Reputation: 1104

How to convert string to char * in C++?

How can I convert a string in to char * I was using the following method, but it doesn't work.

At runtime it gives me the following error:

Run-Time Check Failure #3 - The variable 'url' is being used without being initialized.

eventhough I have initialized it as shown in the code bellow. Can you please show me with an example?

    char* url;
    sup = "news"
    sup="http://www."+sup+"yahoo.com";
    strcpy(url, sup.c_str());

I am using Microsoft Visual Studio 2010, C++ in console

Upvotes: 3

Answers (7)

Nitish

Reputation: 31

In my opinion there are two ways of doing this.

Using &-operator (ampersand)

string str = "hello";
char *p;
p = &str[0];

Using c_str() function

string s = "hello";
const char *p; 
p = s.c_str();

I have tested both and they both are working. Kindly correct me if I am wrong.

Upvotes: 3

Luchian Grigore

Reputation: 258548

strcpy doesn't allocate memory for you, you must do it yourself, remembering to leave space for the null termination character:

char* url = new char[sup.length()+1];

char url[sup.length()+1];
//...
strcpy(url, sup.c_str());

In the first case, don't forget to delete[] the array. The second case will only work if your compiler supports C99 variable-length arrays as an extension.

Upvotes: 5

vivek

Reputation: 5219

std::string sup = "news";
sup="http://www."+sup+"yahoo.com";
char* url = new char[sup.length()+1];
url = const_cast<char*>(sup.c_str());
url[sup.length()] = '\0';

std::cout<<url;            //http://www.newsyahoo.com

Notice the '\0' at the end of url.

Upvotes: 0

CashCow

Reputation: 31425

If you really need a writable buffer of char that is copied from the contents of a std::string, you can use std::vector<char>

std::vector<char> urlAsVec( sup.begin(), sup.end() );
urlAsVec.push_back( '\0' );
char * url = &urlAsVec[0]; // pointer you can safely write to

You can also initialize the vector this way:

std::vector<char> urlAsVec( sup.c_str(), sup.c_str() + sup.size() + 1 );

which will write your null-terminator for you too.

Another alternative:

std::vector<char> urlAsVec( sup.size() + 1 );
sup.copy( &urlAsVec[0], sup.size() );

Note that your vector will automatically initialize all its elements to 0 so the null terminator will be there even though sup.copy() doesn't write it. You can use sup.copy( ptr, len) instead of strcpy anyway, which is slightly safer in that you can specify the buffer size (although strncpy would allow that too), although you will still have to write the null terminator manually (or have it already allocated).

For example if you use this:

char url[ BUFLEN ] = {0};
sup.copy( url, BUFLEN-1 );

for some fixed value BUFLEN you will get a copy or partial copy of the source string written into your buffer. Note that my initializer ensures all your unwritten bytes are 0.

Upvotes: 1

Bojan Komazec

Reputation: 9526

strcpy is unsafe as buffer overflow may occur. Use strncpy where you're providing exactly number of bytes to copy. You need to allocate memory for the destination buffer and add trailing \0:

std::string strInsert("news");
std::string sup("http://www.");
sup += strInsert + "yahoo.com";

char* url = new char[sup.length() + 1];
strncpy(url, sup.c_str(), sup.length());
url[sup.length()] = '\0';
// ... use url
delete[] url;

Upvotes: 2