? I will look for them. I did not see anything in the Deitel & Deitel book that was too helpful.

[earl]

Hit up amazon for those two. Stroustrup largely invented C++ and wrote one of the definitive books on it; Josuttis wrote a book on the standard library that is a great complement to the former.

[Stephen Hewitt]

Simply define an ordering by implementing operator <. i.e. struct data {     string Length;     string Sequence;     string N_Terminal;     string C_Terminal;       friend bool operator<(const data &L, const data &R)     {        // In this example I'm only sorting by 'Sequence'        return L.Sequence<R.Sequence;     } };   // Now you can sort like this. g_DataList.sort(); Steve

[Harold_Wishes]

So after creating the list, doesn't the sort function take in one parameter of type data? And will the list be printed in sorted order after the sort function is invoked?

[Stephen Hewitt]

The list already knows the type of the data it contains. There are two member functions of list called sort: one with no parameters and one which takes a predicate. I'm using the one which takes no parameters and overloading operator <. Give it a try and see if it works for you. Steve

[Harold_Wishes]

I did check amazon and saw several books by Stroustrup. Do you know the title of the book? -- modified at 21:13 Monday 10th July, 2006

[Harold_Wishes]

Thanks! This really helps because my goal is to actually compare the string length of the Sequence variable and assign that value to int Number as I have it defined below:

struct data //
{
int Number;
string Length;
string Sequence;
string N_Terminal;
string C_Terminal;
friend bool operator<(const data &L, const data &R)
{
return L.number > R.number;
}

};

Then I invoke the sort after the list has been populated with items.

g_DataList.sort();

You will notice I overloaded the > operator instead of the < operator since I need the longest strings to appear first. The thing I do not quite get is how the compiler knows what const data &L and const data &R are when they are passed by reference in the friend function. I'm not making the connection between the comparison and the sort function? Perhaps I do not understand the sorting algorithym. I confess my lack of programming experience even though I have the program working like a champ at this stage. Regards, HRW. :) -- modified at 4:40 Tuesday 11th July, 2006

[Stephen Hewitt]

A friend function is really a global function, not a member function. This would be clearer if the declaration and the definition were separated as is shown below: struct data {     int Number;     string Length;     string Sequence;     string N_Terminal;     string C_Terminal;     friend bool operator<(const data &L, const data &R); // Declaration. };   // Definition. bool operator<(const data &L, const data &R) {     return L.number > R.number; } In this example the function doesn't even need to be made a friend as it doesn't access any protected or private members. With this modification it looks like this: struct data {     int Number;     string Length;     string Sequence;     string N_Terminal;     string C_Terminal; };   bool operator<(const data &L, const data &R) {     return L.number > R.number; } When the compiler sees a < and one or both of the parameters (the expressions to the left and right of the <) is a user defined type it looks for a user defined operator<. It chooses from all the candidates by matching the types in the expression it is compiling with the types of the operator. Steve

[earl]

http://www.amazon.com/gp/product/0201700735/sr=8-1/qid=1152627803/ref=pd\_bbs\_1/002-6564248-4536802?ie=UTF8 http://www.amazon.com/gp/product/0201379260/002-6564248-4536802?n=283155