CODEWITHSUNDEEP
c++algorithmlistfind
aretas_pau
aretas_pau

Reputation: 99

std::list and find(), if item/value is not found using find in list error appears

I am testing std::list with a find() function from . Before I implement it into my program, I want to see if they will work just as I think they do. At first it worked well: ` 

list<int> l;

l.push_back(1);
l.push_back(2);
l.push_back(9);

list<int>::iterator it = find(l.begin(), l.end(), 9);
if (*it == 9)
{
    cout << "Found";
}
else 
{
    cout << "Not Found";
}

In this example, output was "Found", that is right, but when I do it like this: 

list<int> l;
l.push_back(1);
l.push_back(2);
l.push_back(9);

list<int>::iterator it = find(l.begin(), l.end(), 3);

if (*it == 9)
{
    cout << "Found";
}
else 
{
    cout << "Not Found";
}

It doesn't output anything, but it should output "Not Found", i get a "Debug Assertion Failed!" error, it says that "Expression: list iterator not dereferencable", how i can fix this problem?

Upvotes: 0

Views: 1078

Answers (1)

juanchopanza
juanchopanza

Reputation: 227418

std::find returns the end() iterator when something is not found in a container. And de-rererencing the end() iterator as you do here:

if (*it == 9)

is undefined behaviour. To see if you found an element, check against the end iterator:

if (it != l.end())
{
   cout << "found\n";
}

Upvotes: 6

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