Reputation: 47640
Does it works correctly (does nothing) when I use
vector<T> v;
v.erase(v.end());
I want to use something like
v.erase(std::find(...));
Should I if
is it v.end()
or not?
There is no info about it on C++.com and CPPreference
Upvotes: 50
Views: 27089
Reputation: 89364
Since C++ 17, if statements can have an init statement, so erasing an element that may not be present can be simplified. As a bonus, the variable used for the iterator will not be visible outside the scope of the if statement.
if (auto it = std::find(v.begin(), v.end(), value); it != v.end()) v.erase(it);
In addition, std::ranges::find
(or std::ranges::find_if
) can be used since C++ 20, which don't require passing in begin()
and end()
iterators.
if (auto it = std::ranges::find(v, value); it != v.end()) v.erase(it);
The above solutions remove the first occurrence of value
if it exists. To remove all occurrences of a value, std::erase
from <vector>
may be used since C++ 20.
std::erase(v, value);
Upvotes: 0
Reputation: 4367
Have you tried this?
v.erase(remove_if(v.begin(), v.end(), (<your criteria>)), v.end());
Upvotes: 7
Reputation: 279305
The standard doesn't quite spell it out, but v.erase(q)
is defined, "Erases the element pointed to by q
" in [sequence.reqmts]
. This means that q
must actually point to an element, which the end iterator doesn't. Passing in the end iterator is undefined behavior.
Unfortunately, you need to write:
auto it = std::find(...);
if (it != <the part of ... that specifies the end of the range searched>) {
v.erase(it);
}
Of course, you could define:
template typename<Sequence, Iterator>
Iterator my_erase(Sequence &s, Iterator it) {
if (it == s.end()) return it;
return s.erase(it);
}
my_erase(v, std::find(v.begin(), v.end(), whatever));
c.erase()
on an associative container returns void
, so to generalize this template to all containers you need some -> decltype
action.
Upvotes: 45
Reputation: 106579
Erasing end()
(or for that matter, even looking at the target of end()
) is undefined behavior. Undefined behavior is allowed to have any behavior, including "just work" on your platform. That doesn't mean that you should be doing it; it's still undefined behavior, and I'll come bite you in the worst ways when you're least expecting it later on.
Depending on what you're doing, you might want to consider set
or unordered_set
instead of vector
here.
Upvotes: 31