How can I find an element in a set which contains pointers to the elements?

set::find c++
c set::find complexity
find pair in set c++
c++ set of pointers

Edit: I fixed my mistake: I'm using a set and not a vector.

Please consider the following example code:

set<Foo *> set_of_foos;

set_of_foos.insert(new Foo(new Bar("x")));
set_of_foos.insert(new Foo(new Bar("y")));

// The way a "foo" is found is not important for the example.
bool find_foo(Foo *foo) {
  return set_of_foos.end() != set_of_foos.find(foo);

Now when I call:

find_foo(new Foo(new Bar("x")));

the function returns false since what I'm looking for can't be found. The reason is obvious to me: The pointers point to different objects since they are allocated both with a new, resulting in different values of the addresses.

But I want to compare the contents of Foo (i.e. "x" in the above example) and not Foo * itself. Using Boost is not an option as well as modifying Foo.

Do I need to loop through each of the Foo * inside set_of_foos or is there a simpler solution? I tried uniquely serializing the contents of each Foo and replace the set<Foo *> with a map<string, Foo *>, but this seems like a very "hacked" solution and not very efficient.

find_foo(new Foo(new Bar("x"))); does not sound like a good idea - it will most likely (in any scenario) lead to memory leak with that search function.

You could use find_if with a functor:

struct comparator {
    Foo* local;
    comparator(Foo* local_): local(local_) {}
    ~comparator() { /* do delete if needed */ } 
    bool operator()(const Foo* other) { /* compare local with other */ }

bool found = vec.end() != std::find_if(vec.begin(), vec.end(), comparator(new Foo(...)));

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Change your vector to set with your custom comparable function to compare Foo objects.

Should be:

struct ltFoo
  bool operator()(Foo* f, Foo* s) const
    return f->value() < s->value();

set<Foo*, ltFoo> sFoo;
sFoo.insert(new Foo(new Bar("x"));
sFoo.insert(new Foo(new Bar("y"));

if (sFoo.find(new Foo(new Bar("y")) != sFoo.end())
    //not exists

Data Structures: Theory and Practice, As in the 2-word element, one of the words contains pointers, but now a single One possibility is to use elements whose size may be one of a restricted set of  The set::find is a built-in function in C++ STL which returns an iterator to the element which is searched in the set container. If the element is not found, then the iterator points to the position just after the last element in the set. Syntax: set_name.find(element) Parameters: The function accepts one mandatory parameter element which

Do I need to loop through each of the Foo * inside vector_of_foos or is there a simpler solution?

You do need to loop to find what you want, but you can use std::find_if or another "wrapped loop". This is more natural with lambdas in C++0x, but in C++03 I'd just use a regular for loop, possibly wrapped in your own function if you need to do this in more than one place.

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Instead of using std::find, use std::find_if and provide your own predicate. This of course relies in you being able to access the member that holds "x" in Foo.

struct FooBar
  FooBar(Foo* search) : _search(search){}
  bool operator(const Foo* ptr)
    return ptr->{access to member} == _search->{access to member};

  Foo* _search;

vector<Foo*>::iterator it = std::find_if(vec.begin(), vec.end(), FooBar(new Foo(new Bar("x")));

If you can't access the member and you can guarantee that all other members will be the same, you could try a bare memcmp in the above functor rather than "==".

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You may consider also using the Boost Ptr container library. It allows having a list of pointers using standard algorithms, find, etc. as if it contained objects, and automatically releasing the memory used by the pointers upon vector deletion.

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[PDF] Why you shouldn't use set (and what you should use instead) Matt , A set may not contain more than one element with the same key, so insert won't As with all STL containers, you can step through all of the elements in a set using words of storage (a color marker, and pointers to two children and a parent). After this, a for loop is used to dereference the pointer and print all the elements in the array. The pointer is incremented in each iteration of the loop i.e at each loop iteration, the pointer points to the next element of the array. Then that array value is printed. This can be seen in the following code snippet.

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  • Point of detail: std::vector doesn't have a find() member function; did you mean std::find(vector_of_foos.begin(), vector_of_foos.end(), s)?
  • std::vector doesn't have a find member. Are you really using a different container type, or the global function std::find()?
  • changed it in the text, thanks for seeing that. I have a set, looked at the wrong line.
  • It depends how carefully the OP is about calling delete at the relevant times. As long as he iterates through vector_of_foos and invokes delete on each element at some point, and so long as ~Foo() invokes delete on whichever member-variable pointer, everything's ok.
  • Memory leaks and alike is not my real concern. This is just me playing around with C++. This line would be the last of the code so it wouldn't matter. But you're right, in a non-practice example is would be more carefull with delete.
  • @Oli: The OP isn't adding the parameter of find_foo to the vector and this is indeed a leak in any reasonable interpretation of the (pseudo-)code given; it sounds like the example has been generalized too much.
  • @Roger: Ah, yes, I didn't see the find_foo()... I absolutely agree, in that case! However, this isn't an answer to the question, though.
  • The Foo here is definitely leaked. The created pointer is passed only to find_foo(). The definition of find_foo is given, and it does not delete its argument. This looks like the Java idiom, and I suspect the poster needs to learn to use new in C++ only when necessary.
  • This is one answer. Note that it changes the set to sort by a user-defined order instead of by address. There's an implicit assumption that there is never a NULL pointer in or passed to sFoo. And the Foo in the call to find() should not be newed. (Hard to say about the Bar without more info.)
  • This is an interessting approach. The problem is, that there is a dynamic size of how many elements Foo contains e.g. 2 Bar, 4 Bar, ... and I'm not sure, how to get the right match for multiple elements.
  • @DrColossos - -1 for a terribly-written question - your code and stated assumptions change by the minute. Nowhere but here do you mention Foo containing multiple Bars. How can people possibly give sensible answers?
  • You probably don't want your key element to be created on the heap, it's not getting deleted. I updated the example to use a key object on the stack.