sorting through a set of doubles

[pauldepstein]

Suppose x is of the type std::set<double>

I wanted to do a basic sort on x so that the elements were in
numerical order, in the sense that an iterator x.begin() would point
to the least element and so on.

My command std::sort(x.begin(), x.end()); generated compiler errors.
What should I have done?

Thank you,

Paul Epstein

 

[Ioannis Gyftos]

Suppose x is of the type std::set<double>

I wanted to do a basic sort on x so that the elements were in
numerical order, in the sense that an iterator  x.begin() would point
to the least element and so on.

My command std::sort(x.begin(), x.end());  generated compiler errors.
What should I have done?

Thank you,

Paul Epstein

Isn't an std::set already sorted?

What were the compiler errors?

http://www.cppreference.com/cppset/index.html

 

[puzzlecracker]

Suppose x is of the type std::set<double>

I wanted to do a basic sort on x so that the elements were in
numerical order, in the sense that an iterator  x.begin() would point
to the least element and so on.

My command std::sort(x.begin(), x.end());  generated compiler errors.
What should I have done?

Thank you,

Paul Epstein

std:set has a function value_comp, which returns a function used to
sort elements in the set. I presume there is a way to set this to a
different function. Would someone confirm that?

 

[Pascal J. Bourguignon]

Suppose x is of the type std::set<double>

I wanted to do a basic sort on x so that the elements were in
numerical order, in the sense that an iterator x.begin() would point
to the least element and so on.

My command std::sort(x.begin(), x.end()); generated compiler errors.
What should I have done?

std::set<double> x;
std::vector<double> y(x.size());
std::copy(x.begin(),x.end(),y.begin());
std::sort(y.begin(),y.end());

Now y is sorted, and *(y.begin()) is the least element.

 

[Kai-Uwe Bux]

It already does unless you have used a special comparison predicate upon
construction of x.

Nothing.


std:set has a function value_comp, which returns a function used to
sort elements in the set. I presume there is a way to set this to a
different function. Would someone confirm that?

You can only set this comparison function upon construction of the set.
Changing it later (causing a resort) is impossible.


Best

Kai-Uwe Bux

 

[puzzlecracker]

It already does unless you have used a special comparison predicate upon
construction of x.


You can only set this comparison function upon construction of the set.
Changing it later (causing a resort) is impossible.

Best

Kai-Uwe Bux

How do you set a comparison functions for set? via template
instantiation?

 

[Kai-Uwe Bux]

How do you set a comparison functions for set? via template
instantiation?

The answer has two parts:

a) std::set<> has a template parameter for the comparison function. It
defaults to std::less< value_type >. In order to provide your own, you have
to specify the template parameter.

b) In std::set< T, A, C >, the default constructor (and maybe some others)
takes arguments for the allocator and the comparison predicate. They
default to A() and C(). So, if the comparison template parameter has the
right default, you don't need to do anything. However, if you use something
like

std::function< bool( T const &, T const & ) >

for C, you will need to provide an appropriate functor.


Best

Kai-Uwe Bux

 

[Juha Nieminen]

std::set<double> x;
std::vector<double> y(x.size());
std::copy(x.begin(),x.end(),y.begin());
std::sort(y.begin(),y.end());

Now y is sorted, and *(y.begin()) is the least element.

"Now sorted"? The data was *already* sorted to begin with. Why sort it
again?

 

[Pascal J. Bourguignon]

Funny thing, though: the result is exactly the same if you comment out
the last line.

Yes, it's just me being paranoiac. However, we could add a different
comparator either in std::set<> or in std::sort().

 

[James Kanze]

"Iterator is not of random-access kind" would be my bet. And,
of course, no additional sorting of the set is needed, it's
already in the ascending order.

Iterator isn't random access, but also, you can't assign through
it (e.g. *iter = ... isn't legal).