I have strings of the following format: "name.bag.csv" I would like to remove the ".bag" from the string. This is an example of the code that I am trying to run:
csv_file_name = "loololololool.bag.csv";
csv_file_name.erase(csv_file_name.end()-8, 4);
std::cout << csv_file_name << std::endl;
But I get an error on the second line:
no matching function for call to ‘std::__cxx11::basic_string<char>::erase(__gnu_cxx::__normal_iterator<char*, std::__cxx11::basic_string<char> >, int)’
csv_file_name.erase(csv_file_name.end()-8, 4);
It seems to only take one argument. However if I do:
csv_file_name = "loololololool.bag.csv";
csv_file_name.erase(13, 4);
std::cout << csv_file_name << std::endl;
it seems to work fine. Also when I execute
csv_file_name = "loololololool.bag.csv";
csv_file_name.erase(csv_file_name.end()-8);
std::cout << csv_file_name << std::endl;
It deletes the single character as it should.
How can this happen? The csv_file_name.end()-8 must be working as it deletes the single character. And taking two arguments should be working as well. But the combination doesn't? Please help!
You are mixing up iterators and indexes.
string.erase(iter) // delete the character pointed to by the iterator
string.erase(iter1, iter2) // delete the characters between the two iterators
string.erase(index, count) // delete count characters starting at index
csv_file_name.end()-8 is an iterator and there is no version of erase that takes an iterator and an index or count.
Instead of
csv_file_name.erase(csv_file_name.end()-8, 4);
I think you meant
csv_file_name.erase(csv_file_name.size()-8, 4);
But really you should be able to read documentation and figure this stuff out for yourself. You'll be a lot more efficient that way.