I have following code that gets and prints a string.
#include<iostream> #include<conio.h> #include<string> using namespace std; int main() { string str; cout << "Enter a string: "; getline(cin, str); cout << str; getch(); return 0; } But how to count the number of characters in this string using strlen() function?
For C++ strings, there's no reason to use strlen. Just use string::length:
std::cout << str.length() << std::endl; You should strongly prefer this to strlen(str.c_str()) for the following reasons:
Clarity: The length() (or size()) member functions unambiguously give back the length of the string. While it's possible to figure out what strlen(str.c_str()) does, it forces the reader to pause for a bit.
Efficiency: length() and size() run in time O(1), while strlen(str.c_str()) will take Θ(n) time to find the end of the string.
Style: It's good to prefer the C++ versions of functions to the C versions unless there's a specific reason to do so otherwise. This is why, for example, it's usually considered better to use std::sort over qsort or std::lower_bound over bsearch, unless some other factors come into play that would affect performance.
The only reason I could think of where strlen would be useful is if you had a C++-style string that had embedded null characters and you wanted to determine how many characters appeared before the first of them. (That's one way in which strlen differs from string::length; the former stops at a null terminator, and the latter counts all the characters in the string). But if that's the case, just use string::find:
size_t index = str.find(0); if (index == str::npos) index = str.length(); std::cout << index << std::endl; 
strlen on a C++ string by writing strlen(str.c_str()), but this would be extremely poor style. Although C++ absorbs C's standard library, the C++ standard libraries often include more C++-friendly versions of those functions, and there's no reason to use strlen when you could just be using string::length.strlen accepts const char*, and std::string isn't a const char*. That said - it is really not a good idea to use strlen here!size(), and I think historically string used length(). It's probably for backwards compatibility.
str[4] = '\0'?
Function strlen shows the number of character before \0 and using it for std::string may report wrong length.
strlen(str.c_str()); // It may return wrong length. In C++, a string can contain \0 within the characters but C-style-zero-terminated strings can not but at the end. If the std::string has a \0 before the last character then strlen reports a length less than the actual length.
Try to use .length() or .size(), I prefer second one since another standard containers have it.
str.size() 
.size was correct, too. I actually use it instead of length.
Use std::string::size or std::string::length (both are the same).
As you insist to use strlen, you can:
int size = strlen( str.c_str() ); note the usage of std::string::c_str, which returns const char*.
BUT strlen counts untill it hit \0 char and std::string can store such chars. In other words, strlen could sometimes lie for the size.
c_str()documentation from cppreference: Returns a pointer to a null-terminated character array with data equivalent to those stored in the string.
std::string can contain embedded null characters. If you have a std::string foo with the value "foo\0bar", it's .length() is 7, but strlen(foo.c_str()) will return 3.If you really, really want to use strlen(), then
cout << strlen(str.c_str()) << endl; else the use of .length() is more in keeping with C++.
Manually:
int strlen(string s) { int len = 0; while (s[len]) len++; return len; } 
#include<iostream> #include<conio.h> #include<string.h> using namespace std; int main() { char str[80]; int i; cout<<"\n enter string:"; cin.getline(str,80); int n=strlen(str); cout<<"\n lenght is:"<<n; getch(); return 0; } This is the program if you want to use strlen . Hope this helps!
Simply use
int len=str.length();
str.size()?mainmust returnint.strlenexactly?