I'm trying to remove all punctuation characters from a std::string in C++. My current code:
string str_in; string::size_type i, j; cout << "please input string with punctuation character..." << endl; cin >> str_in; for (i = 0, j = 0; i != str_in.size(); ++i) if (!ispunct(str_in[i])) str_in[j++] = str_in[i]; str_in[j] = '\0'; cout << str_in << endl; Is str_in[j] = '\0'; wrong?
If you want to truncate str_in to the first j characters, you can say str_in.resize(j).
If you want to use the standard library you could apply the erase-remove idiom like this:
#include <algorithm> #include <iostream> #include <string> int main() { std::string str_in; std::getline(std::cin, str_in); // Here is where the magic happens... str_in.erase(std::remove_if(str_in.begin(), str_in.end(), ::ispunct), str_in.end()); std::cout << str_in << '\n'; return 0; } 
::ispunct as the predicate results in undefined behavior, since you cannot safely call ::ispunct with a char. (The input to ::ispunct must in the range [0, UCHAR_MAX] or be EOF.)the C++ string type is NOT implemented to be null terminated (although a c_str() call will give you a null terminated string.)
So yes, str_in[j] = '\0' is wrong for at least two reasons:
str_in.length() will not reflect the size of the string you expect with the punctuation removed.cout << str_in;Using the std::string class you should probably not oveeride the same buffer, but probably use a str_out buffer instead which will have the right length after you copy all the wanted (i.e. excluding the punctuation character), OR you should instead adjust the length of the str_in instead of adding the null.
I think str_in[j] = '\0' is wrong when the string has no any punctuation.
Instead of modifying the same string, create a new string (e.g. str_out) and append to that:
str_out += str_in[i]; 
str_in.resize(j)
str_in[j] = '\0'is wrong since that's not an appropriate way to terminate astd::string. What is the question?std::strings are not defined to use a terminator. I think you're trying to delete all punctuation marks from astd::string, so"foo.bar+baz"becomes"foobarbaz". If so, that has nothing to do with termination.