std::string to char*

C++StringChar

C++ Problem Overview


I want to convert a std::string into a char* or char[] data type.

std::string str = "string";
char* chr = str;

Results in: “error: cannot convert ‘std::string’ to ‘char’ ...”.

What methods are there available to do this?

C++ Solutions


Solution 1 - C++

It won't automatically convert (thank god). You'll have to use the method c_str() to get the C string version.

std::string str = "string";
const char *cstr = str.c_str();

Note that it returns a const char *; you aren't allowed to change the C-style string returned by c_str(). If you want to process it you'll have to copy it first:

std::string str = "string";
char *cstr = new char[str.length() + 1];
strcpy(cstr, str.c_str());
// do stuff
delete [] cstr;

Or in modern C++:

std::vector<char> cstr(str.c_str(), str.c_str() + str.size() + 1);

Solution 2 - C++

More details here, and here but you can use

string str = "some string" ;
char *cstr = &str[0];

As of C++11, you can also use the str.data() member function, which returns char *

string str = "some string" ;
char *cstr = str.data();

Solution 3 - C++

If I'd need a mutable raw copy of a c++'s string contents, then I'd do this:

std::string str = "string";
char* chr = strdup(str.c_str());

and later:

free(chr); 

So why don't I fiddle with std::vector or new[] like anyone else? Because when I need a mutable C-style raw char* string, then because I want to call C code which changes the string and C code deallocates stuff with free() and allocates with malloc() (strdup uses malloc). So if I pass my raw string to some function X written in C it might have a constraint on it's argument that it has to allocated on the heap (for example if the function might want to call realloc on the parameter). But it is highly unlikely that it would expect an argument allocated with (some user-redefined) new[]!

Solution 4 - C++

(This answer applies to C++98 only.)

Please, don't use a raw char*.

std::string str = "string";
std::vector<char> chars(str.c_str(), str.c_str() + str.size() + 1u);
// use &chars[0] as a char*

Solution 5 - C++

  • If you just want a C-style string representing the same content:

      char const* ca = str.c_str();
    
  • If you want a C-style string with new contents, one way (given that you don't know the string size at compile-time) is dynamic allocation:

      char* ca = new char[str.size()+1];
      std::copy(str.begin(), str.end(), ca);
      ca[str.size()] = '\0';
    

    Don't forget to delete[] it later.

  • If you want a statically-allocated, limited-length array instead:

      size_t const MAX = 80; // maximum number of chars
      char ca[MAX] = {};
      std::copy(str.begin(), (str.size() >= MAX ? str.begin() + MAX : str.end()), ca);
    

std::string doesn't implicitly convert to these types for the simple reason that needing to do this is usually a design smell. Make sure that you really need it.

If you definitely need a char*, the best way is probably:

vector<char> v(str.begin(), str.end());
char* ca = &v[0]; // pointer to start of vector

Solution 6 - C++

This would be better as a comment on bobobobo's answer, but I don't have the rep for that. It accomplishes the same thing but with better practices.

Although the other answers are useful, if you ever need to convert std::string to char* explicitly without const, const_cast is your friend.

std::string str = "string";
char* chr = const_cast<char*>(str.c_str());

Note that this will not give you a copy of the data; it will give you a pointer to the string. Thus, if you modify an element of chr, you'll modify str.

Solution 7 - C++

Assuming you just need a C-style string to pass as input:

std::string str = "string";
const char* chr = str.c_str();

Solution 8 - C++

To obtain a const char * from an std::string use the c_str() member function :

std::string str = "string";
const char* chr = str.c_str();

To obtain a non-const char * from an std::string you can use the data() member function which returns a non-const pointer since C++17 :

std::string str = "string";
char* chr = str.data();

For older versions of the language, you can use range construction to copy the string into a vector from which a non-const pointer can be obtained :

std::string str = "string";
std::vector<char> str_copy(str.c_str(), str.c_str() + str.size() + 1);
char* chr = str_copy.data();

But beware that this won't let you modify the string contained in str, only the copy's data can be changed this way. Note that it's specially important in older versions of the language to use c_str() here because back then std::string wasn't guaranteed to be null terminated until c_str() was called.

Solution 9 - C++

To be strictly pedantic, you cannot "convert a std::string into a char* or char[] data type."

As the other answers have shown, you can copy the content of the std::string to a char array, or make a const char* to the content of the std::string so that you can access it in a "C style".

If you're trying to change the content of the std::string, the std::string type has all of the methods to do anything you could possibly need to do to it.

If you're trying to pass it to some function which takes a char*, there's std::string::c_str().

Solution 10 - C++

Here is one more robust version from Protocol Buffer

char* string_as_array(string* str)
{
	return str->empty() ? NULL : &*str->begin();
}

// test codes
std::string mystr("you are here");
char* pstr = string_as_array(&mystr);
cout << pstr << endl; // you are here

Solution 11 - C++

Conversion in OOP style

converter.hpp

class StringConverter {
    public: static char * strToChar(std::string str);
};

converter.cpp

char * StringConverter::strToChar(std::string str)
{
    return (char*)str.c_str();
}

usage

StringConverter::strToChar("converted string")

Solution 12 - C++

For completeness' sake, don't forget std::string::copy().

std::string str = "string";
const size_t MAX = 80;
char chrs[MAX];

str.copy(chrs, MAX);

std::string::copy() doesn't NUL terminate. If you need to ensure a NUL terminator for use in C string functions:

std::string str = "string";
const size_t MAX = 80;
char chrs[MAX];

memset(chrs, '\0', MAX);
str.copy(chrs, MAX-1);

Solution 13 - C++

You can make it using iterator.

std::string str = "string";
std::string::iterator p=str.begin();
char* chr = &(*p);

Good luck.

Solution 14 - C++

A safe version of orlp's char* answer using unique_ptr:

std::string str = "string";
auto cstr = std::make_unique<char[]>(str.length() + 1);
strcpy(cstr.get(), str.c_str());

Solution 15 - C++

char* result = strcpy((char*)malloc(str.length()+1), str.c_str());

Solution 16 - C++

Alternatively , you can use vectors to get a writable char* as demonstrated below;

//this handles memory manipulations and is more convenient
string str;
vector <char> writable (str.begin (), str.end) ;
writable .push_back ('\0'); 
char* cstring = &writable[0] //or &*writable.begin () 

//Goodluck  

Solution 17 - C++

This will also work

std::string s;
std::cout<<"Enter the String";
std::getline(std::cin, s);
char *a=new char[s.size()+1];
a[s.size()]=0;
memcpy(a,s.c_str(),s.size());
std::cout<<a;  

Solution 18 - C++

No body ever mentioned sprintf?

std::string s;
char * c;
sprintf(c, "%s", s.c_str());

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
Questionuser1598585View Question on Stackoverflow
Solution 1 - C++orlpView Answer on Stackoverflow
Solution 2 - C++boboboboView Answer on Stackoverflow
Solution 3 - C++Nordic MainframeView Answer on Stackoverflow
Solution 4 - C++ildjarnView Answer on Stackoverflow
Solution 5 - C++Lightness Races in OrbitView Answer on Stackoverflow
Solution 6 - C++hairlessbearView Answer on Stackoverflow
Solution 7 - C++Mark BView Answer on Stackoverflow
Solution 8 - C++François AndrieuxView Answer on Stackoverflow
Solution 9 - C++Rob KView Answer on Stackoverflow
Solution 10 - C++zangwView Answer on Stackoverflow
Solution 11 - C++Patryk MerchelskiView Answer on Stackoverflow
Solution 12 - C++Jeffery ThomasView Answer on Stackoverflow
Solution 13 - C++TS.PARKView Answer on Stackoverflow
Solution 14 - C++EnhexView Answer on Stackoverflow
Solution 15 - C++cegprakashView Answer on Stackoverflow
Solution 16 - C++user5616653View Answer on Stackoverflow
Solution 17 - C++Genocide_HoaxView Answer on Stackoverflow
Solution 18 - C++Felix FengView Answer on Stackoverflow