In C one can declare a char array, create a pointer to a specific location in the array, and dereference it to get a suffix in the specified position:
char str[6] = "SLEEP"; char* pointer = &(str[1]); printf("%s\n", pointer); The above code will print
LEEP which is the suffix of "SLEEP" in position 1. Now is there a way to do something similar in C++ without using the substr method as it produces a completely new string and I'm trying to avoid that.
You can do exactly what you've already done.
char str[6] = "SLEEP"; char* pointer = &(str[1]); printf("%s\n", pointer); If you are using a std::string instead of a raw char buffer (like you should be), then under C++11 std::string is guaranteed to have contiguous storage (21.4.1/5) with an appended NUL terminator (21.4.7.1/1):
std::string str = "SLEEP"; const char* pointer = &str[1]; These guarantees are new to C++11 -- C++03 makes no such guarantees. However all implementations I'm aware of do in fact use contigious storage with an appended NUL terminator. They do this because c_str() is required to return a const pointer to a C-style string. If you want a solution that is guaranteed to be compliant in C++03, then std::vector makes the same contiguity guarantee, even in C++03, but of course there you must apply the NUL terminator on your own:
std::string load = "SLEEP"; vector <char> str; copy (load.begin(), load.end(), back_inserter (str)); load.push_back ('\0'); // Appended NUL terminator const char* pointer = &str [1]; But now we're talking about making copies, obviously.
"SLEEP"[1] == 'L', or "SLEEP"[1] == *("SLEEP" + 1) so for address of "SLEEP"[1] I did, "SLEEP" + 1.
#include <string> #include <iostream> int main() { std::string s("SLEEP"); std::cout << &s[1] << std::endl; } This is guaranteed to work in C++11. In C++03, no such guarantees exist, since implementations are allowed to use copy on write (COW). But I don't recall working with an implementation where this didn't work. If you use C++03 and need the guarantee, you will can use the std::string::c_str():
std::cout << &s.c_str()[1] << std::endl; 
std::string, not a string literal. I interpret the question to be about std::string, but I could be mistaken.std::string here, so what you are showing is not equivalent.you can use the array method if you want. The example I give is C++.
for example MyStr[0] is the pointer to the first element.
#include <string> #include <iostream> int main() { std::string MyStr="abc"; std::cout<<&MyStr[1]; } output: bc
Hope that helps :-)
You can use a string reference class. A string reference class provides string-like functionality but doesn't actually own the underlying storage.
Here's an example using the Boost string_ref class (found in Boost.Utility)
#include <iostream> #include <string> #include <boost/utility/string_ref.hpp> int main(int argc, const char * argv[]) { using namespace std; string str = "SLEEP"; cout << "str: " << str << endl; boost::string_ref sr1(str); cout << "sr1: " << sr1 << endl; boost::string_ref sr2 = sr1.substr(1); cout << "sr2: " << sr2 << endl; return 0; } Note: there in only one actual instance of the string "SLEEP". sr1 and sr2 are string references into str.
