Difference between erase and remove

Question

I am bit confused about the difference between the usage of std::remove algorithm. Specifically I am not able to understand what is being removed when I use this algorithm. I wrote a small test code like this:

std::vector<int> a; a.push_back(1); a.push_back(2);  std::remove(a.begin(), a.end(), 1);   int s = a.size();  std::vector<int>::iterator iter = a.begin(); std::vector<int>::iterator endIter = a.end();  std::cout<<"Using iter...\n"; for(; iter != endIter; ++iter) {     std::cout<<*iter<<"\n"; }  std::cout<<"Using size...\n"; for(int i = 0; i < a.size(); ++i) {     std::cout<<a[i]<<"\n"; } 

The output was 2,2 in both the cases.

However, if I use erase with the remove something like this:

a.erase(std::remove(a.begin(), a.end(), 1), a.end()); 

I get the output as 2.

So my questions are:

(1). Is there any use of std::remove other than using it with erase function.

(2). Even after doing std::remove, why a.size() returns 2 and not 1?

I read the item in Scott Meyer's Effective STL book about the erase-remove idiom. But am still having this confusion.

Answer 1

remove() doesn't actually delete elements from the container -- it only shunts non-deleted elements forwards on top of deleted elements. The key is to realise that remove() is designed to work on not just a container but on any arbitrary forward iterator pair: that means it can't actually delete the elements, because an arbitrary iterator pair doesn't necessarily have the ability to delete elements.

For example, pointers to the beginning and end of a regular C array are forward iterators and as such can be used with remove():

int foo[100];  ...  remove(foo, foo + 100, 42);    // Remove all elements equal to 42 

Here it's obvious that remove() cannot resize the array!