How can I convert an std::string to a char* or a const char*?

Solution 1

If you just want to pass a std::string to a function that needs const char* you can use

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

If you want to get a writable copy, like char *, you can do that with this:

std::string str;
char * writable = new char[str.size() + 1];
std::copy(str.begin(), str.end(), writable);
writable[str.size()] = '\0'; // don't forget the terminating 0

// don't forget to free the string after finished using it
delete[] writable;

Edit: Notice that the above is not exception safe. If anything between the new call and the delete call throws, you will leak memory, as nothing will call delete for you automatically. There are two immediate ways to solve this.


boost::scoped_array will delete the memory for you upon going out of scope:

std::string str;
boost::scoped_array<char> writable(new char[str.size() + 1]);
std::copy(str.begin(), str.end(), writable.get());
writable[str.size()] = '\0'; // don't forget the terminating 0

// get the char* using writable.get()

// memory is automatically freed if the smart pointer goes 
// out of scope


This is the standard way (does not require any external library). You use std::vector, which completely manages the memory for you.

std::string str;
std::vector<char> writable(str.begin(), str.end());

// get the char* using &writable[0] or &*writable.begin()

Solution 2

Given say...

std::string x = "hello";

Getting a `char *` or `const char*` from a `string`

How to get a character pointer that's valid while x remains in scope and isn't modified further

C++11 simplifies things; the following all give access to the same internal string buffer:

const char* p_c_str = x.c_str();
const char* p_data  =;
char* p_writable_data =; // for non-const x from C++17 
const char* p_x0    = &x[0];

      char* p_x0_rw = &x[0];  // compiles iff x is not const...

All the above pointers will hold the same value - the address of the first character in the buffer. Even an empty string has a "first character in the buffer", because C++11 guarantees to always keep an extra NUL/0 terminator character after the explicitly assigned string content (e.g. std::string("this\0that", 9) will have a buffer holding "this\0that\0").

Given any of the above pointers:

char c = p[n];   // valid for n <= x.size()
                 // i.e. you can safely read the NUL at p[x.size()]

Only for the non-const pointer p_writable_data and from &x[0]:

p_writable_data[n] = c;
p_x0_rw[n] = c;  // valid for n <= x.size() - 1
                 // i.e. don't overwrite the implementation maintained NUL

Writing a NUL elsewhere in the string does not change the string's size(); string's are allowed to contain any number of NULs - they are given no special treatment by std::string (same in C++03).

In C++03, things were considerably more complicated (key differences highlighted):


    • returns const char* to the string's internal buffer which wasn't required by the Standard to conclude with a NUL (i.e. might be ['h', 'e', 'l', 'l', 'o'] followed by uninitialised or garbage values, with accidental accesses thereto having undefined behaviour).
      • x.size() characters are safe to read, i.e. x[0] through x[x.size() - 1]
      • for empty strings, you're guaranteed some non-NULL pointer to which 0 can be safely added (hurray!), but you shouldn't dereference that pointer.
  • &x[0]

    • for empty strings this has undefined behaviour (21.3.4)
      • e.g. given f(const char* p, size_t n) { if (n == 0) return; ...whatever... } you mustn't call f(&x[0], x.size()); when x.empty() - just use f(, ...).
    • otherwise, as per but:
      • for non-const x this yields a non-const char* pointer; you can overwrite string content
  • x.c_str()

    • returns const char* to an ASCIIZ (NUL-terminated) representation of the value (i.e. ['h', 'e', 'l', 'l', 'o', '\0']).
    • although few if any implementations chose to do so, the C++03 Standard was worded to allow the string implementation the freedom to create a distinct NUL-terminated buffer on the fly, from the potentially non-NUL terminated buffer "exposed" by and &x[0]
    • x.size() + 1 characters are safe to read.
    • guaranteed safe even for empty strings (['\0']).

Consequences of accessing outside legal indices

Whichever way you get a pointer, you must not access memory further along from the pointer than the characters guaranteed present in the descriptions above. Attempts to do so have undefined behaviour, with a very real chance of application crashes and garbage results even for reads, and additionally wholesale data, stack corruption and/or security vulnerabilities for writes.

When do those pointers get invalidated?

If you call some string member function that modifies the string or reserves further capacity, any pointer values returned beforehand by any of the above methods are invalidated. You can use those methods again to get another pointer. (The rules are the same as for iterators into strings).

See also How to get a character pointer valid even after x leaves scope or is modified further below....

So, which is better to use?

From C++11, use .c_str() for ASCIIZ data, and .data() for "binary" data (explained further below).

In C++03, use .c_str() unless certain that .data() is adequate, and prefer .data() over &x[0] as it's safe for empty strings....

...try to understand the program enough to use data() when appropriate, or you'll probably make other mistakes...

The ASCII NUL '\0' character guaranteed by .c_str() is used by many functions as a sentinel value denoting the end of relevant and safe-to-access data. This applies to both C++-only functions like say fstream::fstream(const char* filename, ...) and shared-with-C functions like strchr(), and printf().

Given C++03's .c_str()'s guarantees about the returned buffer are a super-set of .data()'s, you can always safely use .c_str(), but people sometimes don't because:

  • using .data() communicates to other programmers reading the source code that the data is not ASCIIZ (rather, you're using the string to store a block of data (which sometimes isn't even really textual)), or that you're passing it to another function that treats it as a block of "binary" data. This can be a crucial insight in ensuring that other programmers' code changes continue to handle the data properly.
  • C++03 only: there's a slight chance that your string implementation will need to do some extra memory allocation and/or data copying in order to prepare the NUL terminated buffer

As a further hint, if a function's parameters require the (const) char* but don't insist on getting x.size(), the function probably needs an ASCIIZ input, so .c_str() is a good choice (the function needs to know where the text terminates somehow, so if it's not a separate parameter it can only be a convention like a length-prefix or sentinel or some fixed expected length).

How to get a character pointer valid even after x leaves scope or is modified further

You'll need to copy the contents of the string x to a new memory area outside x. This external buffer could be in many places such as another string or character array variable, it may or may not have a different lifetime than x due to being in a different scope (e.g. namespace, global, static, heap, shared memory, memory mapped file).

To copy the text from std::string x into an independent character array:

std::string old_x = x;
// - old_x will not be affected by subsequent modifications to x...
// - you can use `&old_x[0]` to get a writable char* to old_x's textual content
// - you can use resize() to reduce/expand the string
//   - resizing isn't possible from within a function passed only the char* address

std::string old_x = x.c_str(); // old_x will terminate early if x embeds NUL
// Copies ASCIIZ data but could be less efficient as it needs to scan memory to
// find the NUL terminator indicating string length before allocating that amount
// of memory to copy into, or more efficient if it ends up allocating/copying a
// lot less content.
// Example, x == "ab\0cd" -> old_x == "ab".

std::vector<char> old_x(, + x.size());       // without the NUL
std::vector<char> old_x(x.c_str(), x.c_str() + x.size() + 1);  // with the NUL

// (a bit dangerous, as "known" things are sometimes wrong and often become wrong)
char y[N + 1];
strcpy(y, x.c_str());

char y[N + 1];
strncpy(y, x.c_str(), N);  // copy at most N, zero-padding if shorter
y[N] = '\0';               // ensure NUL terminated

char* y = alloca(x.size() + 1);
strcpy(y, x.c_str());

char y[x.size() + 1];
strcpy(y, x.c_str());

char* y = new char[x.size() + 1];
strcpy(y, x.c_str());
//     or as a one-liner: char* y = strcpy(new char[x.size() + 1], x.c_str());
// use y...
delete[] y; // make sure no break, return, throw or branching bypasses this

// see boost shared_array usage in Johannes Schaub's answer

char* y = strdup(x.c_str());
// use y...

Other reasons to want a char* or const char* generated from a string

So, above you've seen how to get a (const) char*, and how to make a copy of the text independent of the original string, but what can you do with it? A random smattering of examples...

  • give "C" code access to the C++ string's text, as in printf("x is '%s'", x.c_str());
  • copy x's text to a buffer specified by your function's caller (e.g. strncpy(callers_buffer, callers_buffer_size, x.c_str())), or volatile memory used for device I/O (e.g. for (const char* p = x.c_str(); *p; ++p) *p_device = *p;)
  • append x's text to an character array already containing some ASCIIZ text (e.g. strcat(other_buffer, x.c_str())) - be careful not to overrun the buffer (in many situations you may need to use strncat)
  • return a const char* or char* from a function (perhaps for historical reasons - client's using your existing API - or for C compatibility you don't want to return a std::string, but do want to copy your string's data somewhere for the caller)
    • be careful not to return a pointer that may be dereferenced by the caller after a local string variable to which that pointer pointed has left scope
    • some projects with shared objects compiled/linked for different std::string implementations (e.g. STLport and compiler-native) may pass data as ASCIIZ to avoid conflicts

Solution 3

Use the .c_str() method for const char *.

You can use &mystring[0] to get a char * pointer, but there are a couple of gotcha's: you won't necessarily get a zero terminated string, and you won't be able to change the string's size. You especially have to be careful not to add characters past the end of the string or you'll get a buffer overrun (and probable crash).

There was no guarantee that all of the characters would be part of the same contiguous buffer until C++11, but in practice all known implementations of std::string worked that way anyway; see Does &s[0] point to contiguous characters in a std::string?.

Note that many string member functions will reallocate the internal buffer and invalidate any pointers you might have saved. Best to use them immediately and then discard.

Solution 4


C++17 (upcoming standard) changes the synopsis of the template basic_string adding a non const overload of data():

charT* data() noexcept;

Returns: A pointer p such that p + i == &operator for each i in [0,size()].

CharT const * from std::basic_string<CharT>

std::string const cstr = { "..." };
char const * p =; // or .c_str()

CharT * from std::basic_string<CharT>

std::string str = { "..." };
char * p =;


CharT const * from std::basic_string<CharT>

std::string str = { "..." };

CharT * from std::basic_string<CharT>

From C++11 onwards, the standard says:

  1. The char-like objects in a basic_string object shall be stored contiguously. That is, for any basic_string object s, the identity &*(s.begin() + n) == &*s.begin() + n shall hold for all values of n such that 0 <= n < s.size().

  1. const_reference operator[](size_type pos) const;
    reference operator[](size_type pos);

    Returns: *(begin() + pos) if pos < size(), otherwise a reference to an object of type CharT with value CharT(); the referenced value shall not be modified.

  1. const charT* c_str() const noexcept;
    const charT* data() const noexcept;

    Returns: A pointer p such that p + i == &operator[](i) for each i in [0,size()].

There are severable possible ways to get a non const character pointer.

1. Use the contiguous storage of C++11

std::string foo{"text"};
auto p = &*foo.begin();


  • Simple and short
  • Fast (only method with no copy involved)


  • Final '\0' is not to be altered / not necessarily part of the non-const memory.

2. Use std::vector<CharT>

std::string foo{"text"};
std::vector<char> fcv(,;
auto p =;


  • Simple
  • Automatic memory handling
  • Dynamic


  • Requires string copy

3. Use std::array<CharT, N> if N is compile time constant (and small enough)

std::string foo{"text"};
std::array<char, 5u> fca;
std::copy(,, fca.begin());


  • Simple
  • Stack memory handling


  • Static
  • Requires string copy

4. Raw memory allocation with automatic storage deletion

std::string foo{ "text" };
auto p = std::make_unique<char[]>(foo.size()+1u);
std::copy(, + foo.size() + 1u, &p[0]);


  • Small memory footprint
  • Automatic deletion
  • Simple


  • Requires string copy
  • Static (dynamic usage requires lots more code)
  • Less features than vector or array

5. Raw memory allocation with manual handling

std::string foo{ "text" };
char * p = nullptr;
  p = new char[foo.size() + 1u];
  std::copy(, + foo.size() + 1u, p);
  // handle stuff with p
  delete[] p;
catch (...)
  if (p) { delete[] p; }


  • Maximum 'control'


  • Requires string copy
  • Maximum liability / susceptibility for errors
  • Complex

Solution 5

Just see this:

string str1("stackoverflow");
const char * str2 = str1.c_str();

However, note that this will return a const char *.

For a char *, use strcpy to copy it into another char array.

Solution 6

I am working with an API with a lot of functions that get a char* as an input.

I have created a small class to face this kind of problem, and I have implemented the RAII idiom.

class DeepString
        DeepString(const DeepString& other);
        DeepString& operator=(const DeepString& other);
        char* internal_; 
        explicit DeepString( const string& toCopy): 
            internal_(new char[toCopy.size()+1]) 
        ~DeepString() { delete[] internal_; }
        char* str() const { return internal_; }
        const char* c_str()  const { return internal_; }

And you can use it as:

void aFunctionAPI(char* input);

//  other stuff

aFunctionAPI("Foo"); //this call is not safe. if the function modified the 
                     //literal string the program will crash
std::string myFoo("Foo");
aFunctionAPI(myFoo.c_str()); //this is not compiling
aFunctionAPI(const_cast<char*>(myFoo.c_str())); //this is not safe std::string 
                                                //implement reference counting and 
                                                //it may change the value of other
                                                //strings as well.
DeepString myDeepFoo(myFoo);
aFunctionAPI(myFoo.str()); //this is fine

I have called the class DeepString because it is creating a deep and unique copy (the DeepString is not copyable) of an existing string.

Solution 7

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

Solution 8

Converting from c++ std string to C style string is really easy now.

For that we have string::copy function which will easily convert std string to C style string. reference

string::copy functions parameters serially

  1. char string pointer
  2. string size, how many characters will b copied
  3. position, from where character copy will start

Another important thing,

This function does not append a null character at the end of operation. So, we need to put it manually.

Code exam are in below -

// char string
char chText[20];

// c++ string
string text =  "I am a Programmer";

// conversion from c++ string to char string
// this function does not append a null character at the end of operation
text.copy(chText, text.size(), 0);

// we need to put it manually
chText[text.size()] = '\0';

// below statement prints "I am a Programmer"
cout << chText << endl;

Vice Versa, Converting from C style string to C++ std string is lot more easier

There is three ways we can convert from C style string to C++ std string

First one is using constructor,

char chText[20] = "I am a Programmer";
// using constructor
string text(chText);

Second one is using string::assign method

// char string
char chText[20] = "I am a Programmer";

// c++ string
string text;

// convertion from char string to c++ string
// using assign function

Third one is assignment operator(=), in which string class uses operator overloading

// char string
char chText[20] = "I am a Programmer";

// c++ string
// convertion from char string to c++ string using assignment operator overloading
string text = chText;

third one can be also write like below -

// char string
char chText[20] = "I am a Programmer";

// c++ string
string text;

// convertion from char string to c++ string
text = chText;

Solution 9

let's say, string str="stack";

1)converting string to char*

  char* s_rw=&str[0]; 

The above char*(i.e., s_rw) is readable and writeable and points to the base address of the string which needs to be converted to char*

2)Converting string to const char*

   const char* s_r=&str[0];

The above const char* (i.e., s_r) is readable but not writeable and points to the base address of the string.

Solution 10

This is especially useful when passing the underlying char* buffer of a std::string to C calls which expect and write to a char* buffer. This way you get the best of both worlds!: the niceties of the C++ std::string and the usability of it directly with C libraries you are calling from C++.

How to use a modern C++ std::string as a C-style read/writable char* or read-only null-terminated const char*

How can I convert a std::string to a char* or a const char*?

Despite being a really old and highly-upvoted answer, the information I'm about to cover isn't already well-covered, if covered at all, so this is a necessary addition, in particular the part about needing to pre-allocate the underlying C-string using the .resize() method if you'd like to use it as a writable buffer.

All of the usages below require C++11 or later, except for the char* data() call, which requires C++17 or later.

To run and test all example code below, and more, see and run my string__use_std_string_as_a_c_str_buffer.cpp file in my eRCaGuy_hello_world repo.

1. Use a std::string as a read/writable char*

To use a C++ std::string as a C-style writable char* buffer, you MUST first pre-allocate the string's internal buffer to change its .size() by using .resize(). Note that using .reserve() to increase only the .capacity() is NOT sufficient! The community wiki page for std::string::operator[] correctly states:

If pos > size(), the behavior is undefined.

The resize() method is what changes the size, not the reserve() method, which changes only the capacity().


#include <cstring>  // `strcpy()`
#include <iostream>
#include <string>

constexpr size_t BUFFER_SIZE = 100;

std::string str;
str.resize(BUFFER_SIZE);  // pre-allocate the underlying buffer
// check the size
std::cout << "str.size() = " << str.size() << "\n";

For all examples below, assume you have these C-strings:

constexpr char cstr1[] = "abcde ";
constexpr char cstr2[] = "fghijk";

Once you have pre-allocated an underlying buffer which is sufficiently large with resize(), you can then access the underlying buffer as a char* in at least 3 ways:

  1. Technique 1 [best option if using C++11]: array indexing using operator[] to obtain a char, followed by obtaining its address with &. Ex:
    char* c_str;
    c_str = &str[0];
    c_str = &str[5]; 
    // etc.
    // Write these 2 C-strings into a `std::string`'s underlying buffer
    strcpy(&str[0], cstr1);
    strcpy(&str[sizeof(cstr1) - 1], cstr2); // `- 1` to overwrite the first 
                                            // null terminator
    // print the string
    std::cout << str << "\n"; // output: `abcde fghijk`
    What if you have a pointer to a std::string? If you have a ptr to a std::string, it must be dereferenced first with *pstr before you can index into it as an array with the operator[] as &(*pstr)[0], so the syntax above becomes a little more awkward. Here is a full example:
    std::string str2;
    std::string* pstr = &str2;
    c_str = &(*pstr)[0]; // <=== dereference the ptr 1st before indexing into it
    // Or, to make the order of precedence 
    // ( really
    // obvious, you can optionally add extra parenthesis like this:
    c_str = &((*pstr)[0]);
  2. Technique 2 [best option if using C++17]: use the .data() method to obtain a char* directly. Ex:
    char* c_str;
    c_str =;
    c_str = + 5;
    // etc.
    // Write these 2 C-strings into the `std::string`'s underlying buffer
    strcpy(, cstr1);
    strcpy( + (sizeof(cstr1) - 1), cstr2); // `- 1` to overwrite the
                                                     // first null terminator
    // print the string
    std::cout << str << "\n"; // output: `abcde fghijk`
  3. Technique 3 [fine in C++11, but is awkward]: use the .begin() method to obtain an iterator to the first char, and then use the iterator's operator*() dereference method to obtain the iterator's char value_type, and then take the address of that to obtain a char*. Ex:
    char* c_str;
    c_str = &(*str.begin());
    c_str = &(*str.begin()) + 5;
    // etc.
    // Write these 2 C-strings into the `std::string`'s underlying buffer
    strcpy(&(*str.begin()), cstr1);
    strcpy(&(*str.begin()) + (sizeof(cstr1) - 1), cstr2); // `- 1` to overwrite 
                                                          // the first null 
                                                          // terminator
    // print the string
    std::cout << str << "\n"; // output: `abcde fghijk`

Something important to be aware of is that when you call str.resize(100), it reserves at least 100 bytes for the underlying string, sets the size() of the string to 100, and initializes all 100 of those chars to char()--AKA: the default value initialization value for char (see my question here), which is the binary zero null-terminator, '\0'. Therefore, whenever you call str.size() it will return 100 even if the string simply has "hello" in it followed by 95 null-terminators, or zeros. To get the length, or number of non-null-terminators in the string, you'll have to resort to the C function strlen(), like this:

std::cout << strlen(str.c_str()) << "\n"; // prints `12` in the examples above

// instead of:
std::cout << str.size() << "\n"; // prints `100` in the examples above

2. Access a std::string as a read-only, null-terminated const char*

To obtain a readable null-terminated const char* from a std::string, use the .c_str() method. It returns a C-style string that is guaranteed to be null-terminated. Note that the .data() method is NOT the same thing, as it is NOT guaranteed to be null-terminated!


std::string str = "hello world";
printf("%s\n", str.c_str());


  1. (questions on Stack Overflow)

    1. How to convert a std::string to const char* or char*: How to convert a std::string to const char* or char*
    2. Directly write into char* buffer of std::string: Directly write into char* buffer of std::string
    3. Is there a way to get std:string's buffer: Is there a way to get std:string's buffer
  2. (my content)

    1. [my test code] string__use_std_string_as_a_c_str_buffer.cpp
    2. [my Q] See the "Adjacently related" section at the bottom of my question here: What is a call to `char()` as a function in C++?
    3. *****+ [my comments about pre-allocating a buffer in the std::string]: Directly write into char* buffer of std::string
    4. *****+ [my comment on how to pre-allocate storage in a std::string, to be used as a char* buffer] Is there a way to get std:string's buffer
  3. (from the community wiki)


      The elements of a basic_string are stored contiguously, that is, for a basic_string s, &*(s.begin () + n) == &*s.begin() + n for any n in [0, s.size()), or, equivalently, a pointer to s[0] can be passed to functions that expect a pointer to the first element of a null-terminated (since C++11)CharT[] array.


      Returns a reference to the character at specified location pos. No bounds checking is performed. If pos > size(), the behavior is undefined.






Solution 11

Try this

std::string s(reinterpret_cast<const char *>(Data), Size);